Summer 2013 Advanced Algebra & Trigonometry Course Content Teacher Edition

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1 Summer 0 Advanced Algebra & Trigonometr Course Content Teacher Edition Algebra and Trigonometr (Third Edition) Beecher, Penna, Bittinger Addision Wesle (Februar, 007)

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3 Basic Concepts of Algebra R. The Real-Number Sstem R. Integer Eponents, Scientific Notation, and Order of Operations R. Addition, Subtraction, and Multiplication of Polnomials R. Factoring R.5 Rational Epressions R.6 Radical Notation and Rational Eponents R.7 The Basics of Equation Solving SUMMARY AND REVIEW R TEST A P P L I C A T I O N Gina wants to establish a college fund for her newborn daughter that will have accumulated $0,000 at the end of 8 r. If she can count on an interest rate of 6%, compounded monthl, how much should she deposit each month to accomplish this? This problem appears as Eercise 95 in Section R..

4 Chapter R Basic Concepts of Algebra R.. The Real-Number Polnomial Functions Sstem and Modeling Identif various kinds of real numbers. Use interval notation to write a set of numbers. Identif the properties of real numbers. Find the absolute value of a real number. Real Numbers In applications of algebraic concepts, we use real numbers to represent quantities such as distance, time, speed, area, profit, loss, and temperature. Some frequentl used sets of real numbers and the relationships among them are shown below. Real numbers Rational numbers Irrational numbers: 5, p,, 7, , Integers:,,,, 0,,,, Rational numbers that are not integers: 9 7,,,, 8., , Whole numbers: 0,,,, Negative integers:,,, Natural numbers (positive integers):,,, Zero: 0 Numbers that can be epressed in the form pq, where p and q are integers and q 0, are rational numbers. Decimal notation for rational numbers either terminates (ends) or repeats. Each of the following is a rational number. a) for an nonzero integer a a 7 b) 7 7 7,or c) 0.5 Terminating decimal d) Repeating decimal

5 Section R. The Real-Number Sstem The real numbers that are not rational are irrational numbers.decimal notation for irrational numbers neither terminates nor repeats. Each of the following is an irrational number. a) There is no repeating block of digits. 7 and. are rational approimations of the irrational number. b) There is no repeating block of digits. c) 6... Although there is a pattern, there is no repeating block of digits The set of all rational numbers combined with the set of all irrational numbers gives us the set of real numbers. The real numbers are modeled using a number line, as shown below. Each point on the line represents a real number, and ever real number is represented b a point on the line..9 E p * ( ) a (a, b) [ a [a, ) b The order of the real numbers can be determined from the number line. If a number a is to the left of a number b, then a is less than b a b.similarl, a is greater than b a b if a is to the right of b on the number line. For eample, we see from the number line above that 7 7.9,because.9 is to the left of 5 5. Also,, because is to the right of. The statement a b, read a is less than or equal to b, is true if either a b is true or a b is true. The smbol is used to indicate that a member, or element,belongs to a set. Thus if we let represent the set of rational numbers, we can see from the diagram on page that We can also write to indicate that is not an element of the set of rational numbers. When all the elements of one set are elements of a second set, we sa that the first set is a subset of the second set. The smbol is used to denote this. For instance, if we let represent the set of real numbers, we can see from the diagram that (read is a subset of ). Interval Notation Sets of real numbers can be epressed using interval notation.for eample, for real numbers a and b such that a b, the open interval a, b is the set of real numbers between, but not including, a and b. That is, a, b a b. The points a and b are endpoints of the interval. The parentheses indicate that the endpoints are not included in the interval. Some intervals etend without bound in one or both directions. The interval a,, for eample, begins at a and etends to the right without bound. That is, a, a. The bracket indicates that a is included in the interval.

6 Chapter R Basic Concepts of Algebra The various tpes of intervals are listed below. Intervals: Tpes, Notation, and Graphs INTERVAL TYPE NOTATION NOTATION GRAPH Open Closed Half-open Half-open Open Half-open Open Half-open a, b a, b a, b a, b a, a,, b, b SET a b a b a b a b a a b b ( ) a b [ ] a b [ ) a b ( ] a b ( a [ a ) b ] b The interval,, graphed below, names the set of all real numbers,. EXAMPLE Write interval notation for each set and graph the set. a) 5 b).7 c) 5 d) 5 Solution a) 5, 5; b).7.7,; c) 5 5, ; 5 0 5

7 Section R. The Real-Number Sstem 5 d) ; 5, Properties of the Real Numbers The following properties can be used to manipulate algebraic epressions as well as real numbers. Properties of the Real Numbers For an real numbers a, b, and c: a b b a ab ba and a b c a b c abc abc a 0 0 a a a a a a 0 a a a and Commutative properties of addition and multiplication Associative properties of addition and multiplication Additive identit propert Additive inverse propert Multiplicative identit propert a a a a a 0 Multiplicative inverse propert ab c ab ac Distributive propert Note that the distributive propert is also true for subtraction since ab c ab c ab ac ab ac. EXAMPLE State the propert being illustrated in each sentence. a) b) 5 m n 5 m n c) 0 d) e) a b a b Solution SENTENCE PROPERTY a) Commutative propert of multiplication: ab ba b) 5 m n 5 m n Associative propert of addition: a b c a b c c) 0 Additive inverse propert: a a 0 d) Multiplicative identit propert: a a a e) a b a b Distributive propert: ab c ab ac

8 6 Chapter R Basic Concepts of Algebra Absolute Value The number line can be used to provide a geometric interpretation of absolute value. The absolute value of a number a, denoted a, is its distance from 0 on the number line. For eample, 5 5, because the distance of 5 from 0 is 5. Similarl,, because the distance of from 0 is. Absolute Value For an real number a, a a, a, if a 0, if a 0. When a is nonnegative, the absolute value of a is a.when a is negative, the absolute value of a is the opposite, or additive inverse, of a. Thus, a is never negative; that is, for an real number a, a 0. Absolute value can be used to find the distance between two points on the number line. a b a b b a Distance Between Two Points on the Number Line For an real numbers a and b, the distance between a and b is a b,or equivalentl, b a. GCM EXAMPLE Find the distance between and. Solution The distance is 5 5, or equivalentl, 5 5. We can also use the absolute-value operation on a graphing calculator to find the distance between two points. On man graphing calculators, absolute value is denoted abs and is found in the MATH NUM menu and also in the CATALOG. abs () abs (()) 5 5

9 Section R. The Real-Number Sstem 7 R. Eercise Set In Eercises 0, consider the numbers, 7, 5., 7, 8, 0, 5....,, 5 5,.96, 9,, 5,,.. Which are whole numbers? 8,0,9, 5. Which are integers?, 8, 0,9, 5 7, 5....,. Which are irrational numbers?, 5 5,. Which are natural numbers? 8,9, 5, 5., 7,, 0, 5. Which are rational numbers?.96, 9, 8,, Which are real numbers? All of them 5., 7,.96, 7. Which are rational numbers but not integers?, Which are integers but not whole numbers? 9. Which are integers but not natural numbers?, 0 0. Which are real numbers but not integers? Write interval notation. Then graph the interval Write interval notation for the graph.. 0, 5., , [ ) 9, 5 ( ] ( ) [ ] , h [ ] h, h ( ] h 7. p, 8., q ( p In Eercises 9 6, the following notation is used: the set of natural numbers, the set of whole numbers, the set of integers, the set of rational numbers, the set of irrational numbers, and the set of real numbers. Classif the statement as true or false True 0. 0 True.. False. 0. True. True. 6 False 5 5. False 6. False 7. False 8. True True 0. True. True. False. True. True 5. False 6. False Name the propert illustrated b the sentence Commutative propert of multiplication 8. Associative propert of addition Multiplicative identit propert 5. 5ab 5ab 5. z z Distributive propert ] q Answers to Eercises 0 0, 50, and 5 can be found on p. IA-.

10 8 Chapter R Basic Concepts of Algebra Additive inverse propert 5. a b a b Commutative propert of multiplication 55. 6m n 6n m Commutative propert of addition 56. t 0 t Additive identit propert Multiplicative inverse propert Distributive propert Simplif Find the distance between the given pair of points on the number line , , , , , , 5 75., , , , 9 6 Collaborative Discussion and Writing To the student and the instructor: The Collaborative Discussion and Writing eercises are meant to be answered with one or more sentences. These eercises can also be discussed and answered collaborativel b the entire class or b small groups. Because of their open-ended nature, the answers to these eercises do not appear at the back of the book. The are denoted b the words Discussion and Writing. 79. How would ou convince a classmate that division is not associative? 80. Under what circumstances is a a rational number? Snthesis To the student and the instructor: The Snthesis eercises found at the end of ever eercise set challenge students to combine concepts or skills studied in that section or in preceding parts of the tet. Between an two (different) real numbers there are man other real numbers. Find each of the following. Answers ma var. 8. An irrational number between 0. and 0.5 Answers ma var; A rational number between.0 and Answers ma var;.5 8. A rational number between and Answers ma var; An irrational number between 5.99 and 6 Answers ma var; The hpotenuse of an isosceles right triangle with legs of length unit can be used to measure a value for b using the Pthagorean theorem, as shown. c 00 c c c Draw a right triangle that could be used to measure units. 0 Answer to Eercise 85 can be found on p. IA-.

11 Section R. Integer Eponents, Scientific Notation, and Order of Operations 9 R. Integer Eponents, Scientific Notation, and Order of Operations Simplif epressions with integer eponents. Solve problems using scientific notation. Use the rules for order of operations. Integers as Eponents When a positive integer is used as an eponent, it indicates the number of times a factor appears in a product. For eample, 7 means and 5 means 5. For an positive integer n, a n a a a a, n factors where a is the base and n is the eponent. Zero and negative-integer eponents are defined as follows. For an nonzero real number a and an integer m, a 0 and a m. a m EXAMPLE Simplif each of the following. a) 6 0 b). 0 Solution a) 6 0 b). 0 EXAMPLE Write each of the following with positive eponents. a) 5 b) c) Solution a) 5 5 b) c)

12 0 Chapter R Basic Concepts of Algebra The results in Eample can be generalized as follows. For an nonzero numbers a and b and an integers m and n, a m bn n. b a m (A factor can be moved to the other side of the fraction bar if the sign of the eponent is changed.) EXAMPLE Write an equivalent epression without negative eponents: 8. z 0 Solution Since each eponent is negative, we move each factor to the other side of the fraction bar and change the sign of each eponent: 8. z 0 z 0 8 The following properties of eponents can be used to simplif epressions. Proper ties of Eponents For an real numbers a and b and an integers m and n, assuming 0 is not raised to a nonpositive power: a m a n a mn Product rule a m n amn a a m n a mn ab b m a m b m a m am b m a 0 b 0 Quotient rule Power rule Raising a product to a power Raising a quotient to a power EXAMPLE e) 5 9z 8 Simplif each of the following. a) 5 b) 8 6 c) t 5 d) s 5

13 Section R. Integer Eponents, Scientific Notation, and Order of Operations Solution a) 5 5,or 8 8 b) c) t 5 t 5 t 5,or t 5 d) s 5 5 s 5 s 0,or s e) 5 9z 5 8 z z,or 5 6 z 5 6 z Scientific Notation We can use scientific notation to name ver large and ver small positive numbers and to perform computations. Scientific Notation Scientific notation for a number is an epression of the tpe N 0 m, where N 0, N is in decimal notation, and m is an integer. Keep in mind that in scientific notation positive eponents are used for numbers greater than or equal to 0 and negative eponents for numbers between 0 and. EXAMPLE 5 Undergraduate Enrollment. In a recent ear, there were 6,59,000 undergraduate students enrolled in post-secondar institutions in the United States (Source: U.S. National Center for Education Statistics). Convert this number to scientific notation. Solution We want the decimal point to be positioned between the and the 6, so we move it 7 places to the left. Since the number to be converted is greater than 0, the eponent must be positive. 6,59,

14 Chapter R Basic Concepts of Algebra EXAMPLE 6 Mass of a Neutron. The mass of a neutron is about kg. Convert this number to scientific notation. Solution We want the decimal point to be positioned between the and the 6, so we move it 7 places to the right. Since the number to be converted is between 0 and, the eponent must be negative EXAMPLE 7 Convert each of the following to decimal notation. a) b) Solution a) The eponent is negative, so the number is between 0 and. We move the decimal point places to the left b) The eponent is positive, so the number is greater than 0. We move the decimal point 5 places to the right ,000 Most calculators make use of scientific notation. For eample, the number 8,000,000,000,000 might be epressed in one of the was shown below..8e e.86e GCM EXAMPLE 8 Distance to a Star. The nearest star, Alpha Centauri C, is about. light-ears from Earth. One light-ear is the distance that light travels in one ear and is about miles. How man miles is it from Earth to Alpha Centauri C? Epress our answer in scientific notation. Solution This is not scientific notation because.86 w miles Writing scientific notation

15 Section R. Integer Eponents, Scientific Notation, and Order of Operations Order of Operations Recall that to simplif the epression 5, first we multipl and 5 to get 0 and then add to get. Mathematicians have agreed on the following procedure, or rules for order of operations. Rules for Order of Operations. Do all calculations within grouping smbols before operations outside. When nested grouping smbols are present, work from the inside out.. Evaluate all eponential epressions.. Do all multiplications and divisions in order from left to right.. Do all additions and subtractions in order from left to right. GCM EXAMPLE 9 Calculate each of the following a) 85 0 b) 5 Solution a) Doing the calculation within b) parentheses Evaluating the eponential epression Multipling Subtracting Note that fraction bars act as grouping smbols. That is, the given epression is equivalent to We can also enter these computations on a graphing calculator as shown below. 8(5)ˆ0 (0/(86)9)/(ˆ5 ) To confirm that it is essential to include parentheses around the numerator and around the denominator when the computation in Eample 9(b) is entered in a calculator, enter the computation without using these parentheses. What is the result?

16 Chapter R Basic Concepts of Algebra EXAMPLE 0 Compound Interest. If a principal P is invested at an interest rate r, compounded n times per ear, in t ears it will grow to an amount A given b A P r. nnt Suppose that $50 is invested at.6% interest, compounded quarterl. How much is in the account at the end of 8 ears? Solution We have P 50, r.6%, or 0.06, n, and t 8. Substituting, we find that the amount in the account at the end of 8 ears is given b A Net, we evaluate this epression: A Dividing Adding Multipling in the eponent Evaluating the eponential epression Multipling Rounding to the nearest cent The amount in the account at the end of 8 ears is $80.6. Answers to Eercises 5 0 can be found on p. IA-.

17 Chapter R Basic Concepts of Algebra R. Eercise Set Simplif a 0 a a ,or ,or m 5 m 5 8. n9 9. 7,or 0. b b b ,or a 5 5a b b a ba b ab 7 7a 5 b n 5n 00n 5. 5 b 0 b b a 9 a 5 7.,or 8.,or 6 9.,or 0. 5,or 5 8 0a 5 b. 8 5,or. a b,or b a 7 b a a

18 Section R. Integer Eponents, Scientific Notation, and Order of Operations 5. ab 8a b c d c d 5 z z m m 5 0. n n 8n p q r z 7 z 5p 8 q 6 r b 8 c 7 5 a 6 b c 5 Convert to scientific notation , ,670, ,600,000, ,90,000, One cubic inch is approimatel equal to m The United States government collected $,7,000,000,000 in individual income taes in a recent ear (Source: U.S. Internal Revenue Service)..7 0 Convert to decimal notation ,00, ,700, ,960,000,000 80,900,000, The amount of solid waste generated in the United States in a recent ear was tons (Source: Franklin Associates, Ltd.).,900, The mass of a proton is about.67 0 g Compute. Write the answer using scientific notation Solve. Write the answer using scientific notation Distance to Pluto. The distance from Earth to the sun is defined as astronomical unit,or AU.It is about 9 million miles. The average distance from Earth to the planet Pluto is 9 AUs. Find this distance in miles mi 7. Parsecs. One parsec is about.6 light-ears and light-ear is about mi. Find the number of miles in parsec mi 75. Nanowires. A nanometer is m. Scientists have developed optical nanowires to transmit light waves short distances. A nanowire with a diameter of 60 nanometers has been used in eperiments on :the transmission of light (Source: New York Times, Januar 9, 00). Find the diameter of such a wire in meters m 76. itunes. In the first quarter of 00, Apple Computer was selling.7 million songs per week on itunes, its online music service (Source: Apple Computer). At $0.99 per song, what is the revenue during a -week period? $ Chesapeake Ba Bridge-Tunnel. The 7.6-mile-long Chesapeake Ba Bridge-Tunnel was completed in 96. Construction costs were $0 million. Find the average cost per mile. $ Personal Space in Hong Kong. The area of Hong Kong is square miles. It is estimated that the population of Hong Kong will be 9,600,000 in 050. Find the number of square miles of land per person in sq mi 79. Nuclear Disintegration. One gram of radium produces 7 billion disintegrations per second. How man disintegrations are produced in hr?. 0 disintegrations 80. Length of Earth s Orbit. The average distance from the earth to the sun is 9 million mi. About how far does the earth travel in a earl orbit? (Assume a circular orbit.) mi Answers to Eercises 9 and can be found on p. IA-.

19 6 Chapter R Basic Concepts of Algebra Calculate Compound Interest. Use the compound interest formula from Eample 0 in Eercises Round to the nearest cent. 87. Suppose that $5 is invested at 6.%, compounded semiannuall. How much is in the account at the end of 5 r? $ Suppose that $9550 is invested at 5.%, compounded semiannuall. How much is in the account at the end of 7 r? $, Suppose that $6700 is invested at.5%, compounded quarterl. How much is in the account at the end of 6 r? $ Suppose that $875 is invested at 5.8%, compounded quarterl. How much is in the account at the end of 9 r? $ gives the amount S accumulated in a savings plan when a deposit of P dollars is made each month for t ears in an account with interest rate r, compounded monthl. Use this formula for Eercises Marisol deposits $50 in a retirement account each month beginning at age 0. If the investment earns 5% interest, compounded monthl, how much will have accumulated in the account when she retires 7 r later? $70, Gordon deposits $00 in a retirement account each month beginning at age 5. If the investment earns % interest, compounded monthl, how much will have accumulated in the account when Gordon retires at age 65? $8, Gina wants to establish a college fund for her newborn daughter that will have accumulated $0,000 at the end of 8 r. If she can count on an interest rate of 6%, compounded monthl, how much should she deposit each month to accomplish this? $09.79 Collaborative Discussion and Writing 9. Are the parentheses necessar in the epression 5 0 5? Wh or wh not? 9. Is for an negative value(s) of? Wh or wh not? Snthesis Savings Plan. P The formula r t S r 96. Liam wants to have $00,000 accumulated in a retirement account b age 70. If he starts making monthl deposits to the plan at age 0 and can count on an interest rate of.5%, compounded monthl, how much should he deposit each month in order to accomplish this? $9.

20 Section R. Addition, Subtraction, and Multiplication of Polnomials 7 R. Addition, Subtraction, and Multiplication of Polnomials Identif the terms, coefficients, and degree of a polnomial. Add, subtract, and multipl polnomials. Polnomials Polnomials are a tpe of algebraic epression that ou will often encounter in our stud of algebra. Some eamples of polnomials are, 5 7,.a, and z 6 5. All but the first are polnomials in one variable. Polnomials in One Variable A polnomial in one variable is an epression of the tpe a n n a n n a a a 0, where n is a nonnegative integer and a n,..., a 0 are real numbers, called coefficients.the parts of a polnomial separated b plus signs are called terms.the leading coefficient is a n, and the constant term is a 0. If a n 0, the degree of the polnomial is n. The polnomial is said to be written in descending order,because the eponents decrease from left to right. EXAMPLE Solution so the terms are Identif the terms of the polnomial 7.5. Writing plus signs between the terms, we have ,, 7.5,, and. A polnomial, like, consisting of onl a nonzero constant term has degree 0. It is agreed that the polnomial consisting onl of 0 has no degree.

21 8 Chapter R Basic Concepts of Algebra EXAMPLE Find the degree of each polnomial. a) b) 9 5 c) 7 Solution POLYNOMIAL DEGREE a) 9 b) 5 5 c) Algebraic epressions like ab 8 and are polnomials in several variables. The degree of a term is the sum of the eponents of the variables in that term. The degree of a polnomial is the degree of the term of highest degree. EXAMPLE Find the degree of the polnomial 7ab a b 8. Solution The degrees of the terms of 7ab a b 8 are, 6, and 0, respectivel, so the degree of the polnomial is 6. A polnomial with just one term, like 9 6, is a monomial. Ifa polnomial has two terms, like, it is a binomial.a polnomial with three terms, like, is a trinomial. Epressions like 5, 9, and 5 are not polnomials, because the cannot be written in the form a n n a n n a a 0,where the eponents are all nonnegative integers and the coefficients are all real numbers. Addition and Subtraction If two terms of an epression have the same variables raised to the same powers, the are called like terms, or similar terms.we can combine, or collect, like terms using the distributive propert. For eample, and 5 are like terms and We add or subtract polnomials b combining like terms. EXAMPLE Add or subtract each of the following. a) 5 7 b) 6 9 5

22 Section R. Addition, Subtraction, and Multiplication of Polnomials 9 Solution a) b) We can subtract b adding an opposite: Rearranging using the commutative and associative properties Using the distributive propert Adding the opposite of Combining like terms Multiplication Multiplication of polnomials is based on the distributive propert for eample, Using the distributive propert Using the distributive propert two more times 7. Combining like terms In general, to multipl two polnomials, we multipl each term of one b each term of the other and add the products. EXAMPLE 5 Multipl: 7. Solution We have 7 7 Using the distributive propert Using the distributive propert three more times Combining like terms We can also use columns to organize our work, aligning like terms under each other in the products Multipling b Multipling b Adding

23 0 Chapter R Basic Concepts of Algebra We can find the product of two binomials b multipling the First terms, then the Outer terms, then the Inner terms, then the Last terms. Then we combine like terms, if possible. This procedure is sometimes called FOIL. EXAMPLE 6 Multipl: 7. Solution We have F L F O I L I O We can use FOIL to find some special products. Special Products of Binomials A B A AB B Square of a sum A B A AB B Square of a difference A BA B A B Product of a sum and a difference EXAMPLE 7 Multipl each of the following. a) b) c) Solution a) b) c)

24 0 Chapter R Basic Concepts of Algebra R. Eercise Set Determine the terms and the degree of the polnomial.. 5 7, ; 5,, 7,. m m m m, m,m, ;. a b 7a b 5ab ;6 a b, 7a b,5ab,. 6p q p q pq 5 6p q, p q, pq, 5; 6 Perform the operations indicated z 7 z 8 z z

25 Section R. Addition, Subtraction, and Multiplication of Polnomials a ba 5 0 ab b 5. n n a a b a b ab b 6n n 5n 0n n 5n 8 n n 0 9. a a 5 a a 5 0. b b b 5b. 8. a ba b a 8ab b a 6 a a a a a 9. b b b Collaborative Discussion and Writing. Is the sum of two polnomials of degree n alwas a polnomial of degree n? Wh or wh not?. Eplain how ou would convince a classmate that A B A B. Snthesis Multipl. Assume that all eponents are natural numbers. 5. a n b n a n b n a n b n 6. t a t a 7 t a t a 8 7. a n b n an anbn bn 8. m t 5n 6m m t 5n t 0n a b ab ab m 5. t n t mn mn t mn mn 5. a b c a b c ab ac bc

26 Chapter R Basic Concepts of Algebra R. Factoring Factor polnomials b removing a common factor. Factor polnomials b grouping. Factor trinomials of the tpe b c. Factor trinomials of the tpe a b c, a, using the FOIL method and the grouping method. Factor special products of polnomials. To factor a polnomial, we do the reverse of multipling; that is, we find an equivalent epression that is written as a product. Terms with Common Factors When a polnomial is to be factored, we should alwas look first to factor out a factor that is common to all the terms using the distributive propert. We usuall look for the constant common factor with the largest absolute value and for variables with the largest eponent common to all the terms. In this sense, we factor out the largest common factor. EXAMPLE Factor each of the following. a) b) 0 Solution a) We can alwas check a factorization b multipling: b) There are several factors common to the terms of 0, but is the largest of these Factoring b Grouping In some polnomials, pairs of terms have a common binomial factor that can be removed in a process called factoring b grouping. EXAMPLE Factor: 5 5. Solution We have Grouping; each group of terms has a common factor. 5 Factoring a common factor out of each group 5. Factoring out the common binomial factor

27 Section R. Factoring Trinomials of the Tpe b c Some trinomials can be factored into the product of two binomials. To factor a trinomial of the form b c, we look for binomial factors of the form p q, where p q c and p q b. That is, we look for two numbers p and q whose sum is the coefficient of the middle term of the polnomial, b, and whose product is the constant term, c. When we factor an polnomial, we should alwas check first to determine whether there is a factor common to all the terms. If there is, we factor it out first. EXAMPLE Factor: 5 6. Solution First, we look for a common factor. There is none. Net, we look for two numbers whose product is 6 and whose sum is 5. Since the constant term, 6, and the coefficient of the middle term, 5, are both positive, we look for a factorization of 6 in which both factors are positive. PAIRS OF FACTORS SUMS OF FACTORS, 6 7, 5 The numbers we need are and. The factorization is. We have 5 6. We can check this b multipling: EXAMPLE Factor:. Solution First, we look for a common factor. Each term has a factor of, so we factor it out first: 7. Now we consider the trinomial 7. We look for two numbers whose product is and whose sum is 7. Since the constant term,, is positive and the coefficient of the middle term, 7, is negative, we look for a factorization of in which both factors are negative. PAIRS OF FACTORS SUMS OF FACTORS,, 6 8, 7 The numbers we need are and.

28 Chapter R Basic Concepts of Algebra The factorization of 7 is. We must also include the common factor that we factored out earlier. Thus we have. EXAMPLE 5 Factor: 8. Solution First, we look for a common factor. Each term has a factor of, so we factor it out first: 8 8. Now we consider the trinomial 8. We look for two numbers whose product is 8 and whose sum is. Since the constant term, 8, is negative, one factor will be positive and the other will be negative. PAIRS OF FACTORS SUMS OF FACTORS, 8 7, 8 7,, The numbers we need are and. We might have observed at the outset that since the sum of the factors is, a negative number, we need consider onl pairs of factors for which the negative factor has the greater absolute value. Thus onl the pairs, 8 and, need have been considered. Using the pair of factors and, we see that the factorization of 8 is. Including the common factor, we have 8. Trinomials of the Tpe a b c, a We consider two methods for factoring trinomials of the tpe a b c, a. The FOIL Method We first consider the FOIL method for factoring trinomials of the tpe a b c, a.consider the following multiplication. F O I L To factor 0, we must reverse what we just did. We look for two binomials whose product is this trinomial. The product of the First terms must be. The product of the Outside terms plus the product of

29 Section R. Factoring 5 the Inside terms must be.the product of the Last terms must be 0. We know from the preceding discussion that the answer is 5. In general, however, finding such an answer involves trial and error. We use the following method. To factor trinomials of the tpe a b c, a, using the FOIL method:. Factor out the largest common factor.. Find two First terms whose product is a : a b c. FOIL. Find two Last terms whose product is c: a b c. FOIL. Repeat steps () and () until a combination is found for which the sum of the Outside and Inside products is b: a b c. I O FOIL EXAMPLE 6 Factor: 0 8. Solution. There is no common factor (other than or ).. Factor the first term,. The onl possibilit (with positive coefficients) is. The factorization, if it eists, must be of the form.. Net, factor the constant term, 8. The possibilities are 8, 8,, and. The factors can be written in the opposite order as well: 8, 8,,and.. Find a pair of factors for which the sum of the outside and the inside products is the middle term, 0. Each possibilit should be checked b multipling. Some trials show that the desired factorization is. The Grouping Method The second method for factoring trinomials of the tpe a b c, a, is known as the grouping method,or the ac-method.

30 6 Chapter R Basic Concepts of Algebra To factor a b c, a, using the grouping method:. Factor out the largest common factor.. Multipl the leading coefficient a and the constant c.. Tr to factor the product ac so that the sum of the factors is b. That is, find integers p and q such that pq ac and p q b.. Split the middle term. That is, write it as a sum using the factors found in step (). 5. Factor b grouping. EXAMPLE 7 Factor: 0 8. Solution. Factor out the largest common factor, : Now consider 6 5. Multipl the leading coefficient, 6, and the constant, : 6.. Tr to factor so that the sum of the factors is the coefficient of the middle term, 5. PAIRS OF FACTORS SUMS OF FACTORS,,, 0, 0, 8 5, 8 5, 6, 6 8 ; 8 5. Split the middle term using the numbers found in step (): Finall, factor b grouping: Be sure to include the common factor to get the complete factorization of the original trinomial: 0 8.

31 Section R. Factoring 7 Special Factorizations We reverse the equation A BA B A B to factor a difference of squares. A B A BA B EXAMPLE 8 Factor each of the following. a) 6 b) 9a 5 c) Solution a) b) c) 6 9a 5 a 5 a 5a can be factored further. Because none of these factors can be factored further, we have factored completel. The rules for squaring binomials can be reversed to factor trinomials that are squares of binomials: A AB B A B ; A AB B A B. EXAMPLE 9 a) b) Factor each of the following. Solution A A B B A B a) 8 6 A A B B A B b)

32 8 Chapter R Basic Concepts of Algebra We can use the following rules to factor a sum or a difference of cubes: A B A BA AB B ; A B A BA AB B. These rules can be verified b multipling. EXAMPLE 0 a) b) Factor each of the following. Solution a) 7 9 b) Not all polnomials can be factored into polnomials with integer coefficients. An eample is 7. There are no real factors of 7 whose sum is. In such a case we sa that the polnomial is not factorable, or prime. C ONNECTING THE CONCEPTS A STRATEGY FOR FACTORING A. Alwas factor out the largest common factor first. B. Look at the number of terms. Two terms: Tr factoring as a difference of squares first. Net, tr factoring as a sum or a difference of cubes. Do not tr to factor a sum of squares. Three terms: Tr factoring as the square of a binomial. Net, tr using the FOIL method or the grouping method for factoring a trinomial. Four or more terms: Tr factoring b grouping and factoring out a common binomial factor. C. Alwas factor completel.ifa factor with more than one term can itself be factored further, do so.

33 Section R. Factoring 9 R. Eercise Set Factor out a common factor a a n n 8 a a 6n n 7. ab cb b a c 8. a Factor b grouping a 0a 5a a a a 6 6. t 6t t a 5a a 5 a a t 6t 5 6 Factor the trinomial. 9. p 6p 8 0. w 7w 0 p p w 5w t 8t 5. 7 t t t t 5 9 t t 5 7. n 0n 8 8. a ab b n n a ba b m m 90 7 m 9m 0. n 9n n 7n p p 6 p 5p 9. 6a 9ab 8b 0. 0m 7mn n a ba 7b 5m nm n. a a 6. a a 0 a a 6a 5a Answers to Eercises 5, 5, and 6 66 can be found on p. IA-. Factor the difference of squares.. m. z 8 m 5. 9 m z z a 8b z 8a z 5 ba 5z b z 5 z z 5. 7pq 7p 5. 5ab 5az Factor the square of a binomial z z z z z z a a a aa 6. p 8pq q ab 5a 5b p q 5a b Factor the sum or difference of cubes m 66. n t a 5 a 6 8t z t t z a a 7. t 6 a a z5z 5z z 8 t t t Factor completel a b 5ab ab6a 5b z z z z z Prime Prime 8. m 9n 8. 5t 6 m 8. nm n 5t t a 5 9a 88. b 7 b a a b b n n 9. 9z 9 n z 6 9. z 0z 5 z 5 z

34 0 Chapter R Basic Concepts of Algebra a a 56a 8 a z t a 7 b 5ab 7 0. a 75a p p 9p 8 p p p m 08. 5m 0 m c cd d. 9a 6ab b c.. m m 6 d 8m a 0m 0 b p 6p 6. 5a 8a p p p 6p Collaborative Discussion and Writing 7. Under what circumstances can A B be factored? 8. Eplain how the rule for factoring a sum of cubes can be used to factor a difference of cubes h h 9. h h p q 5p q 5 6p q 5p q 5 Factor. Assume that variables in eponents represent natural numbers. 5. n 5 n n 8 n. n n n n. a b ab a b. bd ad bc ac b ad c 5. 5 m n n 6. 6a t b 5 m n 5 m n a t b a a t b t b Snthesis Factor t t z z z 00. t t t 9 0. aba b a 6a b 9b 0. a 5a 0a 5a t 9 0t 0 6. a5 a5 0a a , or

35 0 Chapter R Basic Concepts of Algebra R.5 Rational Epressions Determine the domain of a rational epression. Simplif rational epressions. Multipl, divide, add, and subtract rational epressions. Simplif comple rational epressions. A rational epression is the quotient of two polnomials. For eample,,, and 5 are rational epressions. 5

36 Section R.5 Rational Epressions The Domain of a Rational Epression The domain of an algebraic epression is the set of all real numbers for which the epression is defined. Since division b zero is not defined, an number that makes the denominator zero is not in the domain of a rational epression. EXAMPLE Find the domain of each of the following. a) b) 5 Solution a) Since is 0 when, the domain of is the set of all real numbers ecept. b) To determine the domain of 5, we first factor the denominator:. 5 5 The factor is 0 when, and the factor 5 is 0 when 5.Since 5 0 when or 5, the domain is the set of all real numbers ecept and 5. We can describe the domains found in Eample using set-builder notation. For eample, we write The set of all real numbers such that is not equal to as { is a real number and. Similarl, we write The set of all real numbers such that is not equal to and is not equal to 5 as is a real number and and 5. Simplifing, Multipling, and Dividing Rational Epressions To simplif rational epressions, we use the fact that a c b c a b c c a b a b.

37 Chapter R Basic Concepts of Algebra EXAMPLE Simplif: 9 6. Solution Removing a factor of Canceling is a shortcut that is often used to remove a factor of. EXAMPLE Simplif each of the following. 6 a) b) Solution a) b) 6 Factoring the denominator Removing a factor of :,or In Eample (b), we saw that 6 and Factoring the numerator and the denominator Factoring the rational epression Factoring the numerator and the denominator Removing a factor of : are equivalent epressions. This means that the have the same value for all numbers that are in both domains. Note that is not in the domain of either epression, whereas is in the domain of but not in

38 Section R.5 Rational Epressions the domain of 6 and thus is not in the domain of both epressions. To multipl rational epressions, we multipl numerators and multipl denominators and, if possible, simplif the result. To divide rational epressions, we multipl the dividend b the reciprocal of the divisor and, if possible, simplif the result that is, a b c d ac bd a and. b c d a b d ad c bc EXAMPLE Multipl or divide and simplif each of the following. a) 9 b) Solution a) 9 9 Multipling the numerators and the denominators Factoring and removing a factor of : b) Factoring and removing a factor of Multipling b the reciprocal of the divisor Adding and Subtracting Rational Epressions When rational epressions have the same denominator, we can add or subtract b adding or subtracting the numerators and retaining the common denominator. If the denominators differ, we must find equivalent rational epressions that have a common denominator. In general, it is most efficient to find the least common denominator (LCD) of the epressions.

39 Chapter R Basic Concepts of Algebra To find the least common denominator of rational epressions, factor each denominator and form the product that uses each factor the greatest number of times it occurs in an factorization. EXAMPLE 5 Add or subtract and simplif each of the following. a) b) Solution a) Factoring the denominators The LCD is, or. b) Adding the numerators The LCD is Multipling each term b to get the LCD Factoring the denominators Multipling each term b to get the LCD Be sure to change the sign of ever term in the numerator of the epression being subtracted: Factoring and removing a 5 factor of : 5

40 Section R.5 Rational Epressions 5 Comple Rational Epressions A comple rational epression has rational epressions in its numerator or its denominator or both. To simplif a comple rational epression: Method. Find the LCD of all the denominators within the comple rational epression. Then multipl b using the LCD as the numerator and the denominator of the epression for. Method. First add or subtract, if necessar, to get a single rational epression in the numerator and in the denominator. Then divide b multipling b the reciprocal of the denominator. a b EXAMPLE 6 Simplif:. a b Solution Method. The LCD of the four rational epressions in the numerator and the denominator is a b. a b a b a b a b b a a b a a b a b b a b a b a b a b a b b a b ab ba a a b b ba a Multipling b using a b a b Factoring and removing a b a factor of : b a

41 6 Chapter R Basic Concepts of Algebra Method. We add in the numerator and in the denominator. a b a b a b b b a a b a b a b a b ab a ab b a a b a b b a ab b a a b b a ab a b b a The LCD is ab. The LCD is a b. b aaba b abb ab ba a a b b ba a We have a single rational epression in both the numerator and the denominator. Multipling b the reciprocal of the denominator Answers to Eercises 8 can be found on p. IA-.

42 6 Chapter R Basic Concepts of Algebra R.5 Eercise Set Find the domain of the rational epression Multipl or divide and, if possible, simplif. 9. r s 0. r s r s r s a a 6 a 7a a a 8 a a 0 a a a 6a 8 a a 6 a a a a 5 a a 6a

43 Section R.5 Rational Epressions 7 m n 5. r s Add or subtract and, if possible, simplif. a b a 5b a a a a b a b a b 8 m n r s a b a a a a 6 a a a a c 8 c c c c c 5 a a a a a a m n a b a b 9 a b a b 9 a b a a z z z z a a a 5. (Note:.) c 5a a a z z a b a b z 8z 8z Answers to Eercises and can be found on p. IA-. a 6. a b b b a (Note: b a a b.) a a b a b a a a b ab a b 5a a b b a b 6a a b b b a 5 a b 6 Simplif. a b b a b ab a b b a a b c 8 a a c c c b c b b c a 5a 0ab b a ba b 6a 9ab b 5 a ba b b b a a ab b a a a b a b a b b a a b ab

44 8 Chapter R Basic Concepts of Algebra a a a a a a a a a a a a a a a a a ab b b a b a a b 5 5 Snthesis Simplif. h h 65. h 66. h h h h h h h h h Collaborative Discussion and Writing 6. When adding or subtracting rational epressions, we can alwas find a common denominator b forming the product of all the denominators. Eplain wh it is usuall preferable to find the least common denominator. 6. How would ou determine which method to use for simplifing a particular comple rational epression? n n n h 68. h 7. Perform the indicated operations and, if possible, simplif n n n n 7. 0 nn n nn n n n n n n n

45 8 Chapter R Basic Concepts of Algebra R.6 Radical Notation and Rational Eponents Simplif radical epressions. Rationalize denominators or numerators in rational epressions. Convert between eponential and radical notation. Simplif epressions with rational eponents. A number c is said to be a square root of a if c a. Thus, is a square root of 9, because 9, and is also a square root of 9, because 9. Similarl, 5 is a third root (called a cube root) of 5, because 5 5. The number 5 has no other real-number cube root.

46 Section R.6 Radical Notation and Rational Eponents 9 nth Root A number c is said to be an nth root of a if c n a. The smbol a denotes the nonnegative square root of a, and the smbol a denotes the real-number cube root of a. The smbol n a denotes the nth root of a, that is, a number whose nth power is a.the smbol n is called a radical, and the epression under the radical is called the radicand. The number n (which is omitted when it is ) is called the inde.eamples of roots for n,, and, respectivel, are 5, 6, and 600. An real number has onl one real-number odd root. An positive number has two square roots, one positive and one negative. Similarl, for an even inde, a positive number has two real-number roots. The positive root is called the principal root. When an epression such as or 6 is used, it is understood to represent the principal (nonnegative) root. To denote a negative root, we use, 6, and so on. GCM EXAMPLE Simplif each of the following. a) 6 b) 6 c) d) e) 6 5 Solution a) 6 6,because 6 6. b) 6 6,because 6 6 and c) 8,because d) 5,because 5. e) 6 is not a real number, because we cannot find a real number that can be raised to the fourth power to get 6. We can generalize Eample (e) and sa that when a is negative and n is even, n a is not a real number. For eample, and 8 are not real numbers. We can find 6 and 6 in Eample using the square-root feature on the kepad of a graphing calculator, and we can use the cube-root feature to find 8. We can use the th-root feature to find higher roots. (6) (6) 6 6 (8) 5 (/) Frac /

47 0 Chapter R Basic Concepts of Algebra Stud Tip The kestrokes for entering the radical epressions in Eample on a graphing calculator are found in the Graphing Calculator Manual that accompanies this tet. When we tr to find 6 on a graphing calculator set in REAL mode, we get an error message indicating that the answer is nonreal. 6 ERR:NONREAL ANS : Quit : Goto Simplifing Radical Epressions Consider the epression. This is equivalent to 9, or. Similarl, 9.This illustrates the first of several properties of radicals, listed below. Properties of Radicals Let a and b be an real numbers or epressions for which the given roots eist. For an natural numbers m and n ( n ):. If n is even, n a n a.. If n is odd, n a n a.. n a n b n ab. n a a. n ( b 0). b n b 5. n a m n a m. EXAMPLE Simplif each of the following. a) 5 b) 5 c) 5 d) 50 7 e) f ) 8 5 g) 6 5 h) 6 6 Solution a) Using Propert b) 5 5 Using Propert c) Using Propert d) Using Propert 7 e) Using Propert Using Propert

48 Section R.6 Radical Notation and Rational Eponents f) Using Propert 5 5 g) Using Propert 6 6 Using Propert cannot be negative, so absolutevalue signs are not needed for it. h) Using Propert 6 6 Using Propert In man situations, radicands are never formed b raising negative quantities to even powers. In such cases, absolute-value notation is not required. For this reason, we will henceforth assume that no radicands are formed b raising negative quantities to even powers. For eample, we will write and a 5 b a ab. Radical epressions with the same inde and the same radicand can be combined (added or subtracted) in much the same wa that we combine like terms. EXAMPLE Perform the operations indicated. a) 8 5 b) 5 Solution a) Using the distributive propert b) Multipling

49 Chapter R Basic Concepts of Algebra An Application The Pthagorean theorem relates the lengths of the sides of a right triangle. The side opposite the triangle s right angle is called the hpotenuse. The other sides are the legs. The Pthagorean Theorem The sum of the squares of the lengths of the legs of a right triangle is equal to the square of the length of the hpotenuse: a b c. c b a EXAMPLE Surveing. A surveor places poles at points A, B, and C in order to measure the distance across a pond. The distances AC and BC are measured as shown. Find the distance AB across the pond. B 5 d A 7 d C (5 7 ) Solution We see that the lengths of the legs of a right triangle are given. Thus we use the Pthagorean theorem to find the length of the hpotenuse: c a b c a b Solving for c The distance across the pond is about 5. d. Rationalizing Denominators or Numerators There are times when we need to remove the radicals in a denominator or a numerator. This is called rationalizing the denominator or rationalizing the numerator. It is done b multipling b in such a wa as to obtain a perfect nth power.

50 Section R.6 Radical Notation and Rational Eponents EXAMPLE 5 Rationalize the denominator of each of the following. 7 a) b) 9 Solution a) b) Pairs of epressions of the form ab cd and ab cd are called conjugates. The product of such a pair contains no radicals and can be used to rationalize a denominator or a numerator. EXAMPLE 6 Rationalize the numerator:. 5 Solution The conjugate of is. A BA B A B Rational Eponents We are motivated to define rational eponents so that the properties for integer eponents hold for them as well. For eample, we must have a / a / a // a a. Thus we are led to define a / to mean a. Similarl, a /n would mean n a.again, if the laws of eponents are to hold, we must have a /n m a m /n a m/n. Thus we are led to define a m/n to mean n a m, or, equivalentl, n a m.

51 Chapter R Basic Concepts of Algebra Rational Eponents For an real number a and an natural numbers m and n, n, for which n a eists, a /n n a, a m/n n a m n a m, and a m/n. a m/n GCM We can use the definition of rational eponents to convert between radical and eponential notation. EXAMPLE 7 Convert to radical notation and, if possible, simplif each of the following. a) 7 / b) 8 5/ c) m /6 d) /5 Solution a) 7 / 7,or 7 b) c) d) 8 5/ 5/ m /6 6 m / , or /5 5 EXAMPLE 8 Convert each of the following to eponential notation. a) 7 5 b) 6 Solution a) b) / 6 /6 / We can use the laws of eponents to simplif eponential and radical epressions. EXAMPLE 9 Simplif and then, if appropriate, write radical notation for each of the following. a) 5/6 / b) 5/ / c) 7 Solution a) b) c) 5/6 / 5/6/ 9/6 / 5/ / 5// 7 7 / 7 / / 7 /6 6 7

52 Section R.6 Radical Notation and Rational Eponents 5 We can add and subtract rational eponents on a graphing calculator. The FRAC feature from the MATH menu allows us to epress the result as a fraction. The addition of the eponents in Eample 9(a) is shown here. 5/6/ Frac / EXAMPLE 0 Write an epression containing a single radical: a / b 5/6. Solution a / b 5/6 a /6 b 5/6 a b 5 /6 6 a b ab

53 Section R.6 Radical Notation and Rational Eponents 5 R.6 Eercise Set Simplif. Assume that variables can represent an real number t 6t 5. b b 6. c c z 9. z 6 z c d 8cd 0. 6c d 6 9c d d m 5 n 0 mn mn Simplif. Assume that no radicands were formed b raising negative quantities to even powers z 0z m n m n m 6 n m n m m 8 5 8a b bab 5 6ab 7 6a a a 9.. 7b b z z 5z

54 6 Chapter R Basic Concepts of Algebra Distance from Airport. An airplane is fling at an altitude of 700 ft. The slanted distance directl to the airport is,00 ft. How far horizontall is the airplane from the airport? About,709.5 ft 6. An equilateral triangle is shown below. (a) h a ; (b) A a a) Find an epression for its height h in terms of a. b) Find an epression for its area A in terms of a. 6. An isosceles right triangle has legs of length s. Find an epression for the length of the hpotenuse in terms of s. c s 6. The diagonal of a square has length 8. Find the length of a side of the square The area of square PQRS is 00 ft, and A, B, C, and D are the midpoints of the sides. Find the area of square ABCD. 50 ft P a a B h a a Q 700 ft,00 ft A C b 60. Bridge Epansion. During a summer heat wave, a -mi bridge epands ft in length. Assuming that the bulge occurs straight up the middle, estimate the height of the bulge. (In realit, bridges are built with epansion joints to control such buckling.) About 0.8 ft Rationalize the denominator S m n v w D R Answers to Eercises 7 76 can be found on p. IA-.

55 Section R.6 Radical Notation and Rational Eponents 7 Rationalize the numerator a b p q a q Convert to radical notation and simplif. 87. / 88. / / / 8 / / a 5/ b / 9. /5 /5 95. m 5/ n 7/ mn m n 96. p 7/6 q /6 pq 6 pq 5 Convert to eponential notation / / / /5 0. / / /6 0. 5/ Simplif and then, if appropriate, write radical notation. 07. a / 08. a / 8a a 5/6 8a / aa 6 / / / b / 5/6 a / b 5/8.. / / a / b /8 ab. m / n 5/ / n mn. 5/ / z / /5 5 z a a a 9. 0., or, or a b b 5. aa 5 a a m m 5 m m 09. b, or b 5. a / a / a / 6. m / m 7/ m 5/ Write an epression containing a single radical and simplif ab ab b a 5 b. a a a 6 a 5. a a a 6 a 5 a. a a a a. Collaborative Discussion and Writing 5. Eplain how ou would convince a classmate that a b is not equivalent to a b, for positive real numbers a and b. 6. Eplain how ou would determine whether is positive or negative without carring out the actual computation. Snthesis Simplif a a a a a/ 0. a b 5/ c /7 5a b / c 6/5 / 8 / a / b 7/9 c /5 Answers to Eercises 8 86 can be found on p. IA-.

56 8 Chapter R Basic Concepts of Algebra R.7 The Basics of Equation Solving Solve linear equations. Solve quadratic equations. An equation is a statement that two epressions are equal. To solve an equation in one variable is to find all the values of the variable that make the equation true. Each of these numbers is a solution of the equation. The set of all solutions of an equation is its solution set. Equations that have the same solution set are called equivalent equations. Linear and Quadratic Equations A linear equation in one variable is an equation that is equivalent to one of the form a b 0, where a and b are real numbers and a 0. A quadratic equation is an equation that is equivalent to one of the form a b c 0, where a, b, and c are real numbers and a 0. The following principles allow us to solve man linear and quadratic equations. Equation-Solving Principles For an real numbers a, b, and c, The Addition Principle: If a b is true, then a c b c is true. The Multiplication Principle: If a b is true, then ac bc is true. The Principle of Zero Products: If ab 0 is true, then a 0 or b 0, and if a 0 or b 0, then ab 0. The Principle of Square Roots: If k, then k or k. First we consider a linear equation. We will use the addition and multiplication principles to solve it.

57 Section R.7 The Basics of Equation Solving 9 EXAMPLE Solve: 6. Solution We begin b using the distributive propert to remove the parentheses Using the distributive propert Combining like terms Using the addition principle to add 6 on both sides Using the addition principle to add, or subtract, on both sides Using the multiplication principle to multipl b or divide b 8, on both sides 8, Simplifing We check the result in the original equation. CHECK: 6? 6 6 The solution is. TRUE Substituting for Now we consider a quadratic equation that can be solved using the principle of zero products. EXAMPLE Solve:. Solution First we write the equation with 0 on one side. 0 0 Subtracting on both sides Factoring 0 or 0 Using the principle of zero products or CHECK: For : For :? The solutions are and. TRUE? 6 TRUE

58 50 Chapter R Basic Concepts of Algebra The principle of square roots can be used to solve some quadratic equations, as we see in the net eample. 9., 0., EXAMPLE Solve: 6 0. Solution We will use the principle of square roots Adding 6 on both sides Dividing b on both sides to isolate or Using the principle of square roots Both numbers check. The solutions are and, or (read plus or minus )..., 5, 7 7

59 50 Chapter R Basic Concepts of Algebra R.7 Eercise Set Solve t , , , 5, t 5t a a m 7 m t n 5 7 n , 5, , 8. t 9t 0 0, n n , 8 8. t t 7 0 9, z z 6, ,. a 8 5a. n 0 n 5. 5, 7 6., , , 9 9. z, 50. t 5 5, , 5. 6z , , 5, 5 Collaborative Discussion and Writing 55. When using the addition and multiplication principles to solve an equation, how do ou determine which number to add or multipl b on both sides of the equation? 56. Eplain how to write a quadratic equation with solutions and.

60 Section R.7 The Basics of Equation Solving 5 Snthesis Solve t 55t ,, 0, 5, 0,, ,, ,,

61 Chapter R Summar and Review 5 Chapter R Summar and Review Important Properties and Formulas Properties of the Real Numbers Commutative: a b b a; ab ba 6 Associative: a b c a b c; abc abc Additive Identit: Additive Inverse: Multiplicative Identit: Multiplicative Inverse: Distributive: Absolute Value For an real number a, a a, a, if a 0, if a 0. a 0 0 a a a a a a 0 a a a a a a a a 0 ab c ab ac Properties of Eponents For an real numbers a and b and an integers m and n, assuming 0 is not raised to a nonpositive power: The Product Rule: a m a n a mn The Quotient Rule: The Power Rule: a m n amn a a m n a mn a 0 Raising a Product to a Power: ab m a m b m Raising a Quotient to a Power: a b m am b m Compound Interest Formula A P r nnt b 0

62 5 Chapter R Basic Concepts of Algebra Special Products of Binomials A B A AB B A B A AB B A BA B A B Pthagorean Theorem c b a b c Sum or Difference of Cubes Properties of Radicals Let a and b be an real numbers or epressions for which the given roots eist. For an natural numbers m and n n : If n is even, n a n a. If n is odd, n a n a. n a n b n ab. n A B A BA AB B A B A BA AB B a b n a n b n a m n a m. b 0. a Equation-Solving Principles The Addition Principle: If a b is true, then a c b c is true. The Multiplication Principle: If a b is true, then ac bc is true. The Principle of Zero Products: If ab 0 is true, then a 0 or b 0, and if a 0 or b 0, then ab 0. The Principle of Square Roots: If k, then k or k. Rational Eponents For an real number a and an natural numbers m and n, n, for which n a eists, a /n n a, a m/n n a m n a m, and a m/n. a m/n

63 The Trigonometric Functions 5. Trigonometric Functions of Acute Angles 5. Applications of Right Triangles 5. Trigonometric Functions of An Angle 5. Radians, Arc Length, and Angular Speed 5.5 Circular Functions: Graphs and Properties 5.6 Graphs of Transformed Sine and Cosine Functions SUMMARY AND REVIEW 5 TEST A P P L I C A T I O N In Comiske Park, the home of the Chicago White So baseball team, the angle of depression from a seat in the last row of the upper deck directl behind the batter to home plate is 9.9, and the angle of depression to the pitcher s mound is.. Find (a) the viewing distance to home plate and (b) the viewing distance to the pitcher s mound. This problem appears as Eample 7 in Section 5..

64 Chapter 5 The Trigonometric Functions. 5. Trigonometric Polnomial Functions and of Acute Modeling Angles Determine the si trigonometric ratios for a given acute angle of a right triangle. Determine the trigonometric function values of 0, 5, and 60. Using a calculator, find function values for an acute angle, and given a function value of an acute angle, find the angle. Given the function values of an acute angle, find the function values of its complement. The Trigonometric Ratios We begin our stud of trigonometr b considering right triangles and acute angles measured in degrees. An acute angle is an angle with measure greater than 0 and less than 90. Greek letters such as (alpha), (beta), (gamma), (theta), and (phi) are often used to denote an angle. Consider a right triangle with one of its acute angles labeled. The side opposite the right angle is called the hpotenuse. The other sides of the triangle are referenced b their position relative to the acute angle. One side is opposite and one is adjacent to. Hpotenuse u Side adjacent to u Side opposite u Hpotenuse u Adjacent to u Figure Hpotenuse u Adjacent to u Figure Opposite u Opposite u The lengths of the sides of the triangle are used to define the si trigonometric ratios: sine (sin), cosecant (csc), cosine (cos), secant (sec), tangent (tan), cotangent (cot). The sine of is the length of the side opposite of the hpotenuse (see Fig. ): divided b the length sin length of side opposite. length of hpotenuse The ratio depends on the measure of angle and thus is a function of. The notation sin actuall means sin, where sin, or sine, is the name of the function. The cosine of is the length of the side adjacent to divided b the length of the hpotenuse (see Fig. ): cos length of side adjacent to. length of hpotenuse The si trigonometric ratios, or trigonometric functions, are defined as follows.

65 Section 5. Trigonometric Functions of Acute Angles Hpotenuse u Adjacent to u Opposite u Trigonometric Function Values of an Acute Angle Let be an acute angle of a right triangle. Then the si trigonometric functions of are as follows: sin side opposite, csc hpotenuse, hpotenuse side opposite cos side adjacent to, sec hpotenuse, hpotenuse side adjacent to, cot side adjacent to tan side opposite. side adjacent to side opposite a u 5 EXAMPLE In the right triangle shown at left, find the si trigonometric function values of (a) and (b). Solution We use the definitions. a), csc hp sin opp, hp opp cos adj, sec hp, adj hp 5 The references to opposite, adjacent, and 5 hpotenuse are relative tan opp, cot adj opp 5 to. adj 5 b), csc hp sin opp, hp 5 opp 5, sec hp cos adj, hp adj, cot adj tan opp adj 5 opp 5 The references to opposite, adjacent, and hpotenuse are relative to. In Eample (a), we note that the value of sin,, is the reciprocal of, the value of csc. Likewise, we see the same reciprocal relationship between the values of cos and sec and between the values of tan and cot.for an angle, the cosecant, secant, and cotangent values are the reciprocals of the sine, cosine, and tangent function values, respectivel. Reciprocal Functions csc, sec, cot sin cos tan

66 Chapter 5 The Trigonometric Functions If we know the values of the sine, cosine, and tangent functions of an angle, we can use these reciprocal relationships to find the values of the cosecant, secant, and cotangent functions of that angle. Stud Tip Success on the net eam can be planned. Include stud time (even if onl 0 minutes a da) in our dail schedule and commit to making this time a priorit. Choose a time when ou are most alert and a setting in which ou can concentrate. You will be surprised how much more ou can learn and retain if stud time is included each da rather than in one long session before the eam. EXAMPLE Given that sin, cos, and tan 5 5, find csc, sec, and cot. Solution Using the reciprocal relationships, we have csc 5, sec 5, sin cos 5 5 and cot tan. Triangles are said to be similar if their corresponding angles have the same measure. In similar triangles, the lengths of corresponding sides are in the same ratio. The right triangles shown below are similar. Note that the corresponding angles are equal and the length of each side of the second triangle is four times the length of the corresponding side of the first triangle. a 0 a 5 b 6 b Let s observe the sine, cosine, and tangent values of in each triangle. Can we epect corresponding function values to be the same? FIRST TRIANGLE sin 5 cos 5 tan SECOND TRIANGLE sin 0 5 cos tan 6 For the two triangles, the corresponding values of sin, cos, and tan are the same. The lengths of the sides are proportional thus the

67 Section 5. Trigonometric Functions of Acute Angles 5 ratios are the same. This must be the case because in order for the sine, cosine, and tangent to be functions, there must be onl one output (the ratio) for each input (the angle ). The trigonometric function values of depend onl on the measure of the angle, not on the size of the triangle. The Si Functions Related We can find the other five trigonometric function values of an acute angle when one of the function-value ratios is known. pthagorean theorem review section.. EXAMPLE If sin 6 7 and is an acute angle, find the other five trigonometric function values of. Solution We know from the definition of the sine function that the ratio 6 opp is. 7 hp Using this information, let s consider a right triangle in which the hpotenuse has length 7 and the side opposite has length 6. To find the length of the side adjacent to, we recall the Pthagorean theorem: b 7 6 a a b c a 6 7 a 6 9 a 9 6 a. We now use the lengths of the three sides to find the other five ratios: sin 6, csc 7, 7 6 b cos, sec 7, or, 7 6 tan 6, or, cot. 6 Function Values of 0, 5, and 60 In Eamples and, we found the trigonometric function values of an acute angle of a right triangle when the lengths of the three sides were known. In most situations, we are asked to find the function values when the measure of the acute angle is given. For certain special angles such as

68 6 Chapter 5 The Trigonometric Functions 0, 5, and 60, which are frequentl seen in applications, we can use geometr to determine the function values. A right triangle with a 5 angle actuall has two 5 angles. Thus the triangle is isosceles, and the legs are the same length. Let s consider such a triangle whose legs have length. Then we can find the length of its hpotenuse, c, using the Pthagorean theorem as follows: c, or c, or c. Such a triangle is shown below. From this diagram, we can easil determine the trigonometric function values of sin 5 opp, hp cos 5 adj, hp tan 5 opp adj It is sufficient to find onl the function values of the sine, cosine, and tangent, since the others are their reciprocals. It is also possible to determine the function values of 0 and 60. A right triangle with 0 and 60 acute angles is half of an equilateral triangle, as shown in the following figure. Thus if we choose an equilateral triangle whose sides have length and take half of it, we obtain a right triangle that has a hpotenuse of length and a leg of length. The other leg has length a, which can be found as follows: a 0 60 a a a a We can now determine the function values of 0 and 60: sin 0 0.5, sin , cos 0 tan , cos , 0.577, tan Since we will often use the function values of 0, 5, and 60, either the triangles that ield them or the values themselves should be memorized.

69 Section 5. Trigonometric Functions of Acute Angles sin cos tan Let s now use what we have learned about trigonometric functions of special angles to solve problems. We will consider such applications in greater detail in Section 5.. EXAMPLE Height of a Hot-air Balloon. As a hot-air balloon began to rise, the ground crew drove. mi to an observation station. The initial observation from the station estimated the angle between the ground and the line of sight to the balloon to be 0.Approimatel how high was the balloon at that point? (We are assuming that the wind velocit was low and that the balloon rose verticall for the first few minutes.) Solution We begin with a drawing of the situation. We know the measure of an acute angle and the length of its adjacent side. h 0. mi Since we want to determine the length of the opposite side, we can use the tangent ratio, or the cotangent ratio. Here we use the tangent ratio: tan 0 opp adj h. tan 0 h. h Substituting; tan h. The balloon is approimatel 0.7 mi, or 696 ft, high.

70 8 Chapter 5 The Trigonometric Functions Function Values of An Acute Angle Historicall, the measure of an angle has been epressed in degrees, minutes, and seconds. One minute, denoted, is such that 60, or. One second, denoted, is such that, or Then 6 degrees, 7 minutes, seconds could be written as 67. This DMS form was common before the widespread use of scientific calculators. Now the preferred notation is to epress fractional parts of degrees in decimal degree form. Although the DMS notation is still widel used in navigation, we will most often use the decimal form in this tet. Most calculators can convert DMS notation to decimal degree notation and vice versa. Procedures among calculators var. Normal Sci Eng Float Radian Degree Func Par Pol Seq Connected Dot Sequential Simul Real abi reˆ θ i Full Horiz G T GCM EXAMPLE 5 Convert 50 to decimal degree notation. Solution We can use a graphing calculator set in DEGREE mode to convert between DMS form and decimal degree form. (See window at left.) To convert DMS form to decimal degree form, we enter 50 using the ANGLE menu for the degree and minute smbols and ALPHA for the smbol representing seconds. Pressing ENTER gives us , rounded to the nearest hundredth of a degree. 5 '0" GCM Without a calculator, we can convert as follows: EXAMPLE 6 Convert 7.8 to DMS notation. Solution To convert decimal degree form to DMS form, we enter 7.8 and access the DMS feature in the ANGLE menu. The result is ; ; DMS 7 0'8"

71 Section 5. Trigonometric Functions of Acute Angles 9 GCM Without a calculator, we can convert as follows: So far we have measured angles using degrees. Another useful unit for angle measure is the radian, which we will stud in Section 5.. Calculators work with either degrees or radians. Be sure to use whichever mode is appropriate. In this section, we use the degree mode. Keep in mind the difference between an eact answer and an approimation. For eample, sin 60. This is eact! But using a calculator, ou get an answer like sin This is an approimation! Calculators generall provide values onl of the sine, cosine, and tangent functions. You can find values of the cosecant, secant, and cotangent b taking reciprocals of the sine, cosine, and tangent functions, respectivel. EXAMPLE 7 Find the trigonometric function value, rounded to four decimal places, of each of the following. a) tan 9.7 b) sec 8 c) sin 809 Solution a) We check to be sure that the calculator is in DEGREE mode. The function value is tan Rounded to four decimal places b) The secant function value can be found b taking the reciprocal of the cosine function value: sec cos 8 c) We enter sin 809.The result is sin We can use the TABLE feature on a graphing calculator to find an angle for which we know a trigonometric function value.

72 0 Chapter 5 The Trigonometric Functions GCM EXAMPLE 8 Find the acute angle, to the nearest tenth of a degree, whose sine value is approimatel 0.0. Solution With a graphing calculator set in DEGREE mode, we first enter the equation sin. With a minimum value of 0 and a step-value of 0., we scroll through the table of values looking for the -value closest to 0.0. X X.6 Y sin We find that.6 is the angle whose sine value is about 0.0. The quickest wa to find the angle with a calculator is to use an inverse function ke. (We first studied inverse functions in Section. and will consider inverse trigonometric functions in Section 6..) First check to be sure that our calculator is in DEGREE mode. Usuall two kes must be pressed in sequence. For this eample, if we press ND SIN.0 ENTER, we find that the acute angle whose sine is 0.0 is approimatel.6006, or.6. 5 ft 6.5 ft EXAMPLE 9 Ladder Safet. A paint crew has purchased new 0-ft etension ladders. The manufacturer states that the safest placement on a wall is to etend the ladder to 5 ft and to position the base 6.5 ft from the wall. (Source: R.D.Werner Co., Inc.) What angle does the ladder make with the ground in this position? Solution We make a drawing and then use the most convenient trigonometric function. Because we know the length of the side adjacent to and the length of the hpotenuse, we choose the cosine function. From the definition of the cosine function, we have cos adj 6.5 ft 0.6. hp 5 ft Using a calculator, we find the acute angle whose cosine is 0.6: Pressing ND COS 0.6 ENTER Thus when the ladder is in its safest position, it makes an angle of about 75 with the ground. Cofunctions and Complements We recall that two angles are complementar whenever the sum of their measures is 90. Each is the complement of the other. In a right triangle,

73 Section 5. Trigonometric Functions of Acute Angles the acute angles are complementar, since the sum of all three angle measures is 80 and the right angle accounts for 90 of this total. Thus if one acute angle of a right triangle is, the other is 90. The si trigonometric function values of each of the acute angles in the triangle below are listed at the right. Note that 5 and 7 are complementar angles since sin cos tan sin cos tan 5.70 csc sec 7.5 cot 7.70 csc 5.5 sec cot Tr this with the acute, complementar angles 0. and 69.7 as well. What pattern do ou observe? Look for this same pattern in Eample earlier in this section. Note that the sine of an angle is also the cosine of the angle s complement. Similarl, the tangent of an angle is the cotangent of the angle s complement, and the secant of an angle is the cosecant of the angle s complement. These pairs of functions are called cofunctions. A list of cofunction identities follows. u 90 u Cofunction Identities sin cos 90, cos sin 90, tan cot 90, cot tan 90, sec csc 90, csc sec 90 EXAMPLE 0 Given that sin , cos , and tan 8 0.9,find the si trigonometric function values of 7. Solution Using reciprocal relationships, we know that csc 8.6, sin 8 sec 8.055, cos 8 and cot tan 8 Since 7 and 8 are complementar, we have sin 7 cos , cos 7 sin , tan 7 cot , cot 7 tan 8 0.9, sec 7 csc 8.6, csc 7 sec

74 Chapter 5 The Trigonometric Functions 5. Eercise Set In Eercises 6, find the si trigonometric function values of the specified angle f a φ u b u e 8. Given that sin, cos, and tan,find csc, sec, and cot. Given a function value of an acute angle, find the other five trigonometric function values. 9. sin 0. cos 0.7. tan. cot. csc.5. sec 7 5. cos 5 6. sin 0 5 Find the eact function value. 7. cos 5 8. tan 0 9. sec sin 5. cot 60. csc 5. sin 0. cos tan 5 6. sec 0 7. csc 0 8. cot Distance Across a River. Find the distance a across the river. 6. m B 5 0 c Grill A 9 a C 6 m Given that sin 5, cos, and tan 5,find csc, sec, and cot. 0. Distance Between Bases. A baseball diamond is actuall a square 90 ft on a side. If a line is drawn from third base to first base, then a right triangle Answers to Eercises 6 can be found on p. IA-.

75 Section 5. Trigonometric Functions of Acute Angles GCM GCM QPH is formed, where QPH is 5.Using a trigonometric function, find the distance from third base to first base. 7. ft P (third) 5 90 ft Convert to decimal degree notation. Round to two decimal places Convert to degrees, minutes, and seconds. Round to the nearest second Find the function value. Round to four decimal places. 55. cos cot tan sin sec cos cos csc sin tan tan cos h H 90 ft Q (first) 67. csc sec cot sin Find the acute angle, to the nearest tenth of a degree, for the given function value. 7. sin tan tan cos sin tan cos sin GCM 79. cot.7 5. Hint: tan cot 80. csc sec cot Find the eact acute angle value. for the given function 8. sin 5 8. cot cos sin tan cos csc tan cot 0 9. sec 5 Use the cofunction and reciprocal identities to complete each of the following. 9. cos 0 70 sin; sec 0 9. sin 6 6 cos; csc tan 5 cot 8; cot sec csc 77; cos.

76 Chapter 5 The Trigonometric Functions 97. Given that sin , cos , tan 65.5, cot , sec 65.66, csc 65.0, find the si function values of Given that sin 8 0.9, cos , tan , cot 8 7.5, sec , csc , find the si function values of Given that sin , cos , and tan 705.9, find the si function values of Given that sin , cos , and tan , find the si function values of Given that sin 8 p, cos 8 q, and tan 8 r,find the si function values of 8 in terms of p, q, and r. Solve. 08. e t 0,000 [.5] [.5] 0. log log [.5]. log 7 [.5] Snthesis. Given that cos 0.965, find csc Given that sec.50, find sin Find the si trigonometric function values of. q 5. Show that the area of this right triangle is bc sin A. q a A 0 97 Collaborative Discussion and Writing 0. Eplain the difference between reciprocal functions and cofunctions. 0. Eplain wh it is not necessar to memorize the function values for both 0 and 60. c b B a C 6. Show that the area of this triangle is ab sin. Skill Maintenance Make a hand-drawn graph of the function. Then check our work using a graphing calculator. 0. f 05. f e / 06. g log 07. h ln b a u Let h the height of the triangle. Then Area, bh where sin h, or h a sin, so Area. ab sin a Answers to Eercises 97 0, 0 07,, and 5 can be found on p. IA-.

77 Section 5. Applications of Right Triangles 5 5. Applications of Right Triangles Solve right triangles. Solve applied problems involving right triangles and trigonometric functions. Solving Right Triangles Now that we can find function values for an acute angle, it is possible to solve right triangles. To solve a triangle means to find the lengths of all sides and the measures of all angles. B EXAMPLE In ABC (shown at left), find a, b, and B, where a and b represent lengths of sides and B represents the measure of B. Here we use standard lettering for naming the sides and angles of a right triangle: Side a is opposite angle A, side b is opposite angle B,where a and b are the legs, and side c, the hpotenuse, is opposite angle C, the right angle. Solution In ABC,we know three of the measures: 06. a A 6.7, a?, B?, b?, C 90, c 06.. A 6.7 b C Since the sum of the angle measures of an triangle is 80 and C 90, the sum of A and B is 90. Thus, B 90 A We are given an acute angle and the hpotenuse. This suggests that we can use the sine and cosine ratios to find a and b, respectivel: sin 6.7 opp hp a 06. and cos 6.7 adj. hp b 06. Solving for a and b,we get a 06. sin 6.7 a 9.5 and b 06. cos 6.7 b 50.. Thus, A 6.7, a 9.5, B 8., b 50., C 90, c 06..

78 6 Chapter 5 The Trigonometric Functions E D d F EXAMPLE In DEF (shown at left), find D and F.Then find d. Solution In DEF,we know three of the measures: D?, d?, E 90, e, F?, f. We know the side adjacent to D and the hpotenuse. This suggests the use of the cosine ratio: cos D adj. hp We now find the angle whose cosine is of a degree,. To the nearest hundredth D Pressing ND COS ENTER Since the sum of D and F is 90,we can find F b subtracting: F 90 D We could use the Pthagorean theorem to find d, but we will use a trigonometric function here. We could use cos F, sin D, or the tangent or cotangent ratios for either D or F.Let s use tan D: tan D opp, or tan d. adj d Then d tan The si measures are D 55.58, d 9, E 90, e, F., f. b South Rim North Rim 6. mi 50 c Applications Right triangles can be used to model and solve man applied problems in the real world. EXAMPLE Hiking at the Grand Canon. A backpacker hiking east along the North Rim of the Grand Canon notices an unusual rock formation directl across the canon. She decides to continue watching the landmark while hiking along the rim. In hr, she has gone 6. mi due east and the landmark is still visible but at approimatel a 50 angle to the North Rim. (See the figure at left.) a) How man miles is she from the rock formation? b) How far is it across the canon from her starting point?

79 Section 5. Applications of Right Triangles 7 Solution a) We know the side adjacent to the 50 angle and want to find the hpotenuse. We can use the cosine function: 6. mi cos 50 c 6. mi c 9.6 mi. cos 50 After hiking 6. mi, she is approimatel 9.6 mi from the rock formation. b) We know the side adjacent to the 50 angle and want to find the opposite side. We can use the tangent function: tan 50 b 6. mi b 6. mi tan mi. Thus it is approimatel 7. mi across the canon from her starting point. Man applications with right triangles involve an angle of elevation or an angle of depression. The angle between the horizontal and a line of sight above the horizontal is called an angle of elevation. The angle between the horizontal and a line of sight below the horizontal is called an angle of depression. For eample, suppose that ou are looking straight ahead and then ou move our ees up to look at an approaching airplane. The angle that our ees pass through is an angle of elevation. If the pilot of the plane is looking forward and then looks down, the pilot s ees pass through an angle of depression. Horizontal Angle of elevation Angle of depression Horizontal B Angle of depression 850 ft Angle of elevation A 75 ft C EXAMPLE Aerial Photograph. An aerial photographer who photographs farm properties for a real estate compan has determined from eperience that the best photo is taken at a height of approimatel 75 ft and a distance of 850 ft from the farmhouse. What is the angle of depression from the plane to the house? Solution When parallel lines are cut b a transversal, alternate interior angles are equal. Thus the angle of depression from the plane to the house,, is equal to the angle of elevation from the house to the plane,

80 8 Chapter 5 The Trigonometric Functions so we can use the right triangle shown in the figure. Since we know the side opposite B and the hpotenuse, we can find b using the sine function. We first find sin : sin sin B Using a calculator in DEGREE mode, we find the acute angle whose sine is approimatel : 75 ft ft. Pressing nd SIN ENTER Thus the angle of depression is approimatel ft h EXAMPLE 5 Cloud Height. To measure cloud height at night, a vertical beam of light is directed on a spot on the cloud. From a point 5 ft awa from the light source, the angle of elevation to the spot is found to be Find the height of the cloud. Solution From the figure, we have tan 67.5 h 5 ft h 5 ft tan 67.5 ft. The height of the cloud is about ft. Some applications of trigonometr involve the concept of direction, or bearing. In this tet we present two was of giving direction, the first below and the second in Eercise Set 5.. Bearing: First-Tpe One method of giving direction, or bearing, involves reference to a north south line using an acute angle. For eample, N55W means 55 west of north and S67E means 67 east of south. N55W N N N N N60E W E W E W E W E S67E S S S S5W S

81 Section 5. Applications of Right Triangles 9 EXAMPLE 6 Distance to a Forest Fire. A forest ranger at point A sights a fire directl south. A second ranger at point B, 7.5 mi east, sights the same fire at a bearing of S7W. How far from A is the fire? N W A 7.5 mi B E 67 d 7 Fire F S S7W Solution We first find the complement of 7: B Angle B is opposite side d in the right triangle. From the figure shown above, we see that the desired distance d is part of a right triangle. We have d tan mi d 7.5 mi tan mi. The forest ranger at point A is about.5 mi from the fire.

82 50 Chapter 5 The Trigonometric Functions EXAMPLE 7 Comiske Park. In the new Comiske Park, the home of the Chicago White So baseball team, the first row of seats in the upper deck is farther awa from home plate than the last row of seats in the old Comiske Park. Although there is no obstructed view in the new park, some of the fans still complain about the present distance from home plate to the upper deck of seats. (Source: Chicago Tribune, September 9, 99) From a seat in the last row of the upper deck directl behind the batter, the angle of depression to home plate is 9.9, and the angle of depression to the pitcher s mound is.. Find (a) the viewing distance to home plate and (b) the viewing distance to the pitcher s mound d d h Pitcher u 60.5 ft Batter u Stud Tip Tutoring is available to students using this tet. The Addison- Wesle Math Tutor Center, staffed b mathematics instructors, can be reached b telephone, fa, or . When ou are having difficult with an eercise, this live tutoring can be a valuable resource. These instructors have a cop of our tet and are familiar with the content objectives in this course. Solution From geometr we know that 9.9 and.. The standard distance from home plate to the pitcher s mound is 60.5 ft. In the drawing, we let d be the viewing distance to home plate, d the viewing distance to the pitcher s mound, h the elevation of the last row, and the horizontal distance from the batter to a point directl below the seat in the last row of the upper deck. We begin b determining the distance. We use the tangent function with 9.9 and.: or tan 9.9 h h tan 9.9 and and h 60.5 tan.. Then substituting tan 9.9 for h in the second equation, we obtain tan tan.. tan. h 60.5

83 Section 5. Applications of Right Triangles 5 Solving for,we get tan 9.9 tan tan. tan 9.9 tan. tan tan. tan. tan 9.9 tan tan tan. tan 9.9 tan We can then find d and using the cosine function: d or cos d d 6.5 cos 9.9 d 9.7 and and cos d d 77 cos. d 0.7. The distance to home plate is about 50 ft,* and the distance to the pitcher s mound is about 0 ft. *In the old Comiske Park, the distance to home plate was onl 50 ft.,, T 6., s 0.8, t 0.

84 Section 5. Applications of Right Triangles 5 5. Eercise Set In Eercises 6, solve the right triangle.. F. 0 D f F 60, d, f 5. 6 d E b A 0 5. P 78, n., p 5. n M p P. N 6. F 7. H h 8 G f H 66, f 6., h.8. C 6 a A 67. c B A.7, a 5.7, c C a B A 5, a 7.,b 7.. R s 6.7 S T 6., s 0.8, t 0. t 0.7 T In Eercises 7 6, solve the right triangle. (Standard lettering has been used.) A c b 7. A 87, a 9.7 B 7, b 0.9,c 9.7 B a C

85 5 Chapter 5 The Trigonometric Functions 8. a.5, b 8. A., B 55.7, c. 9. b 00, c 50 A 77., B.8, a 9 0. B 56.5, c 0.07 A.5, a 0.07, b A 7.58, c 8. B., a 5.7, b.6. B 0.6, a 7.5 A 69., b.8, c 8.0. A 5, b 0 B 55, a 8.0, c 8.8. B 69., b 9. A,, 5., A b a 5. c 99.8 c.0 6., B 7.6, a a 0., c 0. A 0, B 60, b Safet Line to Raft. Each spring Bran uses his vacation time to read his lake propert for the summer. He wants to run a new safet line from point B on the shore to the corner of the anchored diving raft. The current safet line, which runs perpendicular to the shore line to point A, is 0 ft long. He estimates the angle from B to the corner of the raft to be 50.Approimatel how much rope does he need for the new safet line if he allows 5 ft of rope at each end to fasten the rope? About 6. ft Diving raft 0. Height of a Tree. A supervisor must train a new team of loggers to estimate the heights of trees. As an eample, she walks off 0 ft from the base of a tree and estimates the angle of elevation to the tree s peak to be 70.Approimatel how tall is the tree? About 0 ft 70 0 ft. Sand Dunes National Park. While visiting the Sand Dunes National Park in Colorado, Cole approimated the angle of elevation to the top of a sand dune to be 0.After walking 800 ft closer, he guessed that the angle of elevation had increased b 5.Approimatel how tall is the dune he was observing? About 606 ft 0 ft B 50 A Shoreline ft 8. Enclosing an Area. Alicia is enclosing a triangular area in a corner of her fenced rectangular backard for her Labrador retriever. In order for a certain tree to be included in this pen, one side needs to be.5 ft and make a 5 angle with the new side. How long is the new side? About. ft 9. Easel Displa. A marketing group is designing an easel to displa posters advertising their newest products. The want the easel to be 6 ft tall and the back of it to fit flush against a wall. For optimal ee contact, the best angle between the front and back legs of the easel is.how far from the wall should the front legs be placed in order to obtain this angle? About.5 ft. Tee Shirt Design. A new tee shirt design is to have a regular octagon inscribed in a circle, as shown in the figure. Each side of the octagon is to be.5 in. long. Find the radius of the circumscribed circle. About.6 in. r.5 in.

86 Section 5. Applications of Right Triangles 5. Inscribed Pentagon. A regular pentagon is inscribed in a circle of radius 5.8 cm. Find the perimeter of the pentagon. About 9.9 cm. Height of a Weather Balloon. A weather balloon is directl west of two observing stations that are 0 mi apart. The angles of elevation of the balloon from the two stations are 7.6 and 78..How high is the balloon? About. mi 5. Height of a Kite. For a science fair project, a group of students tested different materials used to construct kites. Their instructor provided an instrument that accuratel measures the angle of elevation. In one of the tests, the angle of elevation was 6. with 670 ft of string out. Assuming the string was taut, how high was the kite? About 599 ft 6. Height of a Building. A window washer on a ladder looks at a nearb building 00 ft awa, noting that the angle of elevation of the top of the building is 8.7 and the angle of depression of the bottom of the building is 6.5.How tall is the nearb building? About 5 ft 9. Distance from a Lighthouse. From the top of a lighthouse 55 ft above sea level, the angle of depression to a small boat is..how far from the foot of the lighthouse is the boat? About 75 ft 0. Lightning Detection. In etremel large forests, it is not cost-effective to position forest rangers in towers or to use small aircraft to continuall watch for fires. Since lightning is a frequent cause of fire, lightning detectors are now commonl used instead. These devices not onl give a bearing on the location but also measure the intensit of the lightning. A detector at point Q is situated 5 mi west of a central fire station at point R.The bearing from Q to where lightning hits due south of R is S7.6E. How far is the hit from point R? About 9.5 mi. Lobster Boat. A lobster boat is situated due west of a lighthouse. A barge is km south of the lobster boat. From the barge, the bearing to the lighthouse is N60E. How far is the lobster boat from the lighthouse? About km ft 6.5 km 6 0' North 7. Distance Between Towns. From a hot-air balloon km high, the angles of depression to two towns in line with the balloon are 8. and.5.how far apart are the towns? About 8 km 8. Angle of Elevation. What is the angle of elevation of the sun when a 5-ft mast casts a 0-ft shadow? About 60.. Length of an Antenna. A vertical antenna is mounted atop a 50-ft pole. From a point on level ground 75 ft from the base of the pole, the antenna subtends an angle of 0.5. Find the length of the antenna. About ft 5 ft ft 0 ft 75 ft

87 5 Chapter 5 The Trigonometric Functions Collaborative Discussion and Writing. In this section, the trigonometric functions have been defined as functions of acute angles. Thus the set of angles whose measures are greater than 0 and less than 90 is the domain for each function. What appear to be the ranges for the sine, the cosine, and the tangent functions given this domain?. Eplain in our own words five was in which length c can be determined in this triangle. Which wa seems the most efficient?. Construction of Picnic Pavilions. A construction compan is mass-producing picnic pavilions for national parks, as shown in the figure. The rafter ends are to be sawed in such a wa that the will be vertical when in place. The front is 8 ft high, the back is 6 ft high, and the distance between the front and back is 8 ft. At what angle should the rafters be cut? Cut so that u 79.8 u 6 c 8 ft 6q ft Skill Maintenance Find the distance between the points. 5. 8, and 6, [.] 0, or about. 6. 9, and 0, 0 [.] 0, or about Convert to an eponential equation: log [.] Convert to a logarithmic equation: e t. [.] ln t Snthesis 9. Find h,to the nearest tenth.. 8 ft. Diameter of a Pipe. A V-gauge is used to find the diameter of a pipe. The advantage of such a device is that it is rugged, it is accurate, and it has no moving parts to break down. In the figure, the measure of angle AVB is 5.A pipe is placed in the V-shaped slot and the distance VP is used to estimate the diameter. The line VP is calibrated b listing as its units the corresponding diameters. This, in effect, establishes a function between VP and d. Q A h V d Find a,to the nearest tenth a P a) Suppose that the diameter of a pipe is cm. What is the distance VP? About.96 cm b) Suppose that the distance VP is.9 cm. What is the diameter of the pipe? About.00 cm c) Find a formula for d in terms of VP. d.0 VP d) Find a formula for VP in terms of d. VP 0.98d. Sound of an Airplane. It is a common eperience to hear the sound of a low-fling airplane and look at the wrong place in the sk to see the plane. Suppose that a plane is traveling directl at ou at a speed of B

88 Section 5. Applications of Right Triangles mph and an altitude of 000 ft, and ou hear the sound at what seems to be an angle of inclination of 0.At what angle should ou actuall look in order to see the plane? Consider the speed of sound to be 00 ftsec. Perceived location of plane P 7 Actual location of plane when heard A mountain and a line drawn from the top of the mountain to the horizon, as shown in the figure. The height of Mt. Shasta in California is,6 ft. From the top of Mt. Shasta, one can see the horizon on the Pacific Ocean. The angle formed between a line to the horizon and the vertical is found to be 875.Use this information to estimate the radius of the earth, in miles. R 98 mi 000 ft 0. Measuring the Radius of the Earth. One wa to measure the radius of the earth is to climb to the top of a mountain whose height above sea level is known and measure the angle between a vertical line to the center of the earth from the top of the V u 875 R R,6

89 Section 5. Trigonometric Functions of An Angle Trigonometric Functions of An Angle Find angles that are coterminal with a given angle and find the complement and the supplement of a given angle. Determine the si trigonometric function values for an angle in standard position when the coordinates of a point on the terminal side are given. Find the function values for an angle whose terminal side lies on an ais. Find the function values for an angle whose terminal side makes an angle of 0, 5, or 60 with the -ais. Use a calculator to find function values and angles. Angles, Rotations, and Degree Measure An angle is a familiar figure in the world around us.

90 56 Chapter 5 The Trigonometric Functions An angle is the union of two ras with a common endpoint called the verte. In trigonometr, we often think of an angle as a rotation. To do so, think of locating a ra along the positive -ais with its endpoint at the origin. This ra is called the initial side of the angle. Though we leave that ra fied, think of making a cop of it and rotating it. A rotation counterclockwise is a positive rotation, and a rotation clockwise is a negative rotation. The ra at the end of the rotation is called the terminal side of the angle. The angle formed is said to be in standard position. Verte Terminal side A positive rotation (or angle) Initial side A negative rotation (or angle) The measure of an angle or rotation ma be given in degrees. The Bablonians developed the idea of dividing the circumference of a circle into 60 equal parts, or degrees. If we let the measure of one of these parts be, then one complete positive revolution or rotation has a measure of 60. One half of a revolution has a measure of 80,one fourth of a revolution has a measure of 90, and so on. We can also speak of an angle of measure 60, 5, 0, or 0. The terminal sides of these angles lie in quadrants I, II, IV, and I, respectivel. The negative rotations 0, 0, and 5 represent angles with terminal sides in quadrants IV, III, and II, respectivel

91 Section 5. Trigonometric Functions of An Angle 57 If two or more angles have the same terminal side, the angles are said to be coterminal. To find angles coterminal with a given angle, we add or subtract multiples of 60. For eample, 0, shown above, has the same terminal side as 60, since Thus we sa that angles of measure 60 and 0 are coterminal. The negative rotation that measures 00 is also coterminal with 60 because The set of all angles coterminal with 60 can be epressed as 60 n 60, where n is an integer. Other eamples of coterminal angles shown above are 90 and 70, 90 and 70, 5 and 5, 0 and 0, and 0 and 60. EXAMPLE Find two positive and two negative angles that are coterminal with (a) 5 and (b) 7. Solution a) We add and subtract multiples of 60.Man answers are possible (60) (60) 669 Thus angles of measure,, 09, and 669 are coterminal with 5. b) We have the following: , , , Thus angles of measure 5, 7, 67, and 607 are coterminal with 7.

92 58 Chapter 5 The Trigonometric Functions Angles can be classified b their measures, as seen in the following figure. 90 Right: u 90 Acute: 0 u 90 Obtuse : 90 u 80 Straight: u Recall that two acute angles are complementar if their sum is 90. For eample, angles that measure 0 and 80 are complementar because Two positive angles are supplementar if their sum is 80. For eample, angles that measure 5 and 5 are supplementar because Complementar angles 5 5 Supplementar angles EXAMPLE Find the complement and the supplement of 7.6. Solution We have , Thus the complement of 7.6 is 8.5 and the supplement is r r P(, ) u Trigonometric Functions of Angles or Rotations Man applied problems in trigonometr involve the use of angles that are not acute. Thus we need to etend the domains of the trigonometric functions defined in Section 5. to angles, or rotations, of an size. To do this, we first consider a right triangle with one verte at the origin of a coordinate sstem and one verte on the positive -ais. (See the figure at left.) The other verte is at P, a point on the circle whose center is at the origin and whose radius r is the length of the hpotenuse of the triangle. This triangle is a reference triangle for angle, which is in standard position. Note that is the length of the side opposite and is the length of the side adjacent to.

93 Section 5. Trigonometric Functions of An Angle 59 Recalling the definitions in Section 5., we note that three of the trigonometric functions of angle are defined as follows: sin opp, cos adj, tan opp. adj hp hp r r Since and are the coordinates of the point P and the length of the radius is the length of the hpotenuse, we can also define these functions as follows: sin -coordinate, radius cos -coordinate, radius tan -coordinate. -coordinate We will use these definitions for functions of angles of an measure. The following figures show angles whose terminal sides lie in quadrants II, III, and IV. P(, ) r u P(, ) r u u r P(, ) A reference triangle can be drawn for angles in an quadrant, as shown. Note that the angle is in standard position; that is, it is alwas measured from the positive half of the -ais. The point P, is a point, other than the verte, on the terminal side of the angle. Each of its two coordinates ma be positive, negative, or zero, depending on the location of the terminal side. The length of the radius, which is also the length of the hpotenuse of the reference triangle, is alwas considered positive. Note that r, or r. Regardless of the location of P, we have the following definitions.

94 60 Chapter 5 The Trigonometric Functions P(, ) r u Trigonometric Functions of An Angle Suppose that P, is an point other than the verte on the terminal side of an angle in standard position, and r is the radius, or distance from the origin to P,. Then the trigonometric functions are defined as follows: sin -coordinate, csc radius, -coordinate r radius r cos -coordinate, sec radius, -coordinate r radius r tan -coordinate, cot -coordinate. -coordinate -coordinate Values of the trigonometric functions can be positive, negative, or zero, depending on where the terminal side of the angle lies. The length of the radius is alwas positive. Thus the signs of the function values depend onl on the coordinates of the point P on the terminal side of the angle. In the first quadrant, all function values are positive because both coordinates are positive. In the second quadrant, first coordinates are negative and second coordinates are positive; thus onl the sine and the cosecant values are positive. Similarl, we can determine the signs of the function values in the third and fourth quadrants. Because of the reciprocal relationships, we need to learn onl the signs for the sine, cosine, and tangent functions. Positive: sin Negative: cos, tan Positive: All Negative: None II (, ) (, ) I III (, ) (, ) IV Positive: tan Negative: sin, cos Positive: cos Negative: sin, tan

95 Section 5. Trigonometric Functions of An Angle 6 EXAMPLE Find the si trigonometric function values for each angle shown. a) b) c) (, ) (, ) u r u r (, ) r u Solution a) We first determine r, the distance from the origin 0, 0 to the point,.the distance between 0, 0 and an point, on the terminal side of the angle is r 0 0. Substituting for and for,we find r Using the definitions of the trigonometric functions, we can now find the function values of. We substitute for, for, and 5 for r: sin r cos r tan, sec r, 5 5, cot. As epected, the tangent and the cotangent values are positive and the other four are negative. This is true for all angles in quadrant III. b) We first determine r, the distance from the origin to the point, : r. Substituting for, for, and for r,we find sin r cos r tan, csc r, , csc r,, sec r,, cot.

96 6 Chapter 5 The Trigonometric Functions c) We determine r, the distance from the origin to the point, : r. Substituting for, for, and for r, we find the trigonometric function values of are, csc sin, cos, sec, tan, cot. (, ) (8, ) u q An point other than the origin on the terminal side of an angle in standard position can be used to determine the trigonometric function values of that angle. The function values are the same regardless of which point is used. To illustrate this, let s consider an angle in standard position whose terminal side lies on the line. We can determine two second-quadrant solutions of the equation, find the length r for each point, and then compare the sine, cosine, and tangent function values using each point. If, then. If 8, then 8. For,, r 0 5. For 8,, r Using, and r 5, we find that sin , cos 5, and tan. Using 8, and r 5, we find that sin , cos 8 5, and tan. 8 We see that the function values are the same using either point. An point other than the origin on the terminal side of an angle can be used to determine the trigonometric function values. The trigonometric function values of depend onl on the angle, not on the choice of the point on the terminal side that is used to compute them.

97 Section 5. Trigonometric Functions of An Angle 6 (, ) u The Si Functions Related When we know one of the function values of an angle, we can find the other five if we know the quadrant in which the terminal side lies. The procedure is to sketch a reference triangle in the appropriate quadrant, use the Pthagorean theorem as needed to find the lengths of its sides, and then find the ratios of the sides. EXAMPLE Given that tan and is in the second quadrant, find the other function values. Solution We first sketch a second-quadrant angle. Since tan, Epressing as since is in quadrant II we make the legs lengths and. The hpotenuse must then have length,or. Now we read off the appropriate ratios: sin, or, csc, cos, or, sec, tan, cot. Terminal Side on an Ais An angle whose terminal side falls on one of the aes is a quadrantal angle. One of the coordinates of an point on that side is 0. The definitions of the trigonometric functions still appl, but in some cases, function values will not be defined because a denominator will be 0. EXAMPLE 5 Find the sine, cosine, and tangent values for 90, 80, 70, and 60. Solution We first make a drawing of each angle in standard position and label a point on the terminal side. Since the function values are the same for all points on the terminal side, we choose 0,,, 0, 0,, and, 0 for convenience. Note that r for each choice. (0, ) 90 (, 0) (, 0) (0, )

98 6 Chapter 5 The Trigonometric Functions Then b the definitions we get sin 90, sin 80 0, sin 70, sin 60 0, 0 0 cos 90 0, cos 80, cos 70 0, cos 60, 0 0 tan 90, Not defined tan 80 0, tan 70, Not defined tan In Eample 5, all the values can be found using a calculator, but ou will find that it is convenient to be able to compute them mentall. It is also helpful to note that coterminal angles have the same function values. For eample, 0 and 60 are coterminal; thus, sin 0 0, cos 0, and tan 0 0. EXAMPLE 6 Find each of the following. a) sin 90 b) csc 50 Solution a) We note that 90 is coterminal with 70.Thus, sin 90 sin 70. b) Since , 50 and 80 are coterminal. Thus, csc 50 csc 80 sin 80 0, which is not defined. Trigonometric values can alwas be checked using a calculator. When the value is undefined, the calculator will displa an ERROR message. ERR: DIVIDE BY 0 : Quit : Goto Reference Angles: 0, 5, and 60 We can also mentall determine trigonometric function values whenever the terminal side makes a 0, 5, or 60 angle with the -ais. Consider, for eample, an angle of 50. The terminal side makes a 0 angle with the -ais, since Reference angle P(, ) P(, ) N O N

99 Section 5. Trigonometric Functions of An Angle 65 As the figure shows, ONP is congruent to ONP; therefore, the ratios of the sides of the two triangles are the same. Thus the trigonometric function values are the same ecept perhaps for the sign. We could determine the function values directl from ONP, but this is not necessar. If we remember that in quadrant II, the sine is positive and the cosine and the tangent are negative, we can simpl use the function values of 0 that we alread know and prefi the appropriate sign. Thus, sin 50 sin 0, cos 50 cos 0, and tan 50 tan 0, or. Triangle ONP is the reference triangle and the acute angle called a reference angle. NOP is Reference Angle The reference angle for an angle is the acute angle formed b the terminal side of the angle and the -ais. EXAMPLE 7 Find the sine, cosine, and tangent function values for each of the following. a) 5 b) 780 Solution a) We draw a figure showing the terminal side of a 5 angle. The reference angle is 5 80, or 5. 5 Reference angle Recall from Section 5. that sin 5, cos 5, and tan 5. Also note that in the third quadrant, the sine and the cosine are negative and the tangent is positive. Thus we have sin 5, cos 5, and tan 5.

100 66 Chapter 5 The Trigonometric Functions b) We draw a figure showing the terminal side of a 780 angle. Since ,we know that 780 and 60 are coterminal Reference angle The reference angle for 60 is the acute angle formed b the terminal side of the angle and the -ais. Thus the reference angle for 60 is 60.We know that since 780 is a fourth-quadrant angle, the cosine is positive and the sine and the tangent are negative. Recalling that sin 60, cos 60, and tan 60, we have sin 780, cos 780, and tan 780. cos() /cos(500) tan(8.) Function Values for An Angle When the terminal side of an angle falls on one of the aes or makes a 0, 5, or 60 angle with the -ais, we can find eact function values without the use of a calculator. But this group is onl a small subset of all angles. Using a calculator, we can approimate the trigonometric function values of an angle. In fact, we can approimate or find eact function values of all angles without using a reference angle. EXAMPLE 8 Find each of the following function values using a calculator and round the answer to four decimal places, where appropriate. a) cos b) sec 500 c) tan 8. d) csc 5.75 e) cos 00 f) sin 7509 g) cot 5 Solution Using a calculator set in DEGREE mode, we find the values. a) b) c) cos 0.76 sec cos 500 tan

101 Section 5. Trigonometric Functions of An Angle 67 /sin(5.75) cos(00) sin(7509) d) csc sin 5.75 e) f) g) cos sin cot 5 tan 5 In man applications, we have a trigonometric function value and want to find the measure of a corresponding angle. When onl acute angles are considered, there is onl one angle for each trigonometric function value. This is not the case when we etend the domain of the trigonometric functions to the set of all angles. For a given function value, there is an infinite number of angles that have that function value. There can be two such angles for each value in the range from 0 to 60. To determine a unique answer in the interval 0, 60, the quadrant in which the terminal side lies must be specified. The calculator gives the reference angle as an output for each function value that is entered as an input. Knowing the reference angle and the quadrant in which the terminal side lies, we can find the specified angle. 6. u EXAMPLE 9 Given the function value and the quadrant restriction, find. a) sin 0.8, 90 b) cot 0.6, 70 Solution a) We first sketch the angle in the second quadrant. We use the calculator to find the acute angle (reference angle) whose sine is 0.8. The reference angle is approimatel 6.. We find the angle b subtracting 6. from 80: Thus,. b) We begin b sketching the angle in the fourth quadrant. Because the tangent and cotangent values are reciprocals, we know that u tan We use the calculator to find the acute angle (reference angle) whose tangent is 6.07, ignoring the fact that tan is negative. The reference angle is approimatel We find angle b subtracting from 60: Thus,. 79.5

102 68 Chapter 5 The Trigonometric Functions 5. Eercise Set For angles of the following measures, state in which quadrant the terminal side lies. It helps to sketch the angle in standard position.. 87 III.. IV. 55 III. 0 III I IV III 8. 5 IV 9. 9 II I. 57 II. 5. I Find two positive angles and two negative angles that are coterminal with the given angle. Answers ma var Find the complement and the supplement , , , , ,.69 The terminal side of angle in standard position lies on the given line in the given quadrant. Find sin, cos, and tan. 9. 0; quadrant IV 0. 0; quadrant II. 5 0; quadrant I. 0.8; quadrant III A function value and a quadrant are given. Find the other five function values. Give eact answers.. sin, quadrant III. tan 5, quadrant I 5. cot, quadrant IV 6. cos, quadrant II 5 7. cos, quadrant IV 5 8. sin 5, quadrant III Find the si trigonometric function values for the angle shown (, 5) r (, ) r b f u 8. r (7, ) (9, ) r a Find the reference angle and the eact function value if it eists. 5; 9. cos 50 0 ; 0. sec 5. tan 5 5;. sin 5 5 ;. sin tan 70 Not defined 5. cos 95 5 ; 6. tan 675 5; 7. csc 0 0; 8. sin 00 60; 9. cot 570 0; 50. cos 0 60; 5. tan 0 0; 5. cot 855 5; 5. sec 90 Not defined 5. sin cos csc tan 0 60; 58. cot 80 Not defined Answers to Eercises 8,, and 5 8 can be found on pp. IA- and IA-.

103 Section 5. Trigonometric Functions of An Angle sin 95 5 ; 60. sin 050 0; 6. csc 5 5; 6. sin cos 0 6. tan 80 60; 65. cot sec 5 5; 67. cos sin 5 5; 69. cos tan 0 0 Find the signs of the si trigonometric function values for the given angles Aerial Navigation. In aerial navigation, directions are given in degrees clockwise from north. Thus, east is 90, south is 80, and west is 70. Several aerial directions or bearings are given below. W 70 0 N 50 S 80 E 90 W 70 0 N S 80 5 E N 0 N Use a calculator in Eercises 79 8, but do not use the trigonometric function kes. 79. Given that sin 0.656, cos 0.757, tan 0.869, find the trigonometric function values for Given that sin , cos , tan , find the trigonometric function values for. 8. Given that sin , cos , tan 65.5, find the trigonometric function values for Given that sin , cos , tan , find the trigonometric function values for 5. W 70 S 80 E An airplane flies 50 km from an airport in a direction of 0.How far east of the airport is the plane then? How far south? East: about 0 km; south: 75 km 0 N 0 W E An airplane leaves an airport and travels for 00 mi in a direction of 00.How far north of the airport is the plane then? How far west? North: 50 mi; west: about 87 mi N 0 00 mi 50 km S 80 W E S 80 W S 80 E 90 Answers to Eercises 7 8 can be found on p. IA-.

104 70 Chapter 5 The Trigonometric Functions 85. An airplane travels at 50 kmh for hr in a direction of 8 from Omaha. At the end of this time, how far south of Omaha is the plane? About km 86. An airplane travels at 0 kmh for hr in a direction of 9 from Chicago. At the end of this time, how far north of Chicago is the plane? About 8 km Find the function value. Round to four decimal places. 87. tan cos cot sin sin cos cos tan cos sec csc sin Given the function value and the quadrant restriction, find. FUNCTION VALUE INTERVAL 99. sin , tan , cos , sec , tan , sin 0. 80, csc.080 0, cos , Collaborative Discussion and Writing 07. Wh do the function values of depend onl on the angle and not on the choice of a point on the terminal side? 08. Wh is the domain of the tangent function different from the domains of the sine and the cosine functions? Skill Maintenance Graph the function. Sketch and label an vertical asmptotes. 09. f 0. g 5 Answers to Eercises 09 can be found on p. IA-. Determine the domain and the range of the function.. f 9. g 7 5 Find the zeros of the function. [.],. f [.]. g 6 Find the -intercepts of the graph of the function. 5. f 6. g 6 [.], 0 [.], 0,, 0 Snthesis 7. Valve Cap on a Biccle. The valve cap on a biccle wheel is.5 in. from the center of the wheel. From the position shown, the wheel starts to roll. After the wheel has turned 90,how far above the ground is the valve cap? Assume that the outer radius of the tire is.75 in in. 8. Seats of a Ferris Wheel. The seats of a ferris wheel are 5 ft from the center of the wheel. When ou board the wheel, ou are 5 ft above the ground. After ou have rotated through an angle of 765, how far above the ground are ou? About 5. ft 5 ft 5 ft.75 in..5 in.

105 Section 5. Radians, Arc Length, and Angular Speed 7 5. Radians, Arc Length, and Angular Speed Find points on the unit circle determined b real numbers. Convert between radian and degree measure; find coterminal, complementar, and supplementar angles. Find the length of an arc of a circle; find the measure of a central angle of a circle. Convert between linear speed and angular speed. Another useful unit of angle measure is called a radian. To introduce radian measure, we use a circle centered at the origin with a radius of length. Such a circle is called a unit circle. Its equation is. (, ) circles review section.. Distances on the Unit Circle The circumference of a circle of radius r is r. Thus for the unit circle, where r, the circumference is. If a point starts at A and travels around the circle (Fig. ), it will travel a distance of. If it travels halfwa around the circle (Fig. ), it will travel a distance of, or. p A A B C A d B D A u p Figure Figure Figure Figure If a point C travels 8 of the wa around the circle (Fig. ), it will travel a distance of 8, or. Note that C is of the wa from A to B. If a point D travels 6 of the wa around the circle (Fig. ), it will travel a distance of, or. Note that D is of the wa from A to B. 6

106 7 Chapter 5 The Trigonometric Functions EXAMPLE How far will a point travel if it goes (a), (b), (c) 8, and 5 (d) 6 of the wa around the unit circle? Solution a) of the total distance around the circle is, which is, or. b) The distance will be, which is, or 6. c) The distance will be 8, which is, or. 5 5 d) The distance will be 6, which is, or 5. Think of 5 as. These distances are illustrated in the following figures. 6 q f p A A point ma travel completel around the circle and then continue. For eample, if it goes around once and then continues of the wa around, it will have traveled a distance of, or 5 (Fig. 5). Ever real number determines a point on the unit circle. For the positive number 0, for eample, we start at A and travel counterclockwise a distance of 0. The point at which we stop is the point determined b the number 0. Note that and that 0.6. Thus the point for 0 travels around the unit circle about times (Fig. 6) r 0 A Figure 5 Figure 6

107 Section 5. Radians, Arc Length, and Angular Speed 7 For a negative number, we move clockwise around the circle. Points for and are shown in the figure below. The number 0 determines the point A. w A A d EXAMPLE On the unit circle, mark the point determined b each of the following real numbers. a) b) 7 6 Solution a) Think of 9 as. (See the figure on the left below.) Since 9 0, the point moves counterclockwise. The point goes completel around once and then continues of the wa from A to B. 9 B A k F A B D A b) The number 76 is negative, so the point moves clockwise. From 6 A to B, the distance is, or, so we need to go beond B another distance of 6, clockwise. (See the figure on the right above.) 6

108 7 Chapter 5 The Trigonometric Functions Radian Measure Degree measure is a common unit of angle measure in man everda applications. But in man scientific fields and in mathematics (calculus, in particular), there is another commonl used unit of measure called the radian. Consider the unit circle. Recall that this circle has radius. Suppose we measure, moving counterclockwise, an arc of length, and mark a point T on the circle. T Arc length is u radian r radian 57. If we draw a ra from the origin through T, we have formed an angle. The measure of that angle is radian. The word radian comes from the word radius. Thus measuring radius along the circumference of the circle determines an angle whose measure is radian. One radian is about 57.. Angles that measure radians, radians, and 6 radians are shown below. u radians u radians u 6 radians When we make a complete (counterclockwise) revolution, the terminal side coincides with the initial side on the positive -ais. We then have an angle whose measure is radians, or about 6.8 radians, which is the circumference of the circle: r.

109 Section 5. Radians, Arc Length, and Angular Speed 75 Thus a rotation of 60 ( revolution) has a measure of radians. A half revolution is a rotation of 80, or radians. A quarter revolution is a rotation of 90, or radians, and so on. u 90 q radians.57 radians u 80 p radians. radians u 70 w radians.7 radians u 60 p radians 6.8 radians To convert between degrees and radians, we first note that 60 radians. It follows that 80 radians. To make conversions, we multipl b, noting that: Converting Between Degree and Radian Measure radians radians radians To convert from degree to radian measure, multipl b 80. To convert from radian to degree measure, multipl b 80. radians GCM EXAMPLE a) 0 b) 97.5 Solution Convert each of the following to radians. radians a) Multipling b 0 80 radians, or about.09 radians radians

110 76 Chapter 5 The Trigonometric Functions b) radians 80 radians 97.5 radians radians We also can use a calculator set in RADIAN mode to convert the angle measures. We enter the angle measure followed b (degrees) from the ANGLE menu. (/) r r GCM EXAMPLE Convert each of the following to degrees. a) radians b) 8.5 radians Solution a) radians Multipling b radians b) 8.5 radians 8.5 radians radians 80 radians With a calculator set in DEGREE mode, we can enter the angle measure followed b r (radians) from the ANGLE menu. The radian degree equivalents of the most commonl used angle measures are illustrated in the following figures. 5 f 80 p h q u d 0 A 0 p 60 j w w 5 h j p p A 0 f d q u When a rotation is given in radians, the word radians is optional and is most often omitted. Thus if no unit is given for a rotation, the rotation is understood to be in radians.

111 Section 5. Radians, Arc Length, and Angular Speed 77 We can also find coterminal, complementar, and supplementar angles in radian measure just as we did for degree measure in Section 5.. EXAMPLE 5 Find a positive angle and a negative angle that are coterminal with. Man answers are possible. Solution To f ind angles coterminal with a given angle, we add or subtract multiples of :, p i 6p i Thus, 8 and 6 are two of the man angles coterminal with. EXAMPLE 6 Find the complement and the supplement of 6. Solution Since 90 equals radians, the complement of 6 is, or Since 80 equals radians, the supplement of 6 is Thus the complement of 6 is and the supplement is 56. Arc Length and Central Angles r s s Radian measure can be determined using a circle other than a unit circle. In the figure at left, a unit circle (with radius ) is shown along with another circle (with radius r, r ). The angle shown is a central angle of both circles. From geometr, we know that the arcs that the angle subtends have their lengths in the same ratio as the radii of the circles. The radii of the circles are r and. The corresponding arc lengths are s and s. Thus we have the proportion s s r,

112 78 Chapter 5 The Trigonometric Functions which also can be written as s. s r Now s is the radian measure of the rotation in question. It is common to use a Greek letter, such as, for the measure of an angle or rotation and the letter s for arc length. Adopting this convention, we rewrite the proportion above as s r. In an circle, the measure (in radians) of a central angle, the arc length the angle subtends, and the length of the radius are related in this fashion. Or, in general, the following is true. Radian Measure The radian measure of a rotation is the ratio of the distance s traveled b a point at a radius r from the center of rotation, to the length of the radius r: s r. r u s sr When using the formula, must be in radians and s and r must be epressed in the same unit. EXAMPLE 7 Find the measure of a rotation in radians when a point m from the center of rotation travels m. Solution We have s r m. The unit is understood to be radians. m

113 Section 5. Radians, Arc Length, and Angular Speed 79 r = 5 cm s θ = EXAMPLE 8 Find the length of an arc of a circle of radius 5 cm associated with an angle of radians. Solution We have s r, or s r. Thus s 5 cm, or about 5. cm. Linear Speed and Angular Speed Linear speed is defined to be distance traveled per unit of time. If we use v for linear speed, s for distance, and t for time, then v s t. Similarl, angular speed is defined to be amount of rotation per unit of time. For eample, we might speak of the angular speed of a biccle wheel as 50 revolutions per minute or the angular speed of the earth as radians per da. The Greek letter (omega) is generall used for angular speed. Thus for a rotation and time t, angular speed is defined as. As an eample of how these definitions can be applied, let s consider the refurbished carousel at the Children s Museum in Indianapolis, Indiana. It consists of three circular rows of animals. All animals, regardless of the row, travel at the same angular speed. But the animals in the outer row travel at a greater linear speed than those in the inner rows. What is the relationship between the linear speed v and the angular speed? To develop the relationship we seek, recall that, for rotations measured in radians,. This is equivalent to s r. We divide b time, t,to obtain s t s t v t r t r Dividing b t Now st is linear speed v and t is angular speed. Thus we have the relationship we seek, v r. sr t

114 80 Chapter 5 The Trigonometric Functions Linear Speed in Terms of Angular Speed The linear speed v of a point a distance r from the center of rotation is given b where v r, is the angular speed in radians per unit of time. For the formula v r, the units of distance for v and r must be the same, must be in radians per unit of time, and the units of time for v and must be the same. 600 km 00 km EXAMPLE 9 Linear Speed of an Earth Satellite. An earth satellite in circular orbit 00 km high makes one complete revolution ever 90 min. What is its linear speed? Use 600 km for the length of a radius of the earth. Solution To use the formula v r, we need to know r and : r 600 km 00 km 7600 km, Now, using v r, we have t. 90 min 5 min Radius of earth plus height of satellite We have, as usual, omitted the word radians v 7600 km km km 5. 5 min 5 min min Thus the linear speed of the satellite is approimatel 5 kmmin. EXAMPLE 0 Angular Speed of a Capstan. An anchor is hoisted at a rate of ftsec as the chain is wound around a capstan with a.8-d diameter. What is the angular speed of the capstan?.8 d Capstan Chain Anchor

115 Section 5. Radians, Arc Length, and Angular Speed 8 Solution We will use the formula v r in the form, taking care to use the proper units. Since v is given in feet per second, we need r in feet: Then r d will be in radians per second: v r.8 ftsec.7 ft ft d.7 ft. d Thus the angular speed is approimatel 0.7 radiansec. t ft. sec 0.7sec.7 ft vr The formulas and v r can be used in combination to find distances and angles in various situations involving rotational motion. EXAMPLE Angle of Revolution. A 00 Tundra V8 is traveling at a speed of 65 mph. Its tires have an outside diameter of 0.56 in. Find the angle through which a tire turns in 0 sec in. Solution Recall that t, or. Thus we can find if we know and t. To find, we use the formula v r. The linear speed v of a point on the outside of the tire is the speed of the Tundra, 65 mph. For convenience, we first convert 65 mph to feet per second: v 65 mi hr hr min 580 ft 60 min 60 sec mi 95. ft. sec The radius of the tire is half the diameter. Now r d 0.56 in. 5.8 in. We will convert to feet, since v is in feet per second: r 5.8 in. 5.8 ft in. ft.7 ft. t

116 8 Chapter 5 The Trigonometric Functions Stud Tip The Student s Solutions Manual is an ecellent resource if ou need additional help with an eercise in the eercise sets. It contains worked-out solutions to the odd-numbered eercises in the eercise sets. Using v r, we have so Then in 0 sec, 95. ft, sec.7 ft 95. ftsec.7 ft t sec sec 0 sec 75. Thus the angle, in radians, through which a tire turns in 0 sec is 75.. Answers to Eercises can be found on p. IA-.

117 8 Chapter 5 The Trigonometric Functions 5. Eercise Set For each of Eercises, sketch a unit circle and mark the points determined b the given real numbers.. a) b) c) d) e) f ). a) b) c) 9 d) e) f ). a) b) c) 6 0 d) e) f ) a) b) c) 5 6 d) 5 e) 7 f ) Find two real numbers between and that determine each of the points on the unit circle. 5. M:, ; M N:, ; P:, ; Q Q:, 6 6 P N 6. M:, ; N:, 7 N ; P:, ; M Q:, Q P 5 7

118 Section 5. Radians, Arc Length, and Angular Speed 8 For Eercises 7 and 8, sketch a unit circle and mark the approimate location of the point determined b the given real number. 7. a). b) 7.5 c) d) 0 8. a) 0.5 b).8 c) 7 d) 500 Find a positive angle and a negative angle that are coterminal with the given angle. Answers ma var. 9., 7 0.,., 5., , 8., Find the complement and the supplement Convert to radian measure. Leave the answer in terms of Convert to radian measure. Round the answer to two decimal places Convert to degree measure. Round the answer to two decimal places Certain positive angles are marked here in degrees. Find the corresponding radian measures Certain negative angles are marked here in degrees. Find the corresponding radian measures p Answers to Eercises 7, 8, 5 0, 59, and 60 can be found on pp. IA- and IA-.

119 8 Chapter 5 The Trigonometric Functions Arc Length and Central Angles. Complete the table. Round the answers to two decimal places DISTANCE, S (ARC LENGTH) RADIUS, R ANGLE, 8 ft ft.9 00 cm 5.65 cm 5 6 d. d in.. in. 65. In a circle with a 0-cm radius, an arc cm long subtends an angle of how man radians? how man degrees, to the nearest degree?.; In a circle with a 0-ft diameter, an arc 0 ft long subtends an angle of how man radians? how man degrees, to the nearest degree? ; In a circle with a -d radius, how long is an arc associated with an angle of.6 radians?. d 68. In a circle with a 5-m radius, how long is an arc associated with an angle of. radians? 0.5 m 69. Angle of Revolution. Through how man radians does the minute hand of a clock rotate from :0 P.M.to :0 P.M.?,or about Angle of Revolution. A tire on a 00 Saturn Ion has an outside diameter of.877 in. Through what angle (in radians) does the tire turn while traveling mi? in Linear Speed. A flwheel with a 5-cm diameter is rotating at a rate of 7 radianssec. What is the linear speed of a point on its rim, in centimeters per minute? 50 cmmin 7. Linear Speed. A wheel with a 0-cm radius is rotating at a rate of radianssec. What is the linear speed of a point on its rim, in meters per minute? 5 mmin 7. Angular Speed of a Printing Press. This tet was printed on a four-color web heatset offset press. A clinder on this press has a.7-in. diameter. The linear speed of a point on the clinder s surface is 8. feet per second. What is the angular speed of the clinder, in revolutions per hour? Printers often refer to the angular speed as impressions per hour (IPH). (Source: Scott Coulter, Quebecor World, Taunton, MA) About 8,85 revolutions per hour 7. Linear Speeds on a Carousel. When Alicia and Zoe ride the carousel described earlier in this section, Alicia alwas selects a horse on the outside row, whereas Zoe prefers the row closest to the center. These rows are 9 ft in. and ft in. from the center, respectivel. The angular speed of the carousel is. revolutions per minute. What is the difference, in miles per hour, in the linear speeds of Alicia and Zoe? (Source: The Children s Museum, Indianapolis, IN) 0.9 mph 75. Linear Speed at the Equator. The earth has a 000-mi radius and rotates one revolution ever hr. What is the linear speed of a point on the equator, in miles per hour? 07 mph

120 Section 5. Radians, Arc Length, and Angular Speed Linear Speed of the Earth. The earth is about 9,000,000 mi from the sun and traverses its orbit, which is nearl circular, ever 65.5 das. What is the linear velocit of the earth in its orbit, in miles per hour? 66,659 mph 77. Determining the Speed of a River. A water wheel has a 0-ft radius. To get a good approimation of the speed of the river, ou count the revolutions of the wheel and find that it makes revolutions per minute (rpm). What is the speed of the river, in miles per hour? 0 mph 79. John Deere Tractor. A rear wheel and tire on a John Deere 850 farm tractor has a 9-in. radius. Find the angle (in radians) through which a wheel rotates in sec if the tractor is traveling at a speed of mph. About 9 9 in. 0 ft 78. The Tour de France. Lance Armstrong won the 00 Tour de France biccle race. The wheel of his biccle had a 67-cm diameter. His overall average linear speed during the race was kmh. What was the angular speed of the wheel, in revolutions per hour? (Source: tourdefrancenews.com) About 9,70 revolutionshr Collaborative Discussion and Writing 80. Eplain in our own words wh it is preferable to omit the word, or unit, radians in radian measures. 8. In circular motion with a fied angular speed, the length of the radius is directl proportional to the linear speed. Eplain wh with an eample. 8. Two new cars are each driven at an average speed of 60 mph for an etended highwa test drive of 000 mi. The diameters of the wheels of the two cars are 5 in. and 6 in., respectivel. If the cars use tires of equal durabilit and profile, differing onl b the diameter, which car will probabl need new tires first? Eplain our answer. Skill Maintenance In each of Eercises 8 90, fill in the blanks with the correct terms. Some of the given choices will not be used. inverse a horizontal line a vertical line eponential function logarithmic function natural common logarithm one-to-one a relation vertical asmptote horizontal asmptote even function odd function sine of cosine of tangent of

121 86 Chapter 5 The Trigonometric Functions 8. The domain of a(n) function f is the range of the inverse. [.] one-to-one 8. The is the length of the side adjacent to divided b the length of the hpotenuse. [5.] cosine of 85. The function f a, where is a real number, a 0 and a, is called the, base a. [.] eponential function 86. The graph of a rational function ma or ma not cross a(n). [.5] horizontal asmptote 87. If the graph of a function f is smmetric with respect to the origin, we sa that it is a(n). [.7] odd function 88. Logarithms, base e,are called logarithms. [.] natural 89. If it is possible for a(n) to intersect the graph of a function more than once, then the function is not one-to-one and its is not a function. [.] horizontal line; inverse 90. A(n) is an eponent. [.] logarithm Snthesis 9. On the earth, one degree of latitude is how man kilometers? how man miles? (Assume that the radius of the earth is 600 km, or 000 mi, approimatel.).7 km; 69.8 mi 9. A point on the unit circle has -coordinate 5.What is its -coordinate? Check using a calculator. 9. A mil is a unit of angle measure. A right angle has a measure of 600 mils. Convert each of the following to degrees, minutes, and seconds. a) 00 mils 570 b) 50 mils A grad is a unit of angle measure similar to a degree. A right angle has a measure of 00 grads. Convert each of the following to grads. a) 8 5. grads 5 5 b).86 grads 7 f 95. Angular Speed of a Gear Wheel. One gear wheel turns another, the teeth being on the rims. The wheels have 0-cm and 50-cm radii, and the smaller wheel rotates at 0 rpm. Find the angular speed of the larger wheel, in radians per second..676 radianssec 96. Angular Speed of a Pulle. Two pulles, 50 cm and 0 cm in diameter, respectivel, are connected b a belt. The larger pulle makes revolutions per minute. Find the angular speed of the smaller pulle, in radians per second..09 radianssec 50 cm 50 cm 0 cm 0 cm 97. Distance Between Points on the Earth. To find the distance between two points on the earth when their latitude and longitude are known, we can use a right triangle for an ecellent approimation if the points are not too far apart. Point A is at latitude 870 N, longitude 8575 W; and point B is at latitude 885 N, longitude 8560 W. Find the distance from A to B in nautical miles. (One minute of latitude is one nautical mile.).6 nautical miles 98. Hands of a Clock. At what times between noon and :00 P.M.are the hands of a clock perpendicular? 6.66 min after :00 noon, or about :6: P.M.

122 Section 5.5 Circular Functions: Graphs and Properties Circular Functions: Graphs and Properties Stud Tip Take advantage of the numerous detailed art pieces in this tet. The provide a visual image of the concept being discussed. Taking the time to stud each figure is an efficient wa to learn and retain the concepts. Given the coordinates of a point on the unit circle, find its reflections across the -ais, the -ais, and the origin. Determine the si trigonometric function values for a real number when the coordinates of the point on the unit circle determined b that real number are given. Find function values for an real number using a calculator. Graph the si circular functions and state their properties. The domains of the trigonometric functions, defined in Sections 5. and 5., have been sets of angles or rotations measured in a real number of degree units. We can also consider the domains to be sets of real numbers, or radians, introduced in Section 5.. Man applications in calculus that use the trigonometric functions refer onl to radians. Let s again consider radian measure and the unit circle. We defined radian measure for as s r When r, s. s, or. u (, ) s u The arc length s on the unit circle is the same as the radian measure of the angle. In the figure above, the point, is the point where the terminal side of the angle with radian measure s intersects the unit circle. We can now etend our definitions of the trigonometric functions using domains composed of real numbers, or radians. In the definitions, s can be considered the radian measure of an angle or the measure of an arc length on the unit circle. Either wa, s is a real number. To each real number s, there corresponds an arc length s on the unit circle. Trigonometric functions with domains composed of real numbers are called circular functions.

123 88 Chapter 5 The Trigonometric Functions (, ) s Basic Circular Functions For a real number s that determines a point, on the unit circle: sin s second coordinate, cos s first coordinate, second coordinate tan s 0, first coordinate csc s 0, second coordinate sec s 0, first coordinate first coordinate cot s 0. second coordinate We can consider the domains of trigonometric functions to be real numbers rather than angles. We can determine these values for a specific real number if we know the coordinates of the point on the unit circle determined b that number. As with degree measure, we can also find these function values directl using a calculator. (, ) 5 5 Reflections on the Unit Circle Let s consider the unit circle and a few of its points. For an point, on the unit circle,, we know that and. If we know the - or -coordinate of a point on the unit circle, we can find the other coordinate. If 5, then (, 5 ) Thus, and 5, 5, 5 5 are points on the unit circle. There are two points with an -coordinate of 5. Now let s consider the radian measure and determine the coordinates of the point on the unit circle determined b. We construct a right triangle b dropping a perpendicular segment from the point to the -ais. (, ) s u u u 0 (, ) 60 q

124 Section 5.5 Circular Functions: Graphs and Properties 89 Since 60, we have a 0 60 right triangle in which the side opposite the 0 angle is one half of the hpotenuse. The hpotenuse, or radius, is, so the side opposite the 0 angle is, or. Using the Pthagorean theorem, we can find the other side: q, u. We know that is positive since the point is in the first quadrant. Thus the coordinates of the point determined b are and,or,. We can alwas check to see if a point is on the unit circle b substituting into the equation :. Because a unit circle is smmetric with respect to the -ais, the -ais, and the origin, we can use the coordinates of one point on the unit circle to find coordinates of its reflections. EXAMPLE Each of the following points lies on the unit circle. Find their reflections across the -ais, the -ais, and the origin. a) b), 5, 5 c), Solution a) b) c) E, R E, R,, q, q, E, R E, R,, q, q,

125 90 Chapter 5 The Trigonometric Functions (, 0) p (0, ) q (0, ) q,, u da, q 0 p (, 0) Finding Function Values Knowing the coordinates of onl a few points on the unit circle along with their reflections allows us to find trigonometric function values of the most frequentl used real numbers, or radians. EXAMPLE Find each of the following function values. a) tan b) cos c) sin d) cos 6 e) cot f ) csc 7 Solution We locate the point on the unit circle determined b the rotation, and then find its coordinates using reflection if necessar. a) The coordinates of the point determined b b) The reflection of, across the -ais are,. is,. u q,, f d, Thus, tan. c) The reflection of, across the -ais is,. Thus, cos. d) The reflection of, across the origin is,. A, q u q, A, q o q, Thus, sin. 6 Thus, cos.

126 Section 5.5 Circular Functions: Graphs and Properties 9 e) The coordinates of the point determined b are, 0. f) The coordinates of the point determined b 7 are 0,. (0, ) (, 0) p t Thus, cot, which is not defined. 0 We can also think of cot as the reciprocal of tan.since tan and the reciprocal of 0 is not defined, we know that cot is not defined. 0 0 Thus, csc 7. Normal Sci Eng Float Radian Degree Func Par Pol Seq Connected Dot Sequential Simul Real abi reˆ θi Full Horiz G T Using a calculator, we can find trigonometric function values of an real number without knowing the coordinates of the point that it determines on the unit circle. Most calculators have both degree and radian modes. When finding function values of radian measures, or real numbers, we must set the calculator in RADIAN mode.(see the window at left.) cos(π/5) tan() sin(.9) EXAMPLE Find each of the following function values of radian measures using a calculator. Round the answers to four decimal places. a) cos b) tan 5 c) sin.9 d) sec 7 Solution Using a calculator set in RADIAN mode, we find the values. a) cos b) tan c) sin d) sec cos 7 Note in part (d) that the secant function value can be found b taking the reciprocal of the cosine value. Thus we can enter cos 7 and use the reciprocal ke.

127 9 Chapter 5 The Trigonometric Functions EXPLORING WITH TECHNOLOGY We can graph the unit circle using a graphing calculator. We use PARAMETRIC mode with the following window and let XT cos T and YT sin T. Here we use DEGREE mode. WINDOW Tmin 0 XTcos(T) YTsin(T) Tma 60 Tstep Xmin.5 T 0 Xma.5 X Xscl Y.5 Ymin Yma Yscl Using the trace ke and an arrow ke to move the cursor around the unit circle, we see the T, X, and Y values appear on the screen. What do the represent? Repeat this eercise in RADIAN mode. What do the T, X, and Y values represent? (For more on parametric equations, see Section 9.7.) From the definitions on p. 88, we can relabel an point, on the unit circle as cos s, sin s, where s is an real number. (, ) (cos s, sin s) s (, 0) q, (0, ),, q u f da, q 0 (, 0) (, 0) p p f A, q q u d,, (0, ) q, Graphs of the Sine and Cosine Functions Properties of functions can be observed from their graphs. We begin b graphing the sine and cosine functions. We make a table of values, plot the points, and then connect those points with a smooth curve. It is helpful to first draw a unit circle and label a few points with coordinates. We can either use the coordinates as the function values or find approimate sine and cosine values directl with a calculator.

128 Section 5.5 Circular Functions: Graphs and Properties 9 s sin s cos s s sin s cos s The graphs are as follows. sin s p w p q q p w p s The sine function cos s p w p q q p w p s We can check these graphs using a graphing calculator. The cosine function sin cos π π π π π Xscl π Xscl

129 9 Chapter 5 The Trigonometric Functions The sine and cosine functions are continuous functions. Note in the graph of the sine function that function values increase from 0 at s 0 to at s, then decrease to 0 at s, decrease further to at s, and increase to 0 at. The reverse pattern follows when s decreases from 0 to. Note in the graph of the cosine function that function values start at when s 0, and decrease to 0 at s. The decrease further to at s, then increase to 0 at s, and increase further to at s. An identical pattern follows when s decreases from 0 to. From the unit circle and the graphs of the functions, we know that the domain of both the sine and cosine functions is the entire set of real numbers,,. The range of each function is the set of all real numbers from to,,. Domain and Range of Sine and Cosine Functions The domain of the sine and cosine functions is,. The range of the sine and cosine functions is,. π EXPLORING WITH TECHNOLOGY Another wa to construct the sine and cosine graphs is b considering the unit circle and transferring vertical distances for the sine function and horizontal distances for the cosine function. Using a graphing calculator, we can visualize the transfer of these distances. We use the calculator set in PARAMETRIC and RADIAN modes and let XT cos T and YT sin T for the unit circle centered at, 0 and XT T and YT sin T for the sine curve. Use the following window settings. π Tmin 0 Tma Tstep. Xmin Xma Xscl Ymin Yma Yscl With the calculator set in SIMULTANEOUS mode, we can actuall watch the sine function (in red) unwind from the unit circle (in blue). In the two screens at left, we partiall illustrate this animated procedure. Consult our calculator s instruction manual for specific kestrokes and graph both the sine curve and the cosine curve in this manner. (For more on parametric equations, see Section 9.7.) A function with a repeating pattern is called periodic. The sine and cosine functions are eamples of periodic functions. The values of these functions repeat themselves ever units. In other words, for an s, we have sin s sin s and cos s cos s.

130 Section 5.5 Circular Functions: Graphs and Properties 95 To see this another wa, think of the part of the graph between 0 and and note that the rest of the graph consists of copies of it. If we translate the graph of sin or cos to the left or right units, we will obtain the original graph. We sa that each of these functions has a period of. Periodic Function A function f is said to be periodic if there eists a positive constant p such that fs p fs for all s in the domain of f.the smallest such positive number p is called the period of the function. T s s p The period p can be thought of as the length of the shortest recurring interval. We can also use the unit circle to verif that the period of the sine and cosine functions is. Consider an real number s and the point T that it determines on a unit circle, as shown at left. If we increase s b, the point determined b s is again the point T. Hence for an real number s, sin s sin s and cos s cos s. It is also true that sin s sin s, sin s 6 sin s, and so on. In fact, for an integer k, the following equations are identities: sin s k sin s and cos s k cos s, or sin s sin s k and cos s cos s k. The amplitude of a periodic function is defined as one half of the distance between its maimum and minimum function values. It is alwas positive. Both the graphs and the unit circle verif that the maimum value of the sine and cosine functions is, whereas the minimum value of each is. Thus, the amplitude of the sine function sin Amplitude: p p p p Period: p

131 96 Chapter 5 The Trigonometric Functions and the amplitude of the cosine function is. cos Amplitude: p p p p Period: p EXPLORING WITH TECHNOLOGY Using the TABLE feature on a graphing calculator, compare the -values for sin and sin and for cos and cos. We set TblMin 0 and Tbl. X X 0 Y Y X X 0 Y Y What appears to be the relationship between sin and sin and between cos and cos? Consider an real number s and its opposite, s. These numbers determine points T and T on a unit circle that are smmetric with respect to the -ais. s s T(, ) sin s T(, ) cos s T (, ) sin (s) s T (, ) cos (s) s Because their second coordinates are opposites of each other, we know that for an number s, sin s sin s. Because their first coordinates are the same, we know that for an number s, cos s cos s.

132 Section 5.5 Circular Functions: Graphs and Properties 97 even and odd functions review section.7. Thus we have shown that the sine function is odd and the cosine function is even. The following is a summar of the properties of the sine and cosine functions. C ONNECTING THE CONCEPTS COMPARING THE SINE AND COSINE FUNCTIONS SINE FUNCTION COSINE FUNCTION sin p p p p cos p p p p. Continuous. Period:. Domain: All real numbers. Range:, 5. Amplitude: 6. Odd: sin s sin s. Continuous. Period:. Domain: All real numbers. Range:, 5. Amplitude: 6. Even: cos s cos s Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions To graph the tangent function, we could make a table of values using a calculator, but in this case it is easier to begin with the definition of tangent and the coordinates of a few points on the unit circle. We recall that tan s sin s. cos s (, ) (cos s, sin s) (0, ) sin s,, s tan s cos s (, 0) (, 0) (, 0), (0, ),

133 98 Chapter 5 The Trigonometric Functions The tangent function is not defined when, the first coordinate, is 0. That is, it is not defined for an number s whose cosine is 0: s,, 5.,... We draw vertical asmptotes at these locations (see Fig. below). pw p q q p w p s pw p q q p w p s Figure Figure We also note that tan s 0 at s 0,,,,..., tan s at s... 7,,,,,,... tan s at s... 9, 5,,,.,... We can add these ordered pairs to the graph (see Fig. above) and investigate the values in, using a calculator. Note that the function value is 0 when s 0, and the values increase without bound as s increases toward. The graph gets closer and closer to an asmptote as s gets closer to, but it never touches the line. As s decreases from 0 to, the values decrease without bound. Again the graph gets closer and closer to an asmptote, but it never touches it. We now complete the graph tan s tan π π p w p q q p w p s DOT Mode The tangent function

134 Section 5.5 Circular Functions: Graphs and Properties 99 From the graph, we see that the tangent function is continuous ecept where it is not defined. The period of the tangent function is. Note that although there is a period, there is no amplitude because there are no maimum and minimum values. When cos s 0, tan s is not defined tan s sin scos s. Thus the domain of the tangent function is the set of all real numbers ecept k, where k is an integer. The range of the function is the set of all real numbers. The cotangent function cot s cos ssin s is not defined when, the second coordinate, is 0 that is, it is not defined for an number s whose sine is 0. Thus the cotangent is not defined for s 0,,,,....The graph of the function is shown below. cot s cot /tan π π p w p q q p w p s DOT Mode The cotangent function The cosecant and sine functions are reciprocal functions, as are the secant and cosine functions. The graphs of the cosecant and secant functions can be constructed b finding the reciprocals of the values of the sine and cosine functions, respectivel. Thus the functions will be positive together and negative together. The cosecant function is not defined for those numbers s whose sine is 0. The secant function is not defined for those numbers s whose cosine is 0. In the graphs below, the sine and cosine functions are shown b the gra curves for reference. csc s csc /sin sin s π π p w p q q p w p s DOT Mode The cosecant function

135 500 Chapter 5 The Trigonometric Functions sec /cos cos s sec s π π p w p q q p w p s DOT Mode The secant function The following is a summar of the basic properties of the tangent, cotangent, cosecant, and secant functions. These functions are continuous ecept where the are not defined. C ONNECTING THE CONCEPTS COMPARING THE TANGENT, COTANGENT, COSECANT, AND SECANT FUNCTIONS TANGENT FUNCTION. Period:. Domain: All real numbers ecept k, where k is an integer. Range: All real numbers COSECANT FUNCTION. Period:. Domain: All real numbers ecept, where k is an integer. Range:,, k COTANGENT FUNCTION. Period:. Domain: All real numbers ecept, where k is an integer. Range: All real numbers SECANT FUNCTION k. Period:. Domain: All real numbers ecept k, where k is an integer. Range:,, In this chapter, we have used the letter s for arc length and have avoided the letters and, which generall represent first and second coordinates. Nevertheless, we can represent the arc length on a unit circle b an variable, such as s, t,, or. Each arc length determines a point that can be labeled with an ordered pair. The first coordinate of that ordered pair is the cosine of the arc length, and the second coordinate is the sine of the arc length. The identities we have developed hold no matter what smbols are used for variables for eample, cos s cos s, cos cos, cos cos, and cos t cos t.

136 Section 5.5 Circular Functions: Graphs and Properties Eercise Set The following points are on the unit circle. Find the coordinates of their reflections across (a) the -ais, (b) the -ais, and (c) the origin... 5, 7,..,, The number determines a point on the unit circle with coordinates,. What are the coordinates of the point determined b? 6. A number determines a point on the unit circle with coordinates, 5. What are the coordinates of the point determined b? Find the function value using coordinates of points on the unit circle. Give eact answers. 7. sin 0 8. cos Find the function value using a calculator set in RADIAN mode. Round the answer to four decimal places, where appropriate. 5. tan cos sec sin cot tan..60. cos 6. sin csc sec tan 7 6. cos sin cot 7 Not defined 9. sin cos tan sin cot 7 0. tan 9 6 In Eercises 8, make hand-drawn graphs.. sin 0. csc. a) Sketch a graph of sin. b) B reflecting the graph in part (a), sketch a graph of sin.. cos. tan c) B reflecting the graph in part (a), sketch a graph 5 6 of sin. Same as (b) d) How do the graphs in parts (b) and (c) compare? 5. sec Not defined 6. cos 0 The same. a) Sketch a graph of cos. b) B reflecting the graph in part (a), sketch a graph 7. cos 8. sin of cos. Same as (a) 6 c) B reflecting the graph in part (a), sketch a graph of cos. 9. sin 5 0. cos d) How do the graphs in parts (a) and (b) compare? 6 The same 5. a) Sketch a graph of sin. See Eercise (a). b) B translating, sketch a graph of. sin 5 0. tan Not defined sin. c) B reflecting the graph of part (a), sketch a graph. 0. of sin. Same as (b) cot 5 tan 5 d) How do the graphs of parts (b) and (c) compare? The same Answers to Eercises 6, (a), (b), (a), (c), and 5(b) can be found on p. IA-. 5

137 50 Chapter 5 The Trigonometric Functions 6. a) Sketch a graph of sin. See Eercise 7(a). b) B translating, sketch a graph of sin. c) B reflecting the graph of part (a), sketch a graph of sin. Same as (b) d) How do the graphs of parts (b) and (c) compare? The same 7. a) Sketch a graph of cos. b) B translating, sketch a graph of cos. c) B reflecting the graph of part (a), sketch a graph of cos. Same as (b) d) How do the graphs of parts (b) and (c) compare? The same 8. a) Sketch a graph of cos. See Eercise 7(a). b) B translating, sketch a graph of cos. c) B reflecting the graph of part (a), sketch a graph of cos. Same as (b) d) How do the graphs of parts (b) and (c) compare? The same 9. Of the si circular functions, which are even? Which are odd? 50. Of the si circular functions, which have period? Which have period? Consider the coordinates on the unit circle for Eercises In which quadrants is the tangent function positive? negative? Positive: I, III; negative: II, IV 5. In which quadrants is the sine function positive? negative? Positive: I, II; negative: III, IV 5. In which quadrants is the cosine function positive? negative? Positive: I, IV; negative: II, III 5. In which quadrants is the cosecant function positive? negative? Positive: I, II; negative: III, IV Collaborative Discussion and Writing 55. Describe how the graphs of the sine and cosine functions are related. 56. Eplain wh both the sine and cosine functions are continuous, but the tangent function, defined as sinecosine, is not continuous. Skill Maintenance Graph both functions in the same viewing window and describe how g is a transformation of f. 57. f, g 58. f, g 59. f, g 60. f, g Write an equation for a function that has a graph with the given characteristics. Check using a graphing calculator. 6. The shape of, but reflected across the -ais, shifted right units, and shifted down unit [.7] 6. The shape of, but shrunk verticall b a factor of and shifted up units [.7] Snthesis Complete. (For eample, sin sin.) 6. cos cos 6. sin sin 65. sin k, k sin 66. cos k, k cos 67. sin sin 68. cos cos 69. cos cos 70. cos cos 7. sin sin 7. sin sin 7. Find all numbers that satisf the following. Check using a graphing calculator. a) sin k, k b) cos, k c) sin 0, k k k 7. Find f g and g f, where f and g cos. Use a graphing calculator to determine the domain, the range, the period, and the amplitude of the function. 75. sin 76. cos Answers to Eercises 6(b), 7(a), 7(b), 8(b), 9, 50, 57 60, and 7 76 can be found on p. IA-.

138 Section 5.5 Circular Functions: Graphs and Properties 50 Determine the domain of the function. 77. f cos 78. g sin 79. f sin k, k an integer. 80. g log sin cos Graph. 8. sin 8. sin 8. sin cos 8. cos 85. One of the motivations for developing trigonometr with a unit circle is that ou can actuall see sin and cos on the circle. Note in the figure at right that AP sin and OA cos. It turns out that ou can also see the other four trigonometric functions. Prove each of the following. a) BD tan b) OD sec c) OE csc d) CE cot Answers to Eercises 77 and can be found on pp. IA- and IA-. C u O A B 86. Using graphs, determine all numbers that satisf sin cos. 87. Using a calculator, consider sin, where is between 0 and. As approaches 0, this function approaches a limiting value. What is it? P D E

139 5 Chapter 5 The Trigonometric Functions Chapter 5 Summar and Review Important Properties and Formulas Trigonometric Function Values of an Acute Angle Let be an acute angle of a right triangle. The si trigonometric functions of are as follows: sin opp, cos adj, tan opp, hp hp adj csc hp, sec hp, cot adj. opp adj opp Hpotenuse u Adjacent to u Opposite u Reciprocal Functions csc, sec, sin cos cot tan Function Values of Special Angles sin 0 cos 0 tan 0 Not defined Cofunction Identities u 90 u sin cos 90, cos sin 90, tan cot 90, cot tan 90, sec csc 90, csc sec 90 (continued)

140 Chapter 5 Summar and Review 5 Trigonometric Functions of An Angle If P, is an point on the terminal side of an angle in standard position, and r is the distance from the origin to P,, where r, then sin, cos, tan, r r P(, ) r u csc r, sec r, cot. Signs of Function Values The signs of the function values depend onl on the coordinates of the point P on the terminal side of an angle. Positive: sin Negative: cos, tan II (, ) Positive: All Negative: None (, ) I Basic Circular Functions For a real number s that determines a point on the unit circle: (, ) s sin s, cos s, tan s., III (, ) (, ) IV Positive: tan Negative: sin, cos Positive: cos Negative: sin, tan Sine is an odd function: sin s sin s. Cosine is an even function: cos s cos s. Radian Degree Equivalents Linear Speed in Terms of Angular Speed v r 80 5 p 5 f h q u d 0 A 0 p 60 j w 5 70 Transformations of Sine and Cosine Functions To graph A sin B C D and A cos B C D :. Stretch or shrink the graph horizontall according to B. Period B. Stretch or shrink the graph verticall according to A.(Amplitude A). Translate the graph horizontall according to CB. Phase shift B C. Translate the graph verticall according to D.

141 5 Chapter 5 The Trigonometric Functions Review Eercises. Find the si trigonometric function values of.. Given that sin 9, find the other five 0 trigonometric function values. Find the eact function value, if it eists.. cos 5 [5.]. cot 60 [5.] 5. cos 95 [5.] 6. sin 50 [5.] 7. sec tan 600 [5.] [5.] Not defined 9. csc 60 [5.] 0. cot 5 [5.]. Convert.7 to degrees, minutes, and seconds. Round to the nearest second. [5.] 6. Convert 77 to decimal degree notation. Round to two decimal places. [5.] 7.56 Find the function value. Round to four decimal places.. tan 8 [5.] 0.5. sec 7.9 [5.].5 5. cos 8 6. sin 5 [5.] [5.] cot. 8. sin 556. [5.].59 [5.] Find in the interval indicated. Round the answer to the nearest tenth of a degree. 9. cos 0.90, 80, 70 [5.] tan.0799, 0, 90 [5.] 7. u Find the eact acute angle, in degrees, given the function value sin [5.] 60. tan [5.] 60. cos [5.] 5. sec [5.] 0 5. Given that sin , cos , and tan , find the si function values for 0.9. Answers to Eercises,, 5, and 5 7 can be found on pp. IA-5 and IA-6. Solve each of the following right triangles. Standard lettering has been used. 6. a 7., c 8.6 [5.] b.5, A 58., B.9 7. a 0.5, B 5.7 [5.] A 8.8, b 7.9, c One leg of a right triangle bears east. The hpotenuse is 7 m long and bears N57E. Find the perimeter of the triangle. [5.] 78 m 9. An observer s ee is 6 ft above the floor. A mural is being viewed. The bottom of the mural is at floor level. The observer looks down to see the bottom and up 7 to see the top. How tall is the mural? [5.] ft 6 ft For angles of the following measures, state in which quadrant the terminal side lies [5.] II. 65. [5.] I. 9 [5.] IV Find a positive angle and a negative angle that are coterminal with the given angle. Answers ma var.. 65 [5.] 5, 95. [5.], Find the complement and the supplement Find the si trigonometric function values for the angle shown. 7 (, ) r u 7 5

142 8. Given that tan 5 and that the terminal side is in quadrant III, find the other five function values. 9. An airplane travels at 50 mph for hr in a direction of 60 from Minneapolis, Minnesota. At the end of that time, how far south of Minneapolis is the airplane? [5.] About 7 mi 0. On a unit circle, mark and label the points determined b 76,,, and 9. For angles of the following measures, convert to radian measure in terms of, and convert to radian measure not in terms of. Round the answer to two decimal places.. 5. [5.],.5. 0 [5.], Convert to degree measure. Round the answer to two decimal places.. [5.] 70. [5.] [5.] [5.] Find the length of an arc of a circle, given a central angle of and a radius of 7 cm. [5.] 7, or 5.5 cm 8. An arc 8 m long on a circle of radius 8 m subtends an angle of how man radians? how man degrees, to the nearest degree? [5.].5, 9 9. At one time, inside La Madeleine French Baker and Cafe in Houston, Teas, there was one of the few remaining working watermills in the world. The 00-r-old French-built waterwheel had a radius of 7 ft and made one complete revolution in 70 sec. What was the linear speed in feet per minute of a point on the rim? (Source: La Madeleine French Baker and Cafe, Houston, TX) [5.] About 7.9 ftmin 50. An automobile wheel has a diameter of in. If the car travels at a speed of 55 mph, what is the angular velocit, in radians per hour, of a point on the edge of the wheel? [5.] 97,89 radianshr 5, 5 5. The point is on a unit circle. Find the coordinates of its reflections across the -ais, the -ais, and the origin. [5.5],, 5, 5, 5, Find the eact function value, if it eists. 5. cos [5.5] 5. tan 5 [5.5] Answers to Eercises 8, 0, 6, 68, and 69 can be found on p. IA sin 5 [5.5] 55. sin 7 [5.5] 56. tan [5.5] 57. cos [5.5] 6 Find the function value. Round to four decimal places. 58. sin [5.5] cos 75 [5.5] cot 6 6. tan [5.5].8 [5.5] Not defined 7 6. sec. [5.5] cos [5.5] Graph b hand each of the si trigonometric functions from to. 65. What is the period of each of the si trigonometric functions? [5.5] Period of sin, cos, sec, csc: ; period of tan, cot: 66. Complete the following table. Domain of tangent: k, k 67. Complete the following table with the sign of the specified trigonometric function value in each of the four quadrants. Determine the amplitude, the period, and the phase shift of the function, and sketch the graph of the function. Then check the graph using a graphing calculator. 68. sin Chapter 5 Review Eercises cos 6 5 FUNCTION DOMAIN RANGE sine cosine tangent,,,,, FUNCTION I II III IV sine cosine tangent

143 56 Chapter 5 The Trigonometric Functions In Eercises 70 7, without a graphing calculator, match the function with one of the graphs (a) (d), which follow. Then check our work using a graphing calculator. a) c) 6 b) d) 70. cos [5.6] (d) 7. sin [5.6] (a) 7. sin 7. cos [5.6] (c) [5.6] (b) 7. Sketch a graph of cos sin for values of between 0 and. Collaborative Discussion and Writing 75. Compare the terms radian and degree. 76. Describe the shape of the graph of the cosine function. How man maimum values are there of the cosine function? Where do the occur? 77. Does 5 sin 7 have a solution for? Wh or wh not? 78. Eplain the disadvantage of a graphing calculator when graphing a function like Snthesis f sin. 79. For what values of in 0, is sin true? All values 80. Graph sin, and determine the domain, the range, and the period. 8. In the graph below, sin is shown and is shown in red. Epress as a transformation of the graph of. [5.6] sin sin,? π 8. Find the domain of log cos. 8. Given that sin 0.6 and that the terminal side is in quadrant II, find the other basic circular function values. [5.] cos , tan , cot.8, sec.67, csc.676 π Answers to Eercises 7 78, 80, and 8 can be found on p. IA-6.

144 Trigonometric Identities, Inverse Functions, and Equations 6. Identities: Pthagorean and Sum and Difference 6. Identities: Cofunction, Double-Angle, and Half-Angle 6. Proving Trigonometric Identities 6. Inverses of the Trigonometric Functions 6.5 Solving Trigonometric Equations 6 SUMMARY AND REVIEW TEST A P P L I C A T I O N The number of dalight hours in Fairbanks, Alaska, varies from about.9 hr to 0.6 hr (Source:Astronomical Applications Department, U.S. Naval Observator, Washington, DC). The function Hd sin 0.066d.7.5 can be used to approimate the number of dalight hours H on a certain da d in Fairbanks. We can use this function to determine on which da of the ear there will be about 0.5 hr of dalight. This problem appears as Eercise 5 in Eercise Set 6.5.

145 50 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations. 6. Polnomial Identities: Functions Pthagorean and and Modeling Sum and Difference State the Pthagorean identities. Simplif and manipulate epressions containing trigonometric epressions. Use the sum and difference identities to find function values. An identit is an equation that is true for all possible replacements of the variables. The following is a list of the identities studied in Chapter 5. Basic Identities sin, csc, sin sin, csc sin cos cos, cos, sec, tan tan, sec cos tan, cot, tan sin, cot tan cos cot cos sin In this section, we will develop some other important identities. Pthagorean Identities (, ), or (cos s, sin s) s We now consider three other identities that are fundamental to a stud of trigonometr. The are called the Pthagorean identities. Recall that the equation of a unit circle in the -plane is. (, 0) For an point on the unit circle, the coordinates and satisf this equation. Suppose that a real number s determines a point on the unit circle with coordinates,, or cos s, sin s. Then cos s and sin s. Substituting cos s for and sin s for in the equation of the unit circle gives us the identit cos s sin s, Substituting cos s for and sin s for which can be epressed as sin s cos s.

146 Section 6. Identities: Pthagorean and Sum and Difference 5 It is conventional in trigonometr to use the notation sin s sin s.note that sin s sin s. rather than (sin ) (sin )(sin ) sin sin ( ) The identit sin s cos s gives a relationship between the sine and the cosine of an real number s. It is an important Pthagorean identit. EXPLORING WITH TECHNOLOGY Addition of -values provides a unique wa of developing the identit sin cos. First, graph sin and cos. B visuall adding the -values, sketch the graph of the sum, sin cos. Then graph using a graphing calculator and check our sketch. The resulting graph appears to be the line, which is the graph of sin cos. These graphs do not prove the identit, but the do provide a check in the interval shown. sin cos π π π π sin cos, π π

147 5 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations We can divide b sin s on both sides of the preceding identit: sin s. Dividing b sin s sin s cos s sin s sin s Simplifing gives us a second identit: cot s csc s. This equation is true for an replacement of s with a real number for which sin s 0, since we divided b sin s. But the numbers for which sin s 0 (or sin s 0) are eactl the ones for which the cotangent and cosecant functions are not defined. Hence our new equation holds for all real numbers s for which cot s and csc s are defined and is thus an identit. The third Pthagorean identit can be obtained b dividing b cos s on both sides of the first Pthagorean identit: sin s cos s cos s cos s cos s Dividing b cos s tan s sec s. Simplifing The identities we have developed hold no matter what smbols are used for the variables. For eample, we could write sin s cos s, sin cos,or sin cos. Pthagorean Identities sin cos, cot csc, tan sec It is often helpful to epress the Pthagorean identities in equivalent forms. PYTHAGOREAN IDENTITIES sin cos cot csc tan sec EQUIVALENT FORMS sin cos cos sin csc cot cot csc sec tan tan sec Simplifing Trigonometric Epressions We can factor, simplif, and manipulate trigonometric epressions in the same wa that we manipulate strictl algebraic epressions.

148 Section 6. Identities: Pthagorean and Sum and Difference 5 Stud Tip The eamples in each section were chosen to prepare ou for success with the eercise set. Stud the step-b-step annotated solutions of the eamples, noting that substitutions are highlighted in red. The time ou spend understanding the eamples will save ou valuable time when ou do our assignment. EXAMPLE Multipl and simplif: cos tan sec. Solution cos tan sec cos tan cos sec cos sin cos cos cos sin Multipling Recalling the identities tan sin cos and sec and substituting cos Simplifing There is no general procedure for manipulating trigonometric epressions, but it is often helpful to write everthing in terms of sines and cosines, as we did in Eample. We also look for the Pthagorean identit, sin cos, within a trigonometric epression. π cos [tan (/cos )], sin X TblStart π Tbl π/ Y.99 ERROR ERROR X Y π GCM EXAMPLE Factor and simplif: sin cos cos. Solution sin cos cos cos sin cos cos cos Removing a common factor Using sin cos A graphing calculator can be used to perform a partial check of an identit. First, we graph the epression on the left side of the equals sign. Then we graph the epression on the right side using the same screen. If the two graphs are indistinguishable, then we have a partial verification that the equation is an identit. Of course, we can never see the entire graph, so there can alwas be some doubt. Also, the graphs ma not overlap precisel, but ou ma not be able to tell because the difference between the graphs ma be less than the width of a piel. However, if the graphs are obviousl different, we know that a mistake has been made. Consider the identit in Eample : cos tan sec sin. Recalling that sec cos, we enter cos tan cos and sin. To graph, we first select SEQUENTIAL mode. Then we select the line -graph stle for and the path -graph stle, denoted b, for. The calculator will graph first. Then as it graphs, the circular cursor will trace the leading edge of the graph, allowing us to determine whether the graphs coincide. As ou can see in the first screen on the left, the graphs appear to be identical. Thus, cos tan sec sin is most likel an identit. The TABLE feature can also be used to check identities. Note in the table at left that the function values are the same ecept for those values of for which cos 0. The domain of ecludes these values. The domain of

149 5 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations is the set of all real numbers. Thus all real numbers ecept,, 5,... are possible replacements for in the identit. Recall that an identit is an equation that is true for all possible replacements. Suppose that we had simplified incorrectl in Eample and had gotten cos on the right instead of sin. Then two different graphs would have appeared in the window. Thus we would have known that we did not have an identit and that cos tan sec cos. cos (tan sec ), cos π π factoring review section R.. EXAMPLE Simplif each of the following trigonometric epressions. cot a) csc sin t sin t b) cos t sin t Solution a) b) cos cot csc sin sin cos sin sin cos cos sin t sin t cos t sin t sin t sin t sin t sin t sin t sin t sin t sin t sin t sin t sin t sin t sin t, or csc t sin t Rewriting in terms of sines and cosines Multipling b the reciprocal The cosine function is even. Substituting sin t for cos t Factoring in both numerator and denominator Simplifing

150 Section 6. Identities: Pthagorean and Sum and Difference 55 We can add and subtract trigonometric rational epressions in the same wa that we do algebraic epressions. rational epressions review section R.5. EXAMPLE Add and simplif: cos tan. sin Solution cos cos sin tan sin sin cos Using cos cos sin sin sin cos cos cos sin sin cos sin sin cos sin sin Multipling b forms of Adding, or sec Simplifing cos tan sin cos Using sin cos When radicals occur, the use of absolute value is sometimes necessar, but it can be difficult to determine when to use it. In Eamples 5 and 6, we will assume that all radicands are nonnegative. This means that the identities are meant to be confined to certain quadrants. EXAMPLE 5 Multipl and simplif: sin cos cos. Solution sin cos cos sin cos sin cos sin sin cos sin EXAMPLE 6 Rationalize the denominator:. tan Solution tan tan tan tan tan tan tan tan

151 56 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations Often in calculus, a substitution is a useful manipulation, as we show in the following eample. 9 u EXAMPLE 7 Epress 9 as a trigonometric function of without using radicals b letting tan. Assume that 0. Then find sin and cos. Solution We have 9 9 tan Substituting tan for 9 9 tan 9 tan 9 sec Factoring Using tan sec sec sec. For 0, sec 0, so sec sec. We can epress 9 sec as 9 sec. In a right triangle, we know that sec is hpotenuseadjacent, when is one of the acute angles. Using the Pthagorean theorem, we can determine that the side opposite is. Then from the right triangle, we see that sin and cos. 9 9 Sum and Difference Identities We now develop some important identities involving sums or differences of two numbers (or angles), beginning with an identit for the cosine of the difference of two numbers. We use the letters u and v for these numbers. Let s consider a real number u in the interval, and a real number v in the interval 0,. These determine points A and B on the unit circle, as shown below. The arc length s is u v, and we know that 0 s. Recall that the coordinates of A are cos u, sin u, and the coordinates of B are cos v, sin v. u s (cos v, sin v) B v (cos u, sin u) A (, 0)

152 Section 6. Identities: Pthagorean and Sum and Difference 57 distance formula review section.. (cos s, sin s) A π s B (, 0) cos ( ), cos cos sin sin π Using the distance formula, we can write an epression for the distance AB: AB cos u cos v sin u sin v. This can be simplified as follows: AB cos u cos u cos v cos v sin u sin u sin v sin v sin u cos u sin v cos v cos u cos v sin u sin v cos u cos v sin u sin v. Now let s imagine rotating the circle on page 56 so that point B is at, 0 as shown at left. Although the coordinates of point A are now cos s, sin s, the distance AB has not changed. Again we use the distance formula to write an epression for the distance AB: AB cos s sin s 0. This can be simplified as follows: AB cos s cos s sin s sin s cos s cos s cos s. Equating our two epressions for AB,we obtain cos u cos v sin u sin v cos s. Solving this equation for cos s gives cos s cos u cos v sin u sin v. () But s u v, so we have the equation cos u v cos u cos v sin u sin v. () Formula () above holds when s is the length of the shortest arc from A to B. Given an real numbers u and v, the length of the shortest arc from A to B is not alwas u v. In fact, it could be v u. However, since cos cos, we know that cos v u cos u v. Thus, cos s is alwas equal to cos u v. Formula () holds for all real numbers u and v. That formula is thus the identit we sought: cos u v cos u cos v sin u sin v. Using a graphing calculator, we can graph cos and cos cos sin sin to illustrate this result. The cosine sum formula follows easil from the one we have just derived. Let s consider cos u v. This is equal to cos u v, and b the identit above, we have cos u v cos u v cos u cos v sin u sin v.

153 58 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations π cos ( ), cos cos sin sin cos(5π/) ( (6) ())/ π But cos v cos v and sin v sin v, so the identit we seek is the following: cos u v cos u cos v sin u sin v. Using a graphing calculator, we can graph cos and cos cos sin sin to illustrate this result. EXAMPLE 8 Find cos 5 eactl. Solution We can epress 5 as a difference of two numbers whose sine and cosine values are known: 9, or. Then, using cos u v cos u cos v sin u sin v, we have 5 cos 5 cos cos cos 6 6. We can check using a graphing calculator set in RADIAN mode. sin sin Consider cos. We can use the identit for the cosine of a difference to simplif as follows: cos cos cos sin sin 0 cos sin sin. Thus we have developed the identit This cofunction identit first sin cos. appeared in Section 5.. () This identit holds for an real number. From it we can obtain an identit for the cosine function. We first let be an real number. Then we replace in sin cos with. This gives us sin, cos cos

154 Section 6. Identities: Pthagorean and Sum and Difference 59 which ields the identit cos sin. () Using identities () and () and the identit for the cosine of a difference, we can obtain an identit for the sine of a sum. We start with identit () and substitute u v for : sin cos s in u v cos u v cos u v Identit () Substituting u v for cos u cos v sin u sin v Using the identit for the cosine of a difference sin u cos v cos u sin v. Using identities () and () Thus the identit we seek is sin u v sin u cos v cos u sin v. To find a formula for the sine of a difference, we can use the identit just derived, substituting v for v: sin u v sin u cos v cos u sin v. Simplifing gives us sin u v sin u cos v cos u sin v. sin(05) (()(6))/ EXAMPLE 9 Find sin 05 eactl. Solution We epress 05 as the sum of two measures: Then sin 05 sin 5 60 sin 5 cos 60 cos 5 sin 60 Using sin u v sin u cos v cos u sin v 6. We can check this result using a graphing calculator set in DEGREE mode.

155 50 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations Formulas for the tangent of a sum or a difference can be derived using identities alread established. A summar of the sum and difference identities follows. Sum and Difference Identities sin u cos u v sin u cos v v cos u cos v cos u sin v, sin u sin v, tan u v tan u tan v tan u tan v There are si identities here, half of them obtained b using the signs shown in color. EXAMPLE 0 Find tan 5 eactl. Solution We rewrite 5 as 5 0 and use the identit for the tangent of a difference: tan 5 tan 5 0. tan 5 tan 0 tan 5 tan 0 EXAMPLE Assume that sin and sin and that and are between 0 and. Then evaluate sin. Solution Using the identit for the sine of a sum, we have sin sin cos cos sin cos cos. a 5 To f inish, we need to know the values of cos and cos. Using reference triangles and the Pthagorean theorem, we can determine these values from the diagrams: Cosine values are positive cos 5 and cos. in the first quadrant. b Substituting these values gives us sin 5 5, or

156 Section 6. Identities: Pthagorean and Sum and Difference 5 6. Eercise Set tan cot sin Multipl and simplif. Check our result using a graphing calculator.. sin cos sin cos sin cos. tan cos csc sin sec. cos sin sec csc sin cos. sin cos sec csc 5. sin cos cos 6. tan sec tan 7. sin csc sin csc sin csc 8. sin t sin t cos t Factor and simplif. Check our result using a graphing calculator. 9. sin cos cos cos sin cos 0. tan cot. sin cos tan cot tan cot sin cos sin cos. sin 8 sin. cos cos sin cos cos. cot 6 cot cot 5. sin 7 sin sin sin tan s 5 tan s 5 tan s 5 tan s Simplif and check using a graphing calculator. sin cos 7. cos sin tan 0 sin cos 5 sin 8.,or 5 tan sin 6 cos sin cos sin sin sin cos cos tan t sec t sec t 6 tan t sec t sec t csc. sec cot cos sin tan t tan t sin cos. sin cos cos. sin sin cos sin s cos s cos s sin s sin 8. cos cos cos sin sin sin cos sin cos tan sec tan sec cos cot sin 9 cos 0 cos 5 sin 9 9 cos 5 cos cos 6 cos 0 Simplif and check using a graphing calculator. Assume that all radicands are nonnegative.. sin cos cos sin cos. cos sin sin cos sin. cos sin cos cos sin cos. tan tan sin sin tan sin 5. sin sin sin 6. cos cos sin cos cos sin Rationalize the denominator. sin sin cos cos cos cot 9. cos 0. sin Rationalize the numerator. cos cos.. sin sin cos sin sin.. sin cos 5 cot sin cos sin s cos s sin s cos s 5sin cos 5cos cos tan cos cos sin cot cos sin sin tan sin cos sin cos cot

157 5 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations Use the given substitution to epress the given radical epression as a trigonometric function without radicals. Assume that a 0 and 0. Then find epressions for the indicated trigonometric functions. 5. Let a sin in a. Then find cos and tan. 6. Let tan in. Then find sin and cos. 7. Let sec in 9. Then find sin and cos. 8. Let a sec in a. Then find sin and cos. Use the given substitution to epress the given radical epression as a trigonometric function without radicals. Assume that Let sin in. 50. Let in 6 sec. Use the sum and difference identities to evaluate eactl. Then check using a graphing calculator sin 5. cos tan tan cos sin tan sin cos sin 7 6 First write each of the following as a trigonometric function of a single angle; then evaluate. 57. sin 7 cos cos 7 sin sin cos 8 cos 5 sin 8 sin 5 cos cos 9 cos 5 sin 9 sin 5 cos sin 0 cos 5 cos 0 sin 5 sin tan 0 tan 6. tan tan 0 tan tan 5 tan 6. tan 0.5 tan 5 tan 6. Derive the formula for the tangent of a sum. 6. Derive the formula for the tangent of a difference. Assuming that sin u and sin v 5 5 and that u and v are between 0 and, evaluate each of the following eactl. 65. cos u v tan u v sin u v cos u v 5 5 Assuming that sin 0.69 and cos 0.0 and that both and are first-quadrant angles, evaluate each of the following. 69. tan sin cos cos 0.55 Simplif. 7. sin sin 7. cos cos 75. cos u v cos v sin u v sin v cos u 76. sin u v cos v cos u v sin v sin u Collaborative Discussion and Writing 77. What is the difference between a trigonometric equation that is an identit and a trigonometric equation that is not an identit? Give an eample of each. 78. Wh is it possible to use a graph to disprove that an equation is an identit but not to prove that one is? Skill Maintenance Solve. 79. [.] All real numbers [.] No solution Given that sin and cos 0.857, find the specified function value. 8. sec 59 [5.] tan 59 [5.].665 Snthesis sin cos sin sin Angles Between Lines. One of the identities gives an eas wa to find an angle formed b two lines. Consider two lines with equations l : m b and l : m b. Answers to Eercises 5 8, 5, 5, 6, and 6 can be found on p. IA-7.

158 Section 6. Identities: Pthagorean and Sum and Difference 5 The slopes m and m are the tangents of the angles and that the lines form with the positive direction of the -ais. Thus we have m tan and m tan. To find the measure of, or, we proceed as follows: tan tan m m. m m This formula also holds when the lines are taken in the reverse order. When is acute, tan will be positive. When is obtuse, tan will be negative. Find the measure of the angle from l to l. 8. l :, 8. l :, l : 5 l : 0; the lines are parallel 6,or l :, 86. l : 0, l : 5, or 5 l : Circus Gu Wire. In a circus, a gu wire A is attached to the top of a 0-ft pole. Wire B is used for performers to walk up to the tight wire, 0 ft above the ground. Find the angle between the wires if the are attached to the ground 0 ft from the pole..8 φ A B 0 ft u tan tan tan tan u 0 ft u l u u or f u l 88. Given that f sin, show that f h f cos h sin h h 89. Given that f cos, show that f h f cos h cos h h Show that each of the following is not an identit b finding a replacement or replacements for which the sides of the equation do not name the same number. Then use a graphing calculator to show that the equation is not an identit. sin sin sin 9. sin 5 9. sin sin 9. cos cos cos 6 9. tan cot 95. cos 6 Find the slope of line l, where m is the slope of line l and is the smallest positive angle from l to l. 96. m, m, Line l contains the points, and 5,. Find the slope of line l such that the angle from l to l is Line l contains the points, 7 and,. Line l contains 0, and, 6. Find the smallest positive angle from l to Find an identit for sin. (Hint: sin sin cos.) 0. Find an identit for cos. (Hint:.) Derive the identit. Check using a graphing calculator. 0. sin cos tan 0. tan tan 6 sin 0. cos tan tan tan tan cos sin h sin sin h 05. sin sin sin cos sin sin sin cos cos sin sin cos cos sin sin cos l h h.. Answers to Eercises and 0 0 can be found on pp. IA-7 and IA-8.

159 5 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations 6. Identities: Cofunction, Double-Angle, and Half-Angle Use cofunction identities to derive other identities. Use the double-angle identities to find function values of twice an angle when one function value is known for that angle. Use the half-angle identities to find function values of half an angle when one function value is known for that angle. Simplif trigonometric epressions using the double-angle and half-angle identities. Cofunction Identities Each of the identities listed below ields a conversion to a cofunction. For this reason, we call them cofunction identities. Cofunction Identities sin, cos, sin cos tan, cot, tan cot sec, csc csc sec We verified the first two of these identities in Section 6.. The other four can be proved using the first two and the definitions of the trigonometric functions. These identities hold for all real numbers, and thus, for all angle measures, but if we restrict to values such that 0, or 0, then we have a special application to the acute angles of a right triangle. Comparing graphs can lead to possible identities. On the left below, we see that the graph of sin is a translation of the graph of sin to the left units. On the right, we see the graph of cos. 90 sin ( ) sin cos

160 Section 6. Identities: Cofunction, Double-Angle, and Half-Angle 55 Comparing the graphs, we observe a possible identit: sin. cos The identit can be proved using the identit for the sine of a sum developed in Section 6.. EXAMPLE Prove the identit sin cos. Solution s in sin cos sin 0 cos cos cos sin Using sin u v sin u cos v cos u sin v We now state four more cofunction identities. These new identities that involve the sine and cosine functions can be verified using previousl established identities as seen in Eample. Cofunction Identities for the Sine and Cosine sin cos, cos sin EXAMPLE Solution a) We have tan Find an identit for each of the following. a) tan b) sec 90 cos sin cot. Thus the identit we seek is sin cos tan. cot Using tan sin cos Using cofunction identities

161 56 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations b) We have sec 90. cos 90 csc sin Thus, sec 90 csc. Double-Angle Identities If we double an angle of measure, the new angle will have measure. Double-angle identities give trigonometric function values of in terms of function values of. To develop these identities, we will use the sum formulas from the preceding section. We first develop a formula for sin.recall that sin u v sin u cos v cos u sin v. We will consider a number and substitute it for both u and v in this identit. Doing so gives us sin sin sin cos cos sin sin cos. Our first double-angle identit is thus sin sin cos. We can graph sin, and using the line -graph stle for and the path -graph stle for and see that the appear to have the same graph. We can also use the TABLE feature. sin, sin cos sin cos π π X Y E X Y Double-angle identities for the cosine and tangent functions can be derived in much the same wa as the identit above: cos cos sin, tan tan. tan EXAMPLE Given that tan and is in quadrant II, find each of the following. a) sin b) cos c) tan d) The quadrant in which lies

162 Section 6. Identities: Cofunction, Double-Angle, and Half-Angle 57 Solution B drawing a reference triangle as shown, we find that sin 5 (, ) 5 and u cos. 5 Thus we have the following. a) sin sin cos b) cos cos sin c) tan tan tan Note that tan could have been found more easil in this case b simpl dividing: sin 5 tan 7. cos 5 7 d) Since sin is negative and cos is positive, we know that is in quadrant IV. Two other useful identities for cos can be derived easil, as follows. cos cos sin sin sin sin cos cos sin cos cos cos Double-Angle Identities sin sin cos, tan tan tan cos cos sin sin cos Solving the last two cosine double-angle identities for sin and cos, respectivel, we obtain two more identities: and cos cos sin cos.

163 58 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations Using division and these two identities gives us the following useful identit: tan cos. cos EXAMPLE Find an equivalent epression for each of the following. a) sin in terms of function values of b) cos in terms of function values of or, raised onl to the first power Solution a) sin sin sin cos cos sin sin cos cos cos sin Using sin sin cos and cos cos sin cos sin cos sin sin cos sin We could also substitute cos sin or sin for cos. Each substitution leads to a different result, but all results are equivalent. b) cos cos cos cos cos cos cos cos Half-Angle Identities If we take half of an angle of measure, the new angle will have measure. Half-angle identities give trigonometric function values of in terms of function values of. To develop these identities, we replace with and take square roots. For eample, sin sin cos sin s in cos cos cos Substituting for. Taking square roots The formula is called a half-angle formula. The use of and depends on the quadrant in which the angle lies. Half-angle identities for the Solving the identit cos sin for sin

164 Section 6. Identities: Cofunction, Double-Angle, and Half-Angle 59 cosine and tangent functions can be derived in a similar manner. Two additional formulas for the half-angle tangent identit are listed below. Half-Angle Identities sin cos tan cos cos cos cos sin cos cos sin,, tan( π/8) (() EXAMPLE 5 Find tan 8 eactl. Then check the answer using a graphing calculator in RADIAN mode. Solution tan 8 tan The identities that we have developed are also useful for simplifing trigonometric epressions. EXAMPLE 6 Simplif each of the following. sin cos a) b) sin cos cos Solution a) We can obtain sin cos in the numerator b multipling the epression b : sin cos cos sin cos tan. sin cos sin cos cos cos sin cos Using sin sin cos

165 550 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations b) We have sin cos cos cos Using sin cos, or sin cos cos. We can check this result using a graph or a table. cos sin cos, π π X Y X Y Tbl π/ Answers to Eercises 8 can be found on p. IA-8.

166 550 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations 6. Eercise Set. Given that sin and cos ,find each of the following. a) The other four function values for 0 b) The si function values for 5. Given that sin and cos, find eact answers for each of the following. a) The other four function values for b) The si function values for 5. Given that sin and that the terminal side is in quadrant II, find eact answers for each of the following. a) The other function values for b) The si function values for c) The si function values for. Given that cos 5 and that the terminal side is in quadrant IV, find eact answers for each of the following. a) The other function values for b) The si function values for c) The si function values for Find an equivalent epression for each of the following. 5. sec 6. cot 7. tan 8. csc Find the eact value of sin, cos, tan, and the quadrant in which lies. 9. sin, in quadrant I 5 sin, cos 7, tan ; II 5 5 7

167 Section 6. Identities: Cofunction, Double-Angle, and Half-Angle 55. tan 75. Given that sin 0.6 and is in quadrant I, find each of the following using identities.. sin 0.6. cos cos 5, in quadrant I sin. cos 0 cos 9 tan 0 69, 69, 9 ;, in quadrant III 5 sin, cos 7, tan ; II 5. sin sin In Eercises 7 0, use a graphing calculator to determine which of the following epressions asserts an identit. Then derive the identit algebraicall. cos 7. cos sin a) cos b) cos sin cos 5. tan 5, in quadrant II sin 8 III. tan 5 0 cos 6 tan 0 89, 89, 6 ;, in quadrant II sin,, ; IV. sin 0 0 cos 9 tan , in quadrant IV tan IV 5. Find an sin equivalent 0 epression cos 5, 5 for, cos in terms ; of function values of. 6. Find an equivalent epression for sin in terms of function values of,, or, raised onl to the first power. Use the half-angle identities to evaluate eactl. 7. cos 5 8. tan sin.5 0. cos 8 5 sin 5 cos sin 7 II c) cot d) π cot 8. cos a) sin csc tan b) sin cos c) cos sin d) cos sin 9. cos a) cos b) tan c) cos sin d) sin 0. sin cos a) cos b) sin c) sin d) sin cos Simplif. Check our results using a graphing calculator.. cos cos π. cos sin cos. sin cos sin. sin cos sin sec 5. cos sec sin cos 6. cot sin cos 7. cos sin cos cos sin cos 8 8. sin cos sin cos π sin cot sin (cot ) sin π π π π π Collaborative Discussion and Writing 9. Discuss and compare the graphs of sin, sin, and sin. Answers to Eercises 5, 6, and 7 0 can be found on pp. IA-8 and IA-9.

168 55 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations 0. Find all errors in the following: Skill Maintenance sin cos sin cos cos 8 sin cos cos sin 8 sin cos. In Eercises 8, answer True or False.. cos sin. sec tan [6.] True [6.] False. sin cos. cot csc [6.] False [6.] True 5. csc cot 6. tan sec [6.] False [6.] True 7. sin cos 8. sec tan [6.] False [6.] True Consider the following functions (a) (f ). Without graphing them, answer questions 9 5 below. a) b) c) d) e) f) f sin f cos f sin f cos 9. Which functions have a graph with an amplitude of? [5.5] (a), (e) 50. Which functions have a graph with a period of? [5.5] (b), (c), (f) 5. Which functions have a graph with a period of? [5.5] (d) 5. Which functions have a graph with a phase shift of? [5.5] (e) Snthesis 5. Given that cos , find the si function values for. f sin f cos 8 Simplif. Check our results using a graphing calculator cos cot sin 56. sin sec cos cos sin sin cos cos tan Find sin, cos, and tan under the given conditions. 58. cos 7, 59. tan, Nautical Mile. Latitude is used to measure north south location on the earth between the equator and the poles. For eample, Chicago has latitude N. (See the figure.) In Great Britain, the nautical mile is defined as the length of a minute of arc of the earth s radius. Since the earth is flattened slightl at the poles, a British nautical mile varies with latitude. In fact, it is given, in feet, b the function where N 6066 cos, is the latitude in degrees. Equator cos sin 57. sin sin cot 90 N 90 S sin sin tan N a) What is the length of a British nautical mile at Chicago? ft b) What is the length of a British nautical mile at the North Pole? 6097 ft 0 cos cot Answers to Eercises 5, 58, and 59 can be found on p. IA-9.

169 Section 6. Identities: Cofunction, Double-Angle, and Half-Angle 55 c) Epress N in terms of cos onl. That is, do not use the double angle. N cos 6. Acceleration Due to Gravit. The acceleration due to gravit is often denoted b g in a formula such as S gt,where S is the distance that an object falls in time t.the number g relates to motion near the earth s surface and is usuall considered constant. In fact, however, g is not constant, but varies slightl with latitude. If stands for latitude, in degrees, g is given with good approimation b the formula g sin sin, where g is measured in meters per second per second at sea level. a) Chicago has latitude N. Find g msec b) Philadelphia has latitude 0N. Find g msec c) Epress g in terms of sin onl. That is, eliminate the double angle. g sin sin 6 WISCONSIN M I C H I G A N CANADA 7 VT. NEW YORK MASS. CONN. Chicago PENNSYLVANIA Philadelphia 0 INDIANA OHIO N.J. ILLINOIS MD. DEL. W.VA. 8 KENTUCKY VIRGINIA

170 Section 6. Proving Trigonometric Identities Proving Trigonometric Identities Prove identities using other identities. The Logic of Proving Identities We outline two algebraic methods for proving identities. Method. Start with either the left or the right side of the equation and obtain the other side. For eample, suppose ou are tring to prove that the equation P Q is an identit. You might tr to produce a string of statements R, R,... or T, T,... like the following, which start with P and end with Q or start with Q and end with P: P R or Q T R T Q P. Method. Work with each side separatel until ou obtain the same epression. For eample, suppose ou are tring to prove that P Q is an identit. You might be able to produce two strings of statements like the following, each ending with the same statement S. P R Q T R T S S.

171 55 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations The number of steps in each string might be different, but in each case the result is S. A first step in learning to prove identities is to have at hand a list of the identities that ou have alread learned. Such a list is in the back endpapers of this tet. Ask our instructor which ones ou are epected to memorize. The more identities ou prove, the easier it will be to prove new ones. A list of helpful hints follows. Hints for Proving Identities. Use method or above.. Work with the more comple side first.. Carr out an algebraic manipulations, such as adding, subtracting, multipling, or factoring.. Multipling b can be helpful when rational epressions are involved. 5. Converting all epressions to sines and cosines is often helpful. 6. Tr something! Put our pencil to work and get involved. You will be amazed at how often this leads to success. Proving Identities In what follows, method is used in Eamples and method is used in Eamples and 5. Stud Tip Forming a small stud group (not more than three or four students) can be helpful when learning to prove identities. Your skills with this topic can be greatl improved in group discussions. Retention of the skills can be maimized when ou eplain the material to someone else. EXAMPLE Prove the identit sin sin cos. Solution Let s use method. We begin with the right side and obtain the left side: sin cos sin sin cos cos Squaring sin cos Recalling the identit sin cos and substituting sin. Using sin sin cos We could also begin with the left side and obtain the right side: sin sin cos Using sin sin cos sin sin cos cos Replacing with sin cos sin cos. Factoring

172 Section 6. Proving Trigonometric Identities 555 EXAMPLE Prove the identit. Solution We use method, starting with the left side. Note that the left side involves sec t, whereas the right side involves cos t, so it might be wise to make use of a basic identit that involves these two epressions: sec t cos t. Substituting cos t for sec t We started with the left side and obtained the right side, so the proof is complete. EXAMPLE sin sin Solution sin sin sec t t sec t sec t t sec t cos cos cos t cos t t cos t Multipling t t cos t t Prove the identit cos cos cos t t cos t t cos cos cos t sec. sin cos sin cos sin cos cos cos sin cos cos sin cos cos cos sin cos cos sin cos cos sec Using double-angle identities Simplifing Multipling cos b, or cos cos Subtracting Using a Pthagorean identit Recalling a basic identit

173 556 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations EXAMPLE Prove the identit sin tan tan sin. Solution For this proof, we are going to work with each side separatel using method. We tr to obtain the same epression on each side. In actual practice, ou might work on one side for a while, then work on the other side, and then go back to the first side. In other words, ou work back and forth until ou arrive at the same epression. Let s start with the right side: tan sin sin cos sin sin cos sin cos cos sin sin cos cos sin cos cos sin sin cos sin tan sin sin cos Recalling the identit tan sin and cos substituting Multipling b in order to subtract Carring out the subtraction Factoring Recalling the identit cos sin and substituting sin. cos At this point, we stop and work with the left side, sin tan, of the original identit and tr to end with the same epression that we ended with on the right side: Recalling the identit tan sin and cos substituting sin. cos We have obtained the same epression from each side, so the proof is complete.

174 Section 6. Proving Trigonometric Identities 557 EXAMPLE 5 Solution Prove the identit cot csc sin. cos We are again using method, beginning with the left side: cot csc cos sin sin cos. Adding sin Using basic identities At this point, we stop and work with the right side of the original identit: sin sin cos cos cos cos sin cos cos sin cos sin Multipling b cos. Simplifing sin Using sin cos The proof is complete since we obtained the same epression from each side. Answers to Eercises and can be found on p. IA-9.

175 Section 6. Proving Trigonometric Identities Eercise Set Prove each of the following identities.. sec sin tan cos sec sin tan cos cos. sin cos sin cos sin cos + cos sin + cos + sin cos sin cos

176 558 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations cos. sin sin cos tan sec. cot csc tan 5. cot 0 tan cot sin cos sin 6. sec csc sec cos cot 7. cos cos tan cot cos tan 8. sec sec sec tan 9. tan sin cos u v 0. tan u cot v cos u sin v. cos 5 cos sin 5 sin sin. cos sin cos. sin cos sin cos sin tan t tan t tan t. tan t tan t tan t tan sin 5. sin tan cos sin sin 6. cos sin 7. sin sin sin sin 8. cos sec sin 9. tan tan cot sec cos sin 0. tan cos cos. sin csc tan cot. sec csc sin sin. sec tan sin sin. tan cot sec csc sin cos 5. cos cot cot cos tan cot 6. sec csc cos sin 7. sin cos cos sin cot 8. csc csc cot sin 9. sec tan sin 0. sec s tan s tan s sec s In Eercises 6, use a graphing calculator to determine which epression (A) (F) on the right can be used to complete the identit. Then tr to prove that identit algebraicall. cos cot sin cos. A. csc sin cos. cot csc. sin cos. cos 5. cot sin 6. cos sin sin cos B. cos Collaborative Discussion and Writing C. tan cot D. cos sin 7. What restrictions must be placed on the variable in each of the following identities? Wh? a) sin tan tan cos b) sin sin cos c) sin cos sin cos sin 8. Eplain wh tan 50 cannot be simplified using the tangent sum formula, but can be simplified using the sine and cosine sum formulas. E. sin cos F. cos sin Answers to Eercises 6 can be found on pp. IA-9 through IA-.

177 Section 6. Proving Trigonometric Identities 559 Skill Maintenance For each function: a) Graph the function. b) Determine whether the function is one-to-one. c) If the function is one-to-one, find an equation for its inverse. d) Graph the inverse of the function. 9. f 0. f. f, 0. f Solve.. 5 [.] 0, [.], [.], i [.] [.5] [.5] 9 Snthesis Prove the identit ln tan ln cot 50. ln sec tan ln sec tan 5. Prove the identit log cos sin log cos sin log cos Mechanics. The following equation occurs in the stud of mechanics: I cos sin. I cos I sin It can happen that I I. Assuming that this happens, simplif the equation. sin cos 5. Alternating Current. In the theor of alternating current, the following equation occurs: R. Ctan tan Show that this equation is equivalent to R cos cos. C sin 5. Electrical Theor. In electrical theor, the following equations occur: and Assuming that these equations hold, show that and E E t cos E E t cos E E E E P P E t cos cos E t sin sin. P. P Answers to Eercises 9, 9 5, 5, and 5 can be found on p. IA-.

178 560 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations 6. Inverses of the Trigonometric Functions Find values of the inverse trigonometric functions. Simplif epressions such as sin sin and sin sin. Simplif epressions involving compositions such as sin cos without using a calculator. Simplif epressions such as sin arctan ab b making a drawing and reading off appropriate ratios. In this section, we develop inverse trigonometric functions. The graphs of the sine, cosine, and tangent functions follow. Do these functions have inverses that are functions? The do have inverses if the are one-to-one, which means that the pass the horizontal-line test. sin cos p p p p p p p p p p p p tan inverse functions review section.. Note that for each function, a horizontal line (shown in red) crosses the graph more than once. Therefore, none of them has an inverse that is a function. The graphs of an equation and its inverse are reflections of each other across the line. Let s eamine the graphs of the inverses of each of the three functions graphed above. Stud Tip When ou stud a section of a mathematics tet, read it slowl, observing all the details of the corresponding art pieces that are discussed in the paragraphs. Also note the precise color-coding in the art that enhances the learning of the concepts. p p p sin p p p cos p p p tan p p p We can check again to see whether these are graphs of functions b using the vertical-line test. In each case, there is a vertical line (shown in red) that crosses the graph more than once, so each fails to be a function.

179 Section 6. Inverses of the Trigonometric Functions 56 Restricting Ranges to Define Inverse Functions Recall that a function like f does not have an inverse that is a function, but b restricting the domain of f to nonnegative numbers, we have a new squaring function, f, 0, that has an inverse, f.this is equivalent to restricting the range of the inverse relation to eclude ordered pairs that contain negative numbers. 6 5 (, ) 6 5 f(), for 0 f () (, ) In a similar manner, we can define new trigonometric functions whose inverses are functions. We can do this b restricting either the domains of the basic trigonometric functions or the ranges of their inverse relations. This can be done in man was, but the restrictions illustrated below with solid red curves are fairl standard in mathematics. p p p p q sin p cos p tan q p p p Figure p Figure p Figure p For the inverse sine function, we choose a range close to the origin that allows all inputs on the interval, to have function values. Thus we choose the interval, for the range (Fig. ). For the inverse cosine function, we choose a range close to the origin that allows all inputs on the interval, to have function values. We choose the interval 0, (Fig. ). For the inverse tangent function, we choose a range close to the origin that allows all real numbers to have function values. The interval, satisfies this requirement (Fig. ).

180 56 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations Inverse Trigonometric Functions FUNCTION DOMAIN RANGE sin arcsin,where sin cos arccos,where cos tan arctan,where tan,,,, 0,, The notation arcsin arises because the function value,, is the length of an arc on the unit circle for which the sine is. Either of the two kinds of notation above can be read the inverse sine of or the arc sine of or the number (or angle) whose sine is. The notation sin is not eponential notation. It does not mean! sin The graphs of the inverse trigonometric functions are as follows. p sin arcsin p cos arccos p tan arctan q q q q q q Domain: [, ] Range: [q, q] Domain: [, ] Range: [0, p] Domain: (, ) Range: (q, q) EXPLORING WITH TECHNOLOGY Inverse trigonometric functions can be graphed using a graphing calculator. Graph sin using the viewing window,,,, with Xscl and Yscl. Now tr graphing cos and tan. Then use the graphs to confirm the domain and the range of each inverse. The following diagrams show the restricted ranges for the inverse trigonometric functions on a unit circle. Compare these graphs with the graphs above. The ranges of these functions should be memorized. The missing endpoints in the graph of the arctangent function indicate inputs that are not in the domain of the original function.

181 Section 6. Inverses of the Trigonometric Functions 56 q q p 0 q arcsine Range [q, q] arccosine Range [0, p] q arctangent Range (q, q) EXAMPLE Find each of the following function values. a) sin b) cos c) tan Solution a) Another wa to state find sin is to sa find such that sin. In the restricted range,, the onl number with a sine of is. Thus, sin, or 5. (See Fig. below.) q q i d p 0 q Figure Figure 5 Figure 6 q q A q b) The onl number with a cosine of in the restricted range is.thus, cos 0,, or 0.(See Fig. 5 above.) c) The onl number in the restricted range, with a tangent of is 6. Thus, tan is 6, or 0. (See Fig. 6 at left.) We can also use a calculator to find inverse trigonometric function values. On most graphing calculators, we can find inverse function values in either radians or degrees simpl b selecting the appropriate mode. The kestrokes involved in finding inverse function values var with the calculator. Be sure to read the instructions for the particular calculator that ou are using.

182 56 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations GCM EXAMPLE Approimate each of the following function values in both radians and degrees. Round radian measure to four decimal places and degree measure to the nearest tenth of a degree. a) b) c) d) e) cos tan 0.6 sin 0.05 cos.8 csc 8.05 Solution FUNCTION VALUE MODE READOUT ROUNDED a) cos Radian Degree b) tan 0.6 Radian Degree c) sin 0.05 Radian Degree d) cos.8 Radian Degree ERR:DOMAIN ERR:DOMAIN The value.8 is not in,, the domain of the arccosine function. e) The cosecant function is the reciprocal of the sine function: csc 8.05 sin 8.05 Radian Degree

183 Section 6. Inverses of the Trigonometric Functions 565 C ONNECTING THE CONCEPTS DOMAINS AND RANGES The following is a summar of the domains and ranges of the trigonometric functions together with a summar of the domains and ranges of the inverse trigonometric functions. For completeness, we have included the arccosecant, the arcsecant, and the arccotangent, though there is a lack of uniformit in their definitions in mathematical literature. FUNCTION sin cos tan csc sec cot INVERSE FUNCTION sin cos tan csc sec cot DOMAIN All reals,, All reals,, All reals ecept k, k odd All reals ecept All reals ecept k, k odd All reals ecept DOMAIN,, k k All reals, or,,,,, All reals, or, RANGE,, All reals,,,,,, All reals,, RANGE, 0,,,0 0, 0, 0,,

184 566 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations Composition of Trigonometric Functions and Their Inverses Various compositions of trigonometric functions and their inverses often occur in practice. For eample, we might want to tr to simplif an epression such as sin sin sin cot or. composition of functions review section.6. In the epression on the left, we are finding the sine of a number whose sine is. Recall from Section. that if a function f has an inverse that is also a function, then f f, for all in the domain of f, and f f, for all in the domain of f. Thus, if f sin and f sin, then sin sin, for all in the domain of sin, which is an number on the interval,. Similar results hold for the other trigonometric functions. Composition of Trigonometric Functions sin sin, for all in the domain of sin. cos cos, for all in the domain of cos. tan tan, for all in the domain of tan. EXAMPLE Simplif each of the following. a) cos cos b) sin sin.8 Solution a) Since is in,, the domain of cos, it follows that cos cos. b) Since.8 is not in,, the domain of sin, we cannot evaluate this epression. We know that there is no number with a sine of.8. Since we cannot find sin.8, we state that sin sin.8 does not eist.

185 Section 6. Inverses of the Trigonometric Functions 567 Now let s consider an epression like sin sin. We might also suspect that this is equal to for an in the domain of sin, but this is not true unless is in the range of the sin function. Note that in order to define sin, we had to restrict the domain of the sine function. In doing so, we restricted the range of the inverse sine function. Thus, sin sin, for all in the range of sin. Similar results hold for the other trigonometric functions. Special Cases sin sin, for all in the range of sin. cos cos, for all in the range of cos. tan tan, for all in the range of tan. EXAMPLE Simplif each of the following. Solution a) Since 6 is in,, the range of the tan function, we can use tan tan. Thus, tan tan. 6 6 b) Note that is not in,, the range of the sin function. Thus we cannot appl sin sin. Instead we first find sin,which is, and substitute: sin sin sin. Now we find some other function compositions. EXAMPLE 5 6 a) tan tan b) sin sin Simplif each of the following. a) sin tan b) cos sin Solution a) Tan is the number (or angle) in, whose tangent is. That is, tan. Thus, and sin tan sin b) cossin sin cos 0.

186 568 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations Net, let s consider cos sin 5. 5 u b Without using a calculator, we cannot find sin 5. However, we can still evaluate the entire epression b sketching a reference triangle. We are looking for angle such that sin, or sin. Since sin is defined in, and , we know that is in quadrant I. We sketch a reference right triangle, as shown at left. The angle in this triangle is an angle whose sine is 5. We wish to find the cosine of this angle. Since the triangle is a right triangle, we can find the length of the base, b. It is. Thus we know that cos b5, or. Therefore, 5 cos sin 5 5 sin cot EXAMPLE 6 Find.. Solution We draw a right triangle whose legs have lengths and, so that cot. Since the range of cot is, 0 0,, we consider onl positive values of. u, positive We find the length of the hpotenuse and then read off the sine ratio. We get sin cot. In the following eample, we use a sum identit to evaluate an epression. EXAMPLE 7 Evaluate: sin sin 5 cos. Solution Since and cos 5 sin are both angles, the epression is the sine of a sum of two angles, so we use the identit sin u v sin u cos v cos u sin v.

187 Section 6. Inverses of the Trigonometric Functions 569 Thus, sin sin 5 cos sin sin cos cos 5 cos sin sin cos 5. 5 cos sin sin cos 5 Using composition identities Now since, cos sin sin 6 simplifies to cos 6, or. We can illustrate this with a reference triangle in quadrant I. To f ind sin cos 5, we use a reference triangle in quadrant I and 5 determine that the sine of the angle whose cosine is is. 5 sin(sin(/)cos(5/)) (5())/ Our epression now simplifies to 5, or 6. 5 Thus, sin sin 5 5 cos 6.

188 570 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations 6. Eercise Set Find each of the following eactl in radians and degrees., 60. sin. cos,60. tan,5. sin 0 0, 0 5. cos,5 6. sec,5 7. tan 0 0, 0 8. tan, cos,0 0. cot 6, 60. csc,0. sin, cot, 0. tan, sin 6. cos, 0, cos 0,90 8. sin,60 9. sec,60 0. csc, 90 Use a calculator to find each of the following in radians, rounded to four decimal places, and in degrees, rounded to the nearest tenth of a degree.. tan , 0.. cos , 07.. sin , 7.0. sin , cos , tan , csc , sec ,. 9. tan , cot , 8. Answer to Eercise can be found on p. IA-.. sin , cos , State the domains of the inverse sine, inverse cosine, and inverse tangent functions. sin : [, ; cos :, ; tan :,. State the ranges of the inverse sine, inverse cosine, and inverse tangent functions. 5. Angle of Depression. An airplane is fling at an altitude of 000 ft toward an island. The straightline distance from the airplane to the island is d feet. Epress, the angle of depression, as a function of d. sin 000d 6. Angle of Inclination. A gu wire is attached to the top of a 50-ft pole and stretched to a point that is d feet from the bottom of the pole. Epress, the angle of inclination, as a function of d. tan 50d d Evaluate. 7. sin sin tan tan.. 9. cos cos 0. sinsin 5 50 ft d. sin sin. cotcot ft

189 . tan tan. cos cos 5. sin tan 6. cos sin 7. tan cos 8. cos sin 9. sin cos 50. sin tan 5. tan sin 5. cos tan sin sin 7 5. tan tan 6 Find. a 55. a sin cos 59. tan sin p 60. p 6. cos 6. sin cos 5 sin 5 Evaluate. 6. cos sin 5 cos sin tan a cot sin q p p 9 sin sin cos sin sin cos 66. cos sin cos tan cos tan sin sin sin 0.60 cos cos sin 0.75 cos Collaborative Discussion and Writing 69. Eplain in our own words wh the ranges of the inverse trigonometric functions are restricted. 70. How does the graph of sin differ from the graph of sin? Answers to Eercises 57 and 60 can be found on p. IA Section 6. Inverses of the Trigonometric Functions Wh is it that sin, but sin? Skill Maintenance In each of Eercises 7 80, fill in the blank with the correct term. Some of the given choices will not be used. linear speed angular speed angle of elevation angle of depression complementar supplementar similar congruent circular periodic period amplitude acute obtuse quadrantal radian measure 7. A function f is said to be if there eists a positive constant p such that fs p fs for all s in the domain of f. [5.5] periodic 7. The of a rotation is the ratio of the distance s traveled b a point at a radius r from the center of rotation to the length of the radius r. [5.] radian measure 7. Triangles are if their corresponding angles have the same measure. [5.] similar 75. The angle between the horizontal and a line of sight below the horizontal is called a(n). [5.] angle of depression 76. is the amount of rotation per unit of time. [5.] Angular speed 77. Two positive angles are if their sum is 80. [5.] supplementar 78. The of a periodic function is one half of the distance between its maimum and minimum function values. [5.5] amplitude 79. A(n) angle is an angle with measure greater than 0 and less than 90. [5.] acute 80. Trigonometric functions with domains composed of real numbers are called functions. [5.5] circular

190 57 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations Snthesis Prove the identit. 8. sin cos 8. tan cot 8. sin tan 8. tan sin 85. sin cos, for cos tan, for Height of a Mural. An art student s ee is at a point A,looking at a mural of height h,with the bottom of the mural feet above the ee (see accompaning illustration). The ee is feet from the wall. Write an epression for in terms of,, and h.then evaluate the epression when 0 ft, 7 ft, and h 5 ft. h tan ; 8.7 tan h 88. Use a calculator to approimate the following epression: 6 tan tan. 5 9 What number does this epression seem to approimate? Answers to Eercises 8 86 can be found on p. IA-.

191 57 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations 6.5 Solving Trigonometric Equations Solve trigonometric equations. When an equation contains a trigonometric epression with a variable, such as cos, it is called a trigonometric equation. Some trigonometric equations are identities, such as sin cos. Now we consider equations, such as cos, that are usuall not identities. As we have done for other tpes of equations, we will solve such equations b finding all values for that make the equation true.

192 Section 6.5 Solving Trigonometric Equations 57 EXAMPLE Solve: cos. Solution We first solve for cos : cos cos. q, o q, i (, 0) The solutions are numbers that have a cosine of. To find them, we use the unit circle (see Section 5.5). There are just two points on the unit circle for which the cosine is, as shown in the figure at left. The are the points corresponding to and. These numbers, plus an multiple of, are the solutions: k and k, where k is an integer. In degrees, the solutions are 0 k 60 and 0 k 60, where k is an integer. To check the solution to cos, we can graph cos and on the same set of aes and find the first coordinates of the points of intersection. Using as the Xscl facilitates our reading of the solutions. First, let s graph these equations on the interval from 0 to, as shown in the figure on the left below. The onl solutions in 0, are and. cos, cos, π π 0 π (, ) π (, ) Xscl π π π 8 π π Yscl 8 π π Xscl π π Net, let s change the viewing window to,,, and graph again. Since the cosine function is periodic, there is an infinite number of solutions. A few of these appear in the graph on the right above. From the graph, we see that the solutions are k and k, where k is an integer.

193 57 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations EXAMPLE Solve: sin. Solution We begin b solving for sin : sin sin sin. S, q, q F A, q (, 0), q G Again, we use the unit circle to find those numbers having a sine of or. The solutions are where k is an integer. In degrees, the solutions are 0 k 60, 50 k 60, 0 k 60, and 0 k 60, where k is an integer. The general solutions listed above could be condensed using odd as well as even multiples of : 6 k 5 5 and k, 6 7 k, k, k, and k, or, in degrees, 0 k 80 and 50 k 80, where k is an integer. Let s do a partial check using a graphing calculator, checking onl the solutions in 0,. We graph sin and and note that the solutions in 0, are 6, 56, 76, and 6. sin, 5 π (, 6 ) 0 π (, 6 ) π (, 6 ) 7 π (, 6 ) Xscl 6 π π In most applications, it is sufficient to find just the solutions from 0 to or from 0 to 60.We then remember that an multiple of, or 60, can be added to obtain the rest of the solutions.

194 Section 6.5 Solving Trigonometric Equations 575 We must be careful to find all solutions in 0, trigonometric equations involving double angles. when solving f,, p cos(0.6) (, 0), 5p j, (, 0) EXAMPLE Solve tan in the interval 0,. Solution We first solve for tan : tan tan. We are looking for solutions to the equation for which 0. Multipling b, we get 0, which is the interval we use when solving tan. Using the unit circle, we find points in 0, for which tan.these values of are as follows:,,, and. Thus the desired values of in 0, are each of these values divided b. Therefore,,,, and Calculators are needed to solve some trigonometric equations. Answers can be found in radians or degrees, depending on the mode setting. EXAMPLE Solve in 0, 60. cos.08 Solution We have cos.08 cos 0.08 cos 0.6. Using a calculator set in DEGREE mode (see window at left), we find that the reference angle, cos 0.6, is. Since cos is positive, the solutions are in quadrants I and IV. The solutions in 0, 60 are and

195 576 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations EXAMPLE 5 Solve cos u cos u in 0, 60. Algebraic Solution We use the principle of zero products: cos u cos u cos u cos u 0 cos u cos u 0 cos u 0 or cos u 0 cos u or cos u cos u or cos u. Thus, u 60, 00 or u 80. The solutions in 0, 60 are 60, 80, and 00. Graphical Solution We can use either the Intersect method or the Zero method to solve trigonometric equations. Here we illustrate b solving the equation using both methods. We set the calculator in DEGREE mode. Intersect Method. We graph the equations cos and cos and use the INTERSECT feature to find the first coordinates of the points of intersection. cos, cos 0 60 Intersection X 60 Y.5 Xscl 60 The leftmost solution is 60.Using the INTERSECT feature two more times, we find the other solutions, 80 and 00.

196 Section 6.5 Solving Trigonometric Equations 577 Zero Method. We write the equation in the form cos u cos u 0. Then we graph cos cos and use the ZERO feature to determine the zeros of the function. cos cos 0 60 Zero X 60 Y 0 Xscl 60 The leftmost zero is 60.Using the ZERO feature two more times, we find the other zeros, 80 and 00.The solutions in 0, 60 are 60, 80, and sin sin Xscl π π EXAMPLE 6 Solve sin sin 0 in 0,. Solution We factor and use the principle of zero products: sin sin 0 sin sin 0 Factoring sin 0 or sin 0 sin 0 or sin 0, The solutions in 0, are 0,, and. or.

197 578 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations If a trigonometric equation is quadratic but difficult or impossible to factor, we use the quadratic formula. sin(0.96) EXAMPLE 7 Solve 0 sin sin 7 0 in 0, 60. Solution This equation is quadratic in sin with a 0, b, and c 7. Substituting into the quadratic formula, we get sin b b ac a 80 0 Using the quadratic formula Substituting sin.696 or sin Since sine values are never greater than, the first of the equations has no solution. Using the other equation, we find the reference angle to be 5.. Since sin is negative, the solutions are in quadrants III and IV. Thus the solutions in 0, 60 are and Trigonometric equations can involve more than one function. cos tan, tan 0 π π Xscl cos tan tan EXAMPLE 8 Solve cos tan tan in 0,. Solution Using a graphing calculator, we can determine that there are si solutions. If we let Xscl, the solutions are read more easil. In the figures at left, we show the Intersect and Zero methods of solving graphicall. Each illustrates that the solutions in 0, are 5 7 0,, and,,,. We can verif these solutions algebraicall, as follows: cos tan tan cos tan tan 0 tan cos 0 tan 0 or cos 0 0 Xscl π π 0, Thus, 0,,,, 5, and 7. cos cos or 5 7,,,.

198 Section 6.5 Solving Trigonometric Equations 579 When a trigonometric equation involves more than one function, it is sometimes helpful to use identities to rewrite the equation in terms of a single function. EXAMPLE 9 Solve sin cos in 0,. Algebraic Solution We have sin cos sin cos sin sin cos cos sin cos Squaring both sides Using sin cos sin cos 0 sin 0. Using sin cos sin We are looking for solutions to the equation for which 0.Multipling b, we get 0, which is the interval we consider to solve sin 0. These values of are 0,,, and. Thus the desired values of in 0, satisfing this equation are 0,,, and. Now we check these in the original equation sin cos : sin 0 cos 0 0, sin cos 0, sin cos 0, sin cos 0. We find that and do not check, but the other values do. Thus the solutions in 0, are 0 and. When the solution process involves squaring both sides, values are sometimes obtained that are not solutions of the original equation. As we saw in this eample, it is important to check the possible solutions. Graphical Solution We can graph the left side and then the right side of the equation as seen in the first window below. Then we look for points of intersection. We could also rewrite the equation as sin cos 0,graph the left side, and look for the zeros of the function, as illustrated in the second window below. In each window, we see the solutions in 0, as 0 and. 0 0 sin cos, sin cos π Xscl π Xscl π π This eample illustrates a valuable advantage of the calculator that is, with a graphing calculator, etraneous solutions do not appear.

199 580 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations EXAMPLE 0 Solve cos sin in 0,. Algebraic Solution We have cos sin sin sin sin sin 0 s in sin 0 Using the identit cos sin Factoring sin 0 or sin 0 sin 0 or sin Principle of zero products 0, or,. 6 6 All four values check. The solutions in 0, are 0, 6, 56, and. 5 Graphical Solution We graph cos sin and look for the zeros of the function. cos sin The solutions in 0, are 0, 6, 56, and. 0 π π Xscl 6 EXAMPLE Solve tan sec 0 in 0,. Stud Tip Check our solutions to the oddnumbered eercises in the eercise sets with the step-b-step annotated solutions in the Student s Solutions Manual. If ou are still having difficult with the concepts of this section, make time to view the content video that corresponds to the section. Algebraic Solution We have tan sec 0 sec sec 0 sec sec 0 sec sec 0 cos, Using the identit tan sec,or tan sec Factoring sec or sec or cos or 0. Principle of zero products Using the identit cos sec All these values check. The solutions in 0, are 0,, and.

200 Section 6.5 Solving Trigonometric Equations 58 Graphical Solution We graph tan sec, but we enter this equation in the form tan. cos We use the ZERO feature to find zeros of the function. tan sec 0 π Xscl π Sometimes we cannot find solutions algebraicall, but we can approimate them with a graphing calculator. EXAMPLE a) b) The solutions in 0, are 0,, and..5 cos sin cos cot Solve each of the following in 0,. Solution a) In the screen on the left below, we graph.5 and cos and look for points of intersection. In the screen on the right, we graph.5 cos and look for the zeros of the function..5, cos.5 cos 0 π 0 π Intersection X.5996 Y π Xscl Zero X.5996 Y 0 Xscl π We determine the solution in 0, to be approimatel..

201 58 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations b) In the screen on the left, we graph sin cos and cot and determine the points of intersection. In the screen on the right, we graph the function sin cos cot and determine the zeros. sin cos, /tan sin cos /tan 0 π 0 π Intersection X.7657 X Y.7666 Y.957 Zero X.7657 X Y 0 Y 0 Each method leads to the approimate solutions. and 5.66 in 0,.

202 58 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations 6.5 Eercise Set Solve, finding all solutions. Epress the solutions in both radians and degrees.. cos. sin. tan. cos 5. sin 6. tan 7. cos 8. sin Solve, finding all solutions in 0, or 0, 60. Verif our answer using a graphing calculator. 9. cos , sin , 9. 0 p. tan 68.0, 8.0 tan, 0 p. cos. csc 0 5. sin sin 6. cos cos 0 7. cos cos 0 6,,, 6 8. sin 7 sin 6, sin 0, sin Answers to Eercises 8, 5, and 9 can be found on p. IA cos 5 cos 0 0. sin t cos t sin t cos t 0. sin cos sin 0. 5 sin 8 sin 98.8,.7. cos 6 cos 0 9.8, 0.9. tan tan 7 70., cot cot 7., 69.5, 7., sin sin 07.,.78 Solve, finding all solutions in 0,. 7. cos sin 0,,, 6 6 cos sin, 0 p 8. sin cos sin 0 0,,, sin cos sin 0 p 7 9. sin sin 0 0,,, 0. tan sin tan 0 0,. sin cos sin 0 0,. cos sin sin 0 0,,,. sec tan sec tan 0, 7. sin sin cos cos cos 0,,, 5. sin sin cos 0,,

203 Section 6.5 Solving Trigonometric Equations 585 GCM 6. tan sec tan.0, 5.76,, 5 7. sec tan 0,, 5, 7 8. cot tan 6, 56, 76, 6 9. cos sin 6, 5 0. cos sin 6,. sec tan ,.85,.09, cos sin 0.79,.76,.6, sin., cos Solve using a calculator, finding all solutions in 0,. 5. sin., sin No solutions in 7. cos 0, cos cos 50. sin tan 0.,.756 0,, Some graphing calculators can use regression to fit a trigonometric function to a set of data. 5. Sales. Sales of certain products fluctuate in ccles. The data in the following table show the total sales of skis per month for a business in a northern climate. 7 sin a) Using the SINE REGRESSION feature on a graphing calculator, fit a sine function of the form A sin B C D to this set of data. b) Approimate the total sales for December and for Jul. $0,500; $,06. cos sin, MONTH, TOTAL SALES, (IN THOUSANDS) August, 8 $ 0 November, 7 Februar, Ma, 5 7 August, Dalight Hours. The data in the following table give the number of dalight hours for certain das in Fairbanks, Alaska. DAY, NUMBER OF DAYLIGHT HOURS, Januar 0, 0.7 Februar 9, March, 6 0. April 8, Ma, 8.5 June, 6. Jul 7, August, 5.8 September 9, 6.7 October, 7. November, 8 6. December 8, 6.8 Source: Astronomical Applications Department; U.S. Naval Observator, Washington, DC a) Using the SINE REGRESSION feature on a graphing calculator, model these data with an equation of the form A sin B C D. b) Approimate the number of dalight hours in Fairbanks for April ( ), Jul ( 85), and December 5 ( 9). c) Determine on which da of the ear there will be about 0.5 hr of dalight. 6th da (March 5th) Answers to Eercises 5(a) and 5(b) can be found on p. IA-.

204 586 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations Collaborative Discussion and Writing 5. Jan lists her answer to a problem as 6 k, for an integer k,while Jacob lists his answer as 6 k and 76 k, for an integer k. Are their answers equivalent? Wh or wh not? 5. Under what circumstances will a graphing calculator give eact solutions of a trigonometric equation? Skill Maintenance Solve the right triangle. 55. C 0 B [5.] B 5, b 0.7, c 5. b c T [5.] R 5.5, T 7.5,.8 t.7. S R A t Solve [.] [.] h Snthesis Solve in 0,. 59. sin,,, 60. cos,,, 6. tan, 6. sin 7sin , 0.,.0, ln cos e sin 0, 65. sin ln e /k,where k (an integer) 66. e ln sin Temperature During an Illness. The temperature T, in degrees Fahrenheit, of a patient t das into a -da illness is given b Tt 0.6 sin. 8 t Find the times t during the illness at which the patient s temperature was 0.. das, 6.76 das 68. Satellite Location. A satellite circles the earth in such a manner that it is miles from the equator (north or south, height from the surface not considered) t minutes after its launch, where 5000cos. 5 t 0 At what times t in the interval 0, 0, the first hr, is the satellite 000 mi north of the equator?.8 min,.8 min, 0.8 min 69. Nautical Mile. (See Eercise 60 in Eercise Set 6..) In Great Britain, the nautical mile is defined as the length of a minute of arc of the earth s radius. Since the earth is flattened at the poles, a British nautical mile varies with latitude. In fact, it is given, in feet, b the function N 6066 cos, where is the latitude in degrees. At what latitude north is the length of a British nautical mile found to be 600 ft? 6.5N 70. Acceleration Due to Gravit. (See Eercise 6 in Eercise Set 6..) The acceleration due to gravit is often denoted b g in a formula such as S gt, where S is the distance that an object falls in t seconds. The number g is generall considered constant, but in fact it varies slightl with latitude. If stands for latitude, in degrees, an ecellent approimation of g is given b the formula g sin sin, where g is measured in meters per second per second at sea level. At what latitude north does g 9.8? N Solve. 7. cos cos 5 sin 5 7. sin tan tan 7. Suppose that sin 5 cos. Find sin cos. 0.9

205 Chapter 6 Summar and Review 587 Chapter 6 Summar and Review Important Properties and Formulas Basic Identities sin, tan sin, csc cos cos, cot cos, sec sin tan, cot sin sin, cos cos, tan tan Pthagorean Identities sin cos, cot csc, tan sec Sum and Difference Identities sin u v sin u cos v cos u sin v, cos u v cos u cos v sin u sin v, tan u v tan u tan v tan u tan v Double-Angle Identities sin sin cos, cos cos sin sin cos, tan tan tan Half-Angle Identities sin cos tan cos cos cos cos sin cos cos sin,, Cofunction Identities sin, sin cos, cos tan, cos sin cot sec csc

206 588 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations Inverse Trigonometric Functions FUNCTION DOMAIN RANGE sin, cos, 0, tan,,, Composition of Trigonometric Functions The following are true for an in the domain of the inverse function: sin sin, cos cos, tan tan. The following are true for an in the range of the inverse function: sin sin, cos cos, tan tan.

207 588 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations Review Eercises Complete the Pthagorean identit.. cot [6.] csc. sin cos [6.] Multipl and simplif. Check using a graphing calculator.. tan cot tan cot [6.] tan cot cos. cos sec [6.] cos Factor and simplif. Check using a graphing calculator. 5. sec csc csc [6.] csc sec csc 6. sin 7 sin 0 [6.] sin 5sin cos u [6.] 0 cos u00 0 cos u cos u Simplif and check using a graphing calculator. sin 8. [6.] 9. cos cos sec [6.] sin tan 0. [6.] cos cos cos sin sec sin tan sec tan sin cos sin cos. cos sin sin cos cot. [6.] csc csc sin cos. [6.] 6 sin cos. Simplif. Assume the radicand is nonnegative. sin cos sin cos [6.] sin cos sin 5. Rationalize the denominator:. sin cos 6. Rationalize the numerator:. tan 7. Given that tan, epress 9 as a trigonometric function without radicals. Assume that 0. [6.] sec cot Answers to Eercises, 5, and 6 can be found on p. IA-.

208 Chapter 6 Review Eercises 589 Use the sum and difference formulas to write equivalent epressions. You need not simplif. 8. cos 9. tan Simplif: cos 7 cos 6 sin 7 sin 6.. Find cos 65 eactl.. Given that tan and sin and that and are between 0 and, evaluate tan eactl. [6.]. Assume that sin 0.58 and cos 0. and that both and are first-quadrant angles. Evaluate cos. [6.] 0.75 Complete the cofunction identit.. cos 5. cos [6.] sin [6.] sin 6. sin [6.] cos 7. Given that cos 5 and that the terminal side is in quadrant III: a) Find the other function values for. b) Find the si function values for. c) Find the si function values for. 8. Find an equivalent epression for csc. [6.] sec 9. Find tan, cos, and sin and the quadrant in which lies, where cos 5 and is in quadrant III. 0. Find sin eactl. [6.] 8. Given that sin 0.8 and is in quadrant I, find sin, cos, and cos. Simplif and check using a graphing calculator.. sin [6.] cos. sin cos sin [6.]. sin cos sin cos [6.] sin cot 5. [6.] tan cot Answers to Eercises 8, 7, 9,, and 6 can be found on pp. IA- and IA-5. Prove the identit. sin 6. cos cos sin cos 7. cot sin tan sin 8. cos tan sin cos 9. cos tan sin cos In Eercises 0, use a graphing calculator to determine which epression (A) (D) on the right can be used to complete the identit. Then prove the identit algebraicall. 0. csc cos cot csc A. sec cos. B. sin sin cos cot. tan sin cos. sin sin cos C. D. sin sin cos sin Find each of the following eactl in both radians and degrees.. sin 5. cos [6.] 6, 0 [6.] 6, 0 6. tan 7. sin 0 [6.] 0, 0 [6.], 5 Use a calculator to find each of the following in radians, rounded to four decimal places, and in degrees, rounded to the nearest tenth of a degree. 8. cos cot.8 [6.].790, 0.7 [6.] 0.976,.8 Evaluate. 50. cos cos [6.] 5. tan tan 5. sin [6.] 5. cos 7 sin sin 7 Find. 5. cos tan 55. cos sin 5 b [6.] b 9 [6.] [6.] [6.] 7 5

209 590 Chapter 6 Trigonometric Identities, Inverse Functions, and Equations Solve, finding all solutions. Epress the solutions in both radians and degrees. 56. cos 57. tan Solve, finding all solutions in 0,. 58. sin 59. sin sin cos cos cos [6.5],, 6. sin 7 sin 0 [6.5] 0, sin sin sin cos 0 6. csc cot 0 [6.5],, 5, 7 6. sin sin 0 [6.5] 0,,, 6. cos sin [6.5] 7, tan 5 tan sec [6.5] 0.86,.97,.006, 6. Solve using a graphing calculator, finding all solutions in 0,. 66. cos [6.5] sin [6.5] No solution in 0, Collaborative Discussion and Writing 68. Prove the identit cos cos sin in three was: a) Start with the left side and deduce the right (method ). b) Start with the right side and deduce the left (method ). c) Work with each side separatel until ou deduce the same epression (method ). Then determine the most efficient method and eplain wh ou chose that method. 69. Wh are the ranges of the inverse trigonometric functions restricted? Snthesis 70. Find the measure of the angle from l to l : [6.] 08. l : : Find an identit for cos u v involving onl cosines. 7. Simplif:. [6.] cos cos csc sin 7. Find sin, cos, and tan under the given conditions: sin, Prove the following equation to be an identit: ln e sin t sin t. 75. Graph: sec. 76. Show that is not an identit. tan sin cos 77. Solve e cos in 0,. [6.5], l Answers to Eercises 56 59, 68, 69, 7, and 7 76 can be found on pp. IA-5 and IA-6.

210 Applications of Trigonometr 7. The Law of Sines 7. The Law of Cosines 7. Comple Numbers: Trigonometric Form 7. Polar Coordinates and Graphs 7.5 Vectors and Applications 7.6 Vector Operations SUMMARY AND REVIEW TEST 7 A P P L I C A T I O N An eagle flies from its nest 7 mi in the direction northeast, where it stops to rest on a cliff. It then flies 8 mi in the direction S0 W to land on top of a tree. Place an -coordinate sstem so that the origin is the bird s nest, the -ais points east, and the -ais points north. (a) At what point is the cliff located? (b) At what point is the tree located? This problem appears as Eercise 55 in Eercise Set 7.5.

211 59 Chapter 7 Applications of Trigonometr. 7. The Law Polnomial of Sines Functions and Modeling Use the law of sines to solve triangles. Find the area of an triangle given the lengths of two sides and the measure of the included angle. To solve a triangle means to find the lengths of all its sides and the measures of all its angles. We solved right triangles in Section 5.. For review, let s solve the right triangle shown below. We begin b listing the known measures. Z q W w Q Q 7. W 90 Z? q? w? z 6. Since the sum of the three angle measures of an triangle is 80, we can immediatel find the measure of the third angle: Z Then using the tangent and cosine ratios, respectivel, we can find q and w: tan 7. q, or 6. q 6. tan 7..8, and cos 7. 6., or w Now all si measures are known and we have solved triangle QWZ. Q 7. W 90 Z 5.9 w cos 7. q.8 w 7.9 z 6. Solving Oblique Triangles The trigonometric functions can also be used to solve triangles that are not right triangles. Such triangles are called oblique. An triangle, right or oblique, can be solved if at least one side and an other two measures are known. The five possible situations are illustrated on the net page.

212 Section 7. The Law of Sines 595. AAS: Two angles of a triangle and a side opposite one of them are known AAS. ASA: Two angles of a triangle and the included side are known. 7.5 ASA 5. SSA: Two sides of a triangle and an angle opposite one of them are known. (In this case, there ma be no solution, one solution, or two solutions. The latter is known as the ambiguous case.). SAS: Two sides of a triangle and the included angle are known. 8q 5.7 0~ SSA SAS 5. SSS: All three sides of the triangle are known SSS The list above does not include the situation in which onl the three angle measures are given. The reason for this lies in the fact that the angle measures determine onl the shape of the triangle and not the size, as shown with the following triangles. Thus we cannot solve a triangle when onl the three angle measures are given

213 596 Chapter 7 Applications of Trigonometr In order to solve oblique triangles, we need to derive the law of sines and the law of cosines. The law of sines applies to the first three situations listed above. The law of cosines, which we develop in Section 7., applies to the last two situations. The Law of Sines C We consider an oblique triangle. It ma or ma not have an obtuse angle. Although we look at onl the acute-triangle case, the derivation of the obtuse-triangle case is essentiall the same. In acute ABC at left, we have drawn an altitude from verte C. It has length h. From ADC,we have b h a sin A h, or h b sin A. b A c D B From BDC, we have sin B h, or h a sin B. a With h b sin A and h a sin B, we now have a sin B b sin A a sin B sin A sin B b sin A sin A sin B Dividing b sin A sin B a. sin A b sin B Simplifing There is no danger of dividing b 0 here because we are dealing with triangles whose angles are never 0 or 80. Thus the sine value will never be 0. If we were to consider altitudes from verte A and verte B in the triangle shown above, the same argument would give us b sin B c sin C a and. sin A c sin C We combine these results to obtain the law of sines. The Law of Sines In an triangle ABC, B a sin A b sin B c. sin C Thus in an triangle, the sides are proportional to the sines of the opposite angles. A c b a C

214 Section 7. The Law of Sines 597 Stud Tip Maimize the learning that ou accomplish during a lecture b preparing for the class. Your time is valuable; let each lecture become a positive learning eperience. Review the lesson from the previous class and read the section that will be covered in the net lecture. Write down questions ou want answered and take an active part in class discussion. Solving Triangles (AAS and ASA) When two angles and a side of an triangle are known, the law of sines can be used to solve the triangle. EXAMPLE In EFG, e.56, E, and G 57. Solve the triangle. Solution We first make a drawing. We know three of the si measures. g F.56 E F? G 57 e.56 f? g? E f 57 G From the figure, we see that we have the AAS situation. We begin b finding F: F We can now find the other two sides, using the law of sines: f sin F e sin E f.56 sin 80 sin.56 sin 80 f sin f 6.58; Substituting Solving for f g sin G e sin E g.56 sin 57 sin.56 sin 57 g sin g 5.6. Thus, we have solved the triangle: E, e.56, F 80, f 6.58, G 57, g 5.6. Substituting Solving for g The law of sines is frequentl used in determining distances.

215 598 Chapter 7 Applications of Trigonometr EXAMPLE Rescue Mission. During a rescue mission, a Marine fighter pilot receives data on an unidentified aircraft from an AWACS plane and is instructed to intercept the aircraft. The diagram shown below appears on the screen, but before the distance to the point of interception appears on the screen, communications are jammed. Fortunatel, the pilot remembers the law of sines. How far must the pilot fl? Z 5 X (Unidentified aircraft) 500 km 7 Y (Pilot) Solution We let represent the distance that the pilot must fl in order to intercept the aircraft and Z represent the point of interception. We first find angle Z: Z Because this application involves the ASA situation, we use the law of sines to determine : sin X z sin Z 500 sin 5 sin sin 5 sin Substituting Solving for Thus the pilot must fl approimatel 76 km in order to intercept the unidentified aircraft. Solving Triangles (SSA) When two sides of a triangle and an angle opposite one of them are known, the law of sines can be used to solve the triangle. Suppose for ABC that b, c, and B are given. The various possibilities are as shown in the eight cases below: five cases when B is acute and three cases when B is obtuse. Note that b c in cases,,, and 6; b c in cases and 7; and b c in cases 5 and 8.

216 Section 7. The Law of Sines 599 Angle B Is Acute Case : No solution b c; side b is too short to reach the base. No triangle is formed. Case : One solution b c; side b just reaches the base and is perpendicular to it. Case : Two solutions b c;an arc of radius b meets the base at two points. (This case is called the ambiguous case.) A A A c b c b c b b B B C B C Case : One solution b c;an arc of radius b meets the base at just one point, other than B. A Case 5: One solution b c;an arc of radius b meets the base at just one point. A c b c b B C B C Angle B Is Obtuse Case 6: No solution b c; side b is too short to reach the base. No triangle is formed. Case 7: No solution b c; an arc of radius b meets the base onl at point B.No triangle is formed. Case 8: One solution b c;an arc of radius b meets the base at just one point. A c b A c b A c b B B B C The eight cases above lead us to three possibilities in the SSA situation: no solution, one solution, or two solutions. Let s investigate these possibilities further, looking for was to recognize the number of solutions.

217 600 Chapter 7 Applications of Trigonometr EXAMPLE No solution. In QRS, q 5, r 8, and Q.6. Solve the triangle. Solution We make a drawing and list the known measures. 8 S 5 Q.6 R? S? q 5 r 8 s? Q.6 s? R? We observe the SSA situation and use the law of sines to find R: q sin Q r sin R 5 8 Substituting sin.6 sin R 8 sin.6 sin R Solving for sin R 5 s in R.87. Since there is no angle with a sine greater than, there is no solution. EXAMPLE One solution. In XYZ,.5, 9.8, and 9.7.Solve the triangle. Solution We make a drawing and organize the given information. X z X X 9.7 Y? Z? z? Y.5 Z We see the SSA situation and begin b finding Y with the law of sines: sin X sin Y Substituting sin 9.7 sin Y 9.8 sin 9.7 s in Y Solving for sin Y.5 s in Y There are two angles less than 80 with a sine of The are 5. and 6.6, to the nearest tenth of a degree. An angle of 6.6 cannot be an angle of this triangle because it alread has an angle of 9.7 and these

218 Section 7. The Law of Sines 60 two angles would total more than 80. Thus, 5. is the onl possibilit for Y. Therefore, Z We now find z: z sin Z sin X z.5 sin.9 sin sin.9 z sin 9.7 z 0.. We now have solved the triangle: X 9.7,.5, Y 5., 9.8, Z.9, z 0.. Substituting Solving for z The net eample illustrates the ambiguous case in which there are two possible solutions. EXAMPLE 5 Two solutions. In ABC, b 5, c 0, and B 9. Solve the triangle. Solution We make a drawing, list the known measures, and see that we again have the SSA situation. B A a C A? B 9 C? a? b 5 c 0 0 sin 0 sin 0 0 We first find C: b sin B c sin C 5 0 sin 9 sin C sin C 0 sin 9 5 Substituting Solving for sin C There are two angles less than 80 with a sine of The are 0 and 0, to the nearest degree. This gives us two possible solutions.

219 60 Chapter 7 Applications of Trigonometr Possible Solution I. If C 0, then A Then we find a: a sin A b sin B a 5 sin sin 9 a 5 sin sin 9 9. These measures make a triangle as shown below; thus we have a solution. Possible Solution II. If C 0, then A Then we find a: a sin A b sin B a 5 sin sin 9 a 5 sin sin 9 6. These measures make a triangle as shown below; thus we have a second solution. B A 5 0 C 9 B 6 0 C 5 0 A Eamples 5 illustrate the SSA situation. Note that we need not memorize the eight cases or the procedures in finding no solution, one solution, or two solutions. When we are using the law of sines, the sine value leads us directl to the correct solution or solutions. The Area of a Triangle The familiar formula for the area of a triangle, A bh, can be used onl when h is known. However, we can use the method used to derive the law of sines to derive an area formula that does not involve the height. Consider a general triangle ABC, with area K, as shown below. B B c h a h c a A D A is acute. C D A A is obtuse. C Note that in the triangle on the right, sin A sin 80 A. Then in each ADB, sin A h c, or h c sin A. Substituting into the formula bh, we get K K bc sin A.

220 Section 7. The Law of Sines 60 An pair of sides and the included angle could have been used. Thus we also have K ab sin C and K ac sin B. The Area of a Triangle The area K of an ABC is one half the product of the lengths of two sides and the sine of the included angle: K. bc sin A ab sin C ac sin B EXAMPLE 6 Area of the Peace Monument. Through the Mentoring in the Cit Program sponsored b Marian College, in Indianapolis, Indiana, children have turned a vacant downtown lot into a monument for peace.* This communit project brought together neighborhood volunteers, businesses, and government in hopes of showing children how to develop positive, nonviolent was of dealing with conflict. A landscape architect used the children s drawings and ideas to design a triangularshaped peace garden. Two sides of the propert, formed b Indiana Avenue and Senate Avenue, measure 8 ft and 0 ft, respectivel, and together form a.7 angle. The third side of the garden, formed b an apartment building, measures 6 ft. What is the area of this propert? 6 ft 8 ft.7 0 ft Solution formula: Since we do not know a height of the triangle, we use the area K K bc sin A 8 ft 0 ft sin.7 K,7 ft. The area of the propert is approimatel,7 ft. *The Indianapolis Star,August 6, 995, p. J8. Alan Da, a landscape architect with Browning Da Mullins Dierdorf, Inc., donated his time to this project.

221 60 Chapter 7 Applications of Trigonometr 7. Eercise Set Solve the triangle, if possible.. B 8, C, b A, a, c or at least approimate, the area of the ard. Find the area of the ard to the nearest square foot. 787 ft. A, C, b 0 B 6, a 7, c 9. A 6.5, a, b B 57., C 86., c 0,or B.6, C 0.9, c. B 8., C 5.6, b. A 6., a., c. 5. C 60, c 0., b. B, A 76, a. 6. A 6.5, a 7., c.5 C 9., B., b c mi, B 7.8, C.6 A 0.6, a 5 mi, b mi 8. a 5 mi, b 5 mi, A.67 No solution 9. b d, c d, C 8.78 B 8.78, A., a.0 d 0. A 9, C 88, b 0 in. B, a 75 in., c 70 in.. a 0.0 cm, b 0.07 cm, A 0. B.7, C 5.0, c 8.0 cm. b.57 km, c.6 km, C 58. A 89, a 5.6 in., b 8. in. No solution. C 6, a 56. m, c. m No solution 5. a 00 m, A.76, C.97 B 5.7, b 0 m, c 8 m 6. B 5, c 5.6 d, b.8 d No solution Find the area of the triangle. 7. B, a 7. ft, c. ft 8. ft 8. A 7, b 0 in., c in. 9 in 9. C 85, a d, b 6 d d 0. C 75.6, a.5 m, b. m.5. B 5., a 6. ft, c 6.7 ft ft. A, b 8. cm, c.7 cm 98.5 cm Solve.. Area of Back Yard. A new homeowner has a triangular-shaped back ard. Two of the three sides measure 5 ft and ft and form an included angle of 5.To determine the amount of fertilizer and grass seed to be purchased, the owner has to know, Answer to Eercise can be found on p. IA-6. m C ft 5 A. Boarding Stable. A rancher operates a boarding stable and temporaril needs to make an etra pen. He has a piece of rope 8 ft long and plans to tie the rope to one end of the barn (S) and run the rope around a tree (T) and back to the barn (Q). The tree is ft from where the rope is first tied, and the rope from the barn to the tree makes an angle of 5 with the barn. Does the rancher have enough rope if he allows ft at each end to fasten the rope? No S 5 ft 5 ft 5. Rock Concert. In preparation for an outdoor rock concert, a stage crew must determine how far apart to place the two large speaker columns on stage (see the figure at the top of the net page). What generall works best is to place them at 50 angles to the center of the front row. The distance from the center of the front row to each of the speakers is 0 ft. How far apart does the crew need to place the speakers on stage? About.86 ft, or ft 0 in. Q B T

222 Section 7. The Law of Sines 605 Speaker ft 0 ft Stage 9. Fire Tower. A ranger in fire tower A spots a fire at a direction of 95.A ranger in fire tower B,located 5 mi at a direction of 05 from tower A,spots the same fire at a direction of 55.How far from tower A is the fire? from tower B? From A: about 5 mi; from B: about 66 mi N B 6. Lunar Crater. Points A and B are on opposite sides of a lunar crater. Point C is 50 m from A.The measure of BAC is determined to be and the measure of ACB is determined to be.what is the width of the crater? About 76. m C Fire 95 N 5 5 mi A Length of Pole. A pole leans awa from the sun at an angle of 7 to the vertical. When the angle of elevation of the sun is 5, the pole casts a shadow 7 ft long on level ground. How long is the pole? About 5 ft 7 P In Eercises 8, keep in mind the two tpes of bearing considered in Sections 5. and Reconnaissance Airplane. A reconnaissance airplane leaves its airport on the east coast of the United States and flies in a direction of 085.Because of bad weather, it returns to another airport 0 km to the north of its home base. For the return trip, it flies in a direction of 8.What is the total distance that the airplane flew? About 67 km New airport 0 km Original airport N B 5 7 ft 85 8 N 0. Lighthouse. A boat leaves lighthouse A and sails 5. km. At this time it is sighted from lighthouse B, 7. km west of A. The bearing of the boat from B is N650E. How far is the boat from B? About. km or 0.6 km. Mackinac Island. Mackinac Island is located 8 mi N0W of Chebogan, Michigan, where the Coast Guard cutter Mackinaw is stationed. A freighter in distress radios the Coast Guard cutter for help. It radios its position as S780E of Mackinac Island and N60W of Chebogan. How far is the freighter from Chebogan? About mi Mackinac Island Chebogan

223 606 Chapter 7 Applications of Trigonometr. Gears. Three gears are arranged as shown in the figure below. Find the angle. r 8 ft r ft f r 6 ft 89 Snthesis 5. Prove the following area formulas for a general triangle ABC with area represented b K. K a sin B sin C sin A K c sin A sin B sin C K b sin C sin A sin B 6. Area of a Parallelogram. Prove that the area of a parallelogram is the product of two adjacent sides and the sine of the included angle. Collaborative Discussion and Writing. Eplain wh the law of sines cannot be used to find the first angle when solving a triangle given three sides.. We considered eight cases of solving triangles given two sides and an angle opposite one of them. Describe the relationship between side b and the height h in each. Skill Maintenance Find the acute angle A, in both radians and degrees, for the given function value. 5. cos A 0. [5.].8, cos A.56 [5.] No angle Convert to decimal degree notation [5.] [5.] Find the absolute value: 5. [R.] 5 Find the values. 0. cos [5.] 6. sin 5 [5.]. sin 00 [5.]. cos [5.]. Multipl: i i. [.] 7. Area of a Quadrilateral. Prove that the area of a quadrilateral is one half the product of the lengths of its diagonals and the sine of the angle between the diagonals. 8. Find d. d 8.8 in. S s in. a d c in. s 50 d b 5 in. Answers to Eercises 5 7 can be found on pp. IA-6 and IA-7.

224 Section 7. The Law of Cosines The Law of Cosines Use the law of cosines to solve triangles. Determine whether the law of sines or the law of cosines should be applied to solve a triangle. The law of sines is used to solve triangles given a side and two angles (AAS and ASA) or given two sides and an angle opposite one of them (SSA). A second law, called the law of cosines, is needed to solve triangles given two sides and the included angle (SAS) or given three sides (SSS). The Law of Cosines To derive this propert, we consider an ABC placed on a coordinate sstem. We position the origin at one of the vertices sa, C and the positive half of the -ais along one of the sides sa, CB. Let, be the coordinates of verte A. Point B has coordinates a, 0 and point C has coordinates 0, 0. (, ) A b c C (0, 0) a (a, 0) B Then cos C, so b b cos C and sin C, so b sin C. b Thus point A has coordinates b cos C, b sin C. (, ), or (b cos C, b sin C) A Net, we use the distance formula to determine c : c a 0, or c b cos C a b sin C 0. Now we multipl and simplif: b c C (0, 0) a B (a, 0) c b cos C ab cos C a b sin C a b sin C cos C ab cos C a b ab cos C. Using the identit sin cos

225 608 Chapter 7 Applications of Trigonometr Had we placed the origin at one of the other vertices, we would have obtained a b c bc cos A or b a c ac cos B. The Law of Cosines In an triangle ABC, C a b c bc cos A, b a c ac cos B, b a or c a b ab cos C. A c B Thus, in an triangle, the square of a side is the sum of the squares of the other two sides, minus twice the product of those sides and the cosine of the included angle. When the included angle is 90, the law of cosines reduces to the Pthagorean theorem. Solving Triangles (SAS) When two sides of a triangle and the included angle are known, we can use the law of cosines to find the third side. The law of cosines or the law of sines can then be used to finish solving the triangle. EXAMPLE Solve ABC if a, c 8, and B 5.. Solution We first label a triangle with the known and unknown measures. b C A? B 5. C? a b? c 8 A 8 5. B We can find the third side using the law of cosines, as follows: b a c ac cos B b 8 8 cos 5. Substituting b b 7. We now have a, b 7, and c 8, and we need to find the other two angle measures. At this point, we can find them in two was. One wa uses the law of sines. The ambiguous case ma arise, however, and we would have to be alert to this possibilit. The advantage of using

226 Section 7. The Law of Cosines 609 the law of cosines again is that if we solve for the cosine and find that its value is negative, then we know that the angle is obtuse. If the value of the cosine is positive, then the angle is acute. Thus we use the law of cosines to find a second angle. Let s find angle A. We select the formula from the law of cosines that contains cos A and substitute: a b c bc cos A cos A cos A cos A cos A A.0. The third angle is now eas to find: Thus, C A.0, a, B 5., b 7, C.8, c 8. Substituting Due to errors created b rounding, answers ma var depending on the order in which the are found. Had we found the measure of angle C first in Eample, the angle measures would have been C. and A 0.7.Variances in rounding also change the answers. Had we used 7. for b in Eample, the angle measures would have been A.5 and C.. Suppose we used the law of sines at the outset in Eample to find b. We were given onl three measures: a, c 8, and B 5.. When substituting these measures into the proportions, we see that there is not enough information to use the law of sines: a, sin A b sin B l sin A b sin 5. b sin B a sin A c sin C l b 8 sin 5. c sin C l 8 sin A. sin C, sin C In all three situations, the resulting equation, after the substitutions, still has two unknowns. Thus we cannot use the law of sines to find b.

227 60 Chapter 7 Applications of Trigonometr Solving Triangles (SSS) When all three sides of a triangle are known, the law of cosines can be used to solve the triangle. EXAMPLE Solve RST if r.5, s.7, and t.8. Solution We sketch a triangle and label it with the given measures. R.8.7 R? S? T? r.5 s.7 t.8 S.5 T Since we do not know an of the angle measures, we cannot use the law of sines. We begin instead b finding an angle with the law of cosines. We choose to find S first and select the formula that contains cos S : s r t rt cos S cos S cos S cos S S Similarl, we find angle R: Then Thus, r s t st cos R cos R cos R cos R R T R 7.80, r.5, S 95.86, s.7, T 6., t.8. Substituting

228 Section 7. The Law of Cosines cm B C cm cm A EXAMPLE Knife Bevel. Knifemakers know that the bevel of the blade (the angle formed at the cutting edge of the blade) determines the cutting characteristics of the knife. A small bevel like that of a straight razor makes for a keen edge, but is impractical for heav-dut cutting because the edge dulls quickl and is prone to chipping. A large bevel is suitable for heav-dut work like chopping wood. Survival knives, being universal in application, are a compromise between small and large bevels. The diagram at left illustrates the blade of a hand-made Randall Model 8 survival knife. What is its bevel? (Source: Randall Made Knives, P.O. Bo 988, Orlando, FL 80) Solution We know three sides of a triangle. We can use the law of cosines to find the bevel, angle A. a b c bc cos A 0.5 cos A cos A 0.5 cos A 8 cos A A.6. Thus the bevel is approimatel.6. C ONNECTING THE CONCEPTS CHOOSING THE APPROPRIATE LAW The following summarizes the situations in which to use the law of sines and the law of cosines. To solve an oblique triangle: Use the law of sines for: AAS ASA SSA Use the law of cosines for: SAS SSS The law of cosines can also be used for the SSA situation, but since the process involves solving a quadratic equation, we do not include that option in the list above.

229 6 Chapter 7 Applications of Trigonometr EXAMPLE In ABC, three measures are given. Determine which law to use when solving the triangle. You need not solve the triangle. a) a, b, c 0 b) a 07, B.8, C 57.6 c) A, C 7, a 8.7 d) B 0, a 960, c 0 e) b 7.6, a 7.9, A 9 f) A 6, B 9, C 80 Solution It is helpful to make a drawing of a triangle with the given information. The triangle need not be drawn to scale. The given parts are shown in color. FIGURE SITUATION LAW TO USE a) C SSS Law of Cosines A B b) C ASA Law of Sines A B c) C AAS Law of Sines A B d) C SAS Law of Cosines Stud Tip The InterAct Math Tutorial software that accompanies this tet provides practice eercises that correlate at the objective level to the odd-numbered eercises in the tet. Each practice eercise is accompanied b an eample and guided solution designed to involve students in the solution process. This software is available in our campus lab or on CD-ROM. A B e) C SSA Law of Sines A B f) C AAA Cannot be solved A B

230 Section 7. The Law of Cosines 6 7. Eercise Set Solve the triangle, if possible.. A 0, b, c a 5, B, C 6. B, a, c 5 b 5, A 0, C 7. a, b, c 0 A 6.8, B.5, C a., b., c 6. A 5.96, B 5.96, C B 70, c 6 m, a 78 m b 75 m, A 95, C 9 6. C.8, a 5. cm, b 7.8 cm c 5. cm, A 0.8, B a 6 m, b 0 m, c m A.5, B 0.75, C B 7.66, a.78 km, c 5.7 km b 9.8 km, A 50.59, C a ft, b ft, c 8 ft No solution 0. A 96, b 5.8 d, c 8. d a 5.5 d, B 8, C 55. a 6. km, b. km, c 9.5 km A 79.9, B 5.55, C 6.5. C 8, a 6 mm, b 9 mm c 5 mm, A 857, B 0. a 60. mi, b 0. mi, C 8.7 c 5.7 mi, A 89., B.0. a. cm, b 5. cm, c 7 cm A 8.7, B.0, C b 0. in., c 7. in., A 5.56 a.9 in., B 6.7, C a 7 d, b 5. d, c.5 d No solution Determine which law applies. Then solve the triangle. 7. A 70, B, b. Law of sines; C 98, a 96.7, c a 5, c 7, B 6 Law of cosines; b, A 9, C 6 9. a., b.7, c.8 Law of cosines; A 7.7, B 5.75, C a.5, b.5, A 58 No solution. A 0., B 9.8, C 00 Cannot be solved. a 60, b 0, C 7 Law of cosines; c, A 9, B. a.6, b 6., c. Law of cosines; A.7, B 07.08, C 9.. B 00, C 80, c 0.9 Law of sines; A 60, a 5.6, b 6.0 Solve. 5. Poachers. A park ranger establishes an observation post from which to watch for poachers. Despite losing her map, the ranger does have a compass and a rangefinder. She observes some poachers, and the rangefinder indicates that the are 500 ft from her position. The are headed toward big game that she knows to be 75 ft from her position. Using her compass, she finds that the poachers azimuth (the direction measured as an angle from north) is 55 and that of the big game is.what is the distance between the poachers and the game? About 67 ft 6. Circus Highwire Act. A circus highwire act walks up an approach wire to reach a highwire. The approach wire is ft long and is currentl anchored so that it forms the maimum allowable angle of 5 with the ground. A greater approach angle causes the aerialists to slip. However, the aerialists find that there is enough room to anchor the approach wire 0 ft back in order to make the approach angle less severe. When this is done, how much farther will the have to walk up the approach wire, and what will the new approach angle be? 6 ft farther, about 8 A Poachers 500 ft 55 N 75 ft Rangers Game C High wire 5 Ground 0 ft B D

231 6 Chapter 7 Applications of Trigonometr 7. In-line Skater. An in-line skater skates on a fitness trail along the Pacific Ocean from point A to point B.As shown below, two streets intersecting at point C also intersect the trail at A and B.In his car, the skater found the lengths of AC and BC to be approimatel 0.5 mi and. mi, respectivel. From a map, he estimates the included angle at C to be 0.How far did he skate from A to B? About.5 mi 0. Survival Trip. A group of college students is learning to navigate for an upcoming survival trip. On a map, the have been given three points at which the are to check in. The map also shows the distances between the points. However, to navigate the need to know the angle measurements. Calculate the angles for them. S.5, T 7., U mi. mi A C B Start T.6 km 50 S 5. km. km U Baseball Bunt. A batter in a baseball game drops a bunt down the first-base line. It rolls ft at an angle of 5 with the base path. The pitcher s mound is 60.5 ft from home plate. How far must the pitcher travel to pick up the ball? (Hint:A baseball diamond is a square.) About 0.8 ft Batter ft Pitcher 9. Ships. Two ships leave harbor at the same time. The first sails N5W at 5 knots (a knot is one nautical mile per hour). The second sails NE at 0 knots. After hr, how far apart are the ships? About 7 nautical mi d 5. Airplanes. Two airplanes leave an airport at the same time. The first flies 50 kmh in a direction of 0.The second flies 00 kmh in a direction of 00.After hr, how far apart are the planes? About 9 km. Slow-pitch Softball. A slow-pitch softball diamond is a square 65 ft on a side. The pitcher s mound is 6 ft from home plate. How far is it from the pitcher s mound to first base? About 6 ft. Isosceles Trapezoid. The longer base of an isosceles trapezoid measures ft. The nonparallel sides measure 0 ft, and the base angles measure 80. a) Find the length of a diagonal. About 6 ft b) Find the area. About ft. Area of Sail. A sail that is in the shape of an isosceles triangle has a verte angle of 8.The angle is included b two sides, each measuring 0 ft. Find the area of the sail. About ft 5. Three circles are arranged as shown in the figure below. Find the length PQ. About.7 cm r. cm P N5W NE r.8 cm Q 09 r. cm

232 Section 7. The Law of Cosines Swimming Pool. A triangular swimming pool measures ft on one side and.8 ft on another side. These sides form an angle that measures 0.8. How long is the other side? About 8.8 ft Collaborative Discussion and Writing 7. Tr to solve this triangle using the law of cosines. Then eplain wh it is easier to solve it using the law of sines. 8. Eplain wh we cannot solve a triangle given SAS with the law of sines. Skill Maintenance Classif the function as linear, quadratic, cubic, quartic, rational, eponential, logarithmic, or trigonometric. 9. f [.] Quartic 0. 7 [.] Linear. sin sin [5.5] Trigonometric. f / [.] Eponential. f [.5] Rational. f 7 [.] Cubic 5. e e [.] Eponential 6. log log [.] Logarithmic 7. f cos [5.5] Trigonometric 8. [.] Quadratic Snthesis. A C Canon Depth. A bridge is being built across a canon. The length of the bridge is 505 ft. From the deepest point in the canon, the angles of a 9 B elevation of the ends of the bridge are 78 and 7. How deep is the canon? About 986 ft 50. Heron s Formula. If a, b, and c are the lengths of the sides of a triangle, then the area K of the triangle is given b K ss as bs c, where s a b c. The number s is called the semiperimeter.prove Heron s formula. (Hint: Use the area formula K bc sin A developed in Section 7..) Then use Heron s formula to find the area of the triangular swimming pool described in Eercise A when 90 a sin ; 5. Area of Isosceles Triangle. Find a formula for the area of an isosceles triangle in terms of the congruent sides and their included angle. Under what conditions will the area of a triangle with fied congruent sides be maimum? 5. Reconnaissance Plane. A reconnaissance plane patrolling at 5000 ft sights a submarine at bearing 5 and at an angle of depression of 5.A carrier is at bearing 05 and at an angle of depression of 60. How far is the submarine from the carrier? About 0,06 ft E E ft 78 z 505 ft h 7 5 N N Answer to Eercise 50 can be found on p. IA-7.

233 66 Chapter 7 Applications of Trigonometr 7. Comple Numbers: Trigonometric Form comple numbers review section.. Graph comple numbers. Given a comple number in standard form, find trigonometric, or polar, notation; and given a comple number in trigonometric form, find standard notation. Use trigonometric notation to multipl and divide comple numbers. Use DeMoivre s theorem to raise comple numbers to powers. Find the nth roots of a comple number. Graphical Representation Just as real numbers can be graphed on a line, comple numbers can be graphed on a plane. We graph a comple number a bi in the same wa that we graph an ordered pair of real numbers a, b. However, in place of an -ais, we have a real ais, and in place of a -ais, we have an imaginar ais. Horizontal distances correspond to the real part of a number. Vertical distances correspond to the imaginar part. EXAMPLE Graph each of the following comple numbers. a) i b) 5i c) i d) i e) Solution Imaginar ais 5 i i 5 5 5i i 5 Real ais Imaginar ais a bi, or (a, b) z (0, 0) Real ais We recall that the absolute value of a real number is its distance from 0 on the number line. The absolute value of a comple number is its distance from the origin in the comple plane. For eample, if z a bi, then using the distance formula, we have z a bi a 0 b 0 a b. Absolute Value of a Comple Number The absolute value of a comple number a bi is a bi a b.

234 Section 7. Comple Numbers: Trigonometric Form 67 abs(i) abs(i) abs(/5i) EXAMPLE Find the absolute value of each of the following. a) i b) i c) Solution a) i b) i c) 5 i i We can check these results using a graphing calculator as shown at left. Note that and Trigonometric Notation for Comple Numbers Now let s consider a nonzero comple number a bi. Suppose that its absolute value is r. Ifwe let be an angle in standard position whose terminal side passes through the point a, b, as shown in the figure, then 5 i cos a r, or a r cos Imaginar ais a bi, or (a, b) and sin b r, or b r sin. r u a b Real ais Stud Tip It is never too soon to begin reviewing for the final eamination. Take a few minutes each week to read the highlighted (blue-screened boed) formulas, theorems, and properties. There is also at least one Connecting the Concepts feature in each chapter. Spend time reviewing the organized information and art in this special feature. Substituting these values for a and b into the a bi notation, we get a bi r cos r sin i rcos i sin. This is trigonometric notation for a comple number a bi. The number r is called the absolute value of a bi, and is called the argument of a bi. Trigonometric notation for a comple number is also called polar notation. Trigonometric Notation for Comple Numbers a bi rcos i sin To find trigonometric notation for a comple number given in standard notation, a bi, we must find r and determine the angle for which sin br and cos ar.

235 68 Chapter 7 Applications of Trigonometr EXAMPLE Find trigonometric notation for each of the following comple numbers. a) i b) i Solution a) We note that a and b. Then r a b, Imaginar ais i, or (, ) sin b, or, r u and Real cos a, or. r ais Since is in quadrant I,, or 5, and we have i cos i sin, or i cos 5 i sin 5. angle(i) angle( ()i) 5 0 b) We see that a and b. Then r, sin, and cos. Since is in quadrant IV,, or 0, and we have i cos i sin, 6 6 or i cos 0 i sin 0. As shown at left, a graphing calculator (in DEGREE mode) can be used to determine angle values in degrees. 6 Imaginar ais u Real ais i, or (, ) In changing to trigonometric notation, note that there are man angles satisfing the given conditions. We ordinaril choose the smallest positive angle.

236 Section 7. Comple Numbers: Trigonometric Form 69 To change from trigonometric notation to standard notation, a bi, we recall that a r cos and b r sin. EXAMPLE Find standard notation, a bi, for each of the following comple numbers. a) b) 8cos 7 7 cos 0 i sin 0 i sin Solution a) Rewriting, we have cos 0 i sin 0 cos 0 sin 0i. Thus, a cos 0 Degree Mode (cos(0)isin(0)) i Radian Mode (8)(cos(7π/)isin(7π/)) i and, so cos 0 i sin 0 i. b) Rewriting, we have 8cos 7 7 i sin 8 cos. 7 8 sin 7 i Thus, and so b sin 0 a 8 cos 7 8 b 8 sin 7, 8 8cos 7 7 i sin i. Multiplication and Division with Trigonometric Notation Multiplication of comple numbers is easier to manage with trigonometric notation than with standard notation. We simpl multipl the absolute values and add the arguments. Let s state this in a more formal manner.

237 60 Chapter 7 Applications of Trigonometr Comple Numbers: Multiplication For an comple numbers r cos i sin and r cos i sin, r cos i sin r cos i sin r r cos i sin. PROOF r cos i sin r cos i sin r r cos cos sin sin r r sin cos cos sin i Now, using identities for sums of angles, we simplif, obtaining r r cos r r sin i, or r r cos i sin, which was to be shown. Degree Mode (cos(0)isin(0))(cos(0) isin(0)) Radian Mode i (cos(π)isin(π))(cos(π/) isin(π/)) 6i EXAMPLE 5 Multipl and epress the answer to each of the following in standard notation. a) cos 0 i sin 0 and cos 0 i sin 0 b) cos i sin and cos sin i Solution a) cos 0 i sin 0 cos 0 i sin 0 cos 0 0 i sin 0 0 cos 60 i sin 60 i b) cos i sin cos sin i cos i sin 6cos 60 i 6i 6 6i i sin

238 Section 7. Comple Numbers: Trigonometric Form 6 EXAMPLE 6 Convert to trigonometric notation and multipl: i i. Solution We first find trigonometric notation: i cos 5 i sin 5, See Eample (a). i cos 0 i sin 0. See Eample (b). Then we multipl: cos 5 i sin 5 cos 0 i sin 0 cos 5 0 i sin 5 0 cos 75 i sin 75 cos 5 i sin has the same terminal side as 5. To divide comple numbers, we divide the absolute values and subtract the arguments. We state this fact below, but omit the proof. Comple Numbers: Division For an comple numbers r cos i sin and r cos i sin, r 0, r cos i sin. r cos i sin r cos i sin r EXAMPLE 7 Divide cos i sin b cos i sin and epress the solution in standard notation. Solution We have cos i sin cos i sin cos cos i sin sin i i 0.

239 6 Chapter 7 Applications of Trigonometr ( i)/( i) i EXAMPLE 8 Convert to trigonometric notation and divide: i. i Solution We first convert to trigonometric notation: i cos 5 i sin 5, See Eample (a). i cos 5 i sin 5. We now divide: cos 5 i sin 5 cos 5 i sin 5 cos 5 5 i sin 5 5 cos 70 i sin 70 0 i i. Powers of Comple Numbers An important theorem about powers and roots of comple numbers is named for the French mathematician Abraham DeMoivre (667 75). Let s consider the square of a comple number rcos i sin : rcos i sin rcos i sin rcos i sin r r cos i sin r cos i sin. Similarl, we see that rcos i sin r r r cos i sin r cos i sin. DeMoivre s theorem is the generalization of these results. DeMoivre s Theorem For an comple number rcos i sin and an natural number n, rcos i sin n r n cos n i sin n.

240 Section 7. Comple Numbers: Trigonometric Form 6 (i)ˆ9 66i ( ()i)ˆ i EXAMPLE 9 Find each of the following. a) i 9 b) i 0 Solution a) We first find trigonometric notation: i cos 5 i sin 5. See Eample (a). Then i 9 cos 5 i sin cos 9 5 i sin 9 5 9/ cos 05 i sin 05 6cos 5 i sin 5 6 i DeMoivre s theorem 05 has the same terminal side as i. b) We first convert to trigonometric notation: i cos 0 i sin 0. See Eample (b). Then i 0 cos 0 i sin cos 00 i sin 00 0cos 60 i sin 60 0 i 5 5i. 00 has the same terminal side as 60. Roots of Comple Numbers As we will see, ever nonzero comple number has two square roots. A nonzero comple number has three cube roots, four fourth roots, and so on. In general, a nonzero comple number has n different nth roots. The can be found using the formula that we now state but do not prove. Roots of Comple Numbers The nth roots of a comple number rcos i sin, r 0, are given b r /ncos n where k 0,,,..., n k n i sin k n n,

241 6 Chapter 7 Applications of Trigonometr i Imaginar ais i Real ais EXAMPLE 0 Find the square roots of i. Solution We first find trigonometric notation: i cos 60 i sin 60. Then n, n, and k 0, ; and cos 60 i sin 60 / /cos k i sin k cos 0 k 80 i sin 0 k 80, k 0,. Thus the roots are cos 0 i sin 0 for k 0 and cos 0 i sin 0 for k, or i and i., k 0, In Eample 0, we see that the two square roots of the number are opposites of each other. We can illustrate this graphicall. We also note that the roots are equall spaced about a circle of radius r in this case, r.the roots are 60, or 80 apart. EXAMPLE Find the cube roots of. Then locate them on a graph. Solution We begin b finding trigonometric notation: cos 0 i sin 0. Imaginar ais q i Then n, n, and k 0,, ; and cos 0 i sin 0 / /cos 0 The roots are 60 k i sin 0 60 k, k 0,,. q i Real ais cos 0 i sin 0, cos 0 i sin 0, and cos 0 i sin 0, or,, and. i i The graphs of the cube roots lie equall spaced about a circle of radius. The roots are 60, or 0 apart. The nth roots of are often referred to as the nth roots of unit. In Eample, we found the cube roots of unit.

242 Section 7. Comple Numbers: Trigonometric Form 65 EXPLORING WITH TECHNOLOGY Using a graphing calculator set in PARAMETRIC mode, we can approimate the nth roots of a number p.we use the following window and let XT p n cos T and YT p n sin T. WINDOW Tmin 0 Tma 60,ifin degree mode, or,ifin radian mode Tstep 60n,or n Xmin, Xma, Xscl Ymin, Yma, Yscl To find the fifth roots of 8, enter XT 8 5 cos T and YT 8 5 sin T.In this case, use DEGREE mode. After the graph has been generated, use the TRACE feature to locate the fifth roots. The T, X, and Y values appear on the screen. What do the represent? XT 8 /5 cos T YT 8 /5 sin T T ,.5 Three of the fifth roots of 8 are approimatel.557, i, and i. Find the other two. Then use a calculator to approimate the cube roots of unit that were found in Eample. Also approimate the fourth roots of 5 and the tenth roots of unit.

243 66 Chapter 7 Applications of Trigonometr 7. Eercise Set Graph the comple number and find its absolute value.. i. i. i. 5 i 5. i 6. 6 i i Epress the indicated number in both standard notation and trigonometric notation Imaginar Imaginar ais ais Real ais Real ais 5. 0cos 70 i sin 70 0i 6. cos 0 i sin cos 8. 5 cos i sin i sin 9. cos i sin i 0. cos i sin i 6. cos 60 i sin 60 i. cos 5 i sin 5 i 5 i 5 i. Imaginar ais. Imaginar ais. 5cos i sin i sin cos Real Real ais ais 5..5cos 5 i sin 5.5cos i sin cos i sin Find trigonometric notation. 6. i sin cos. i. 0 0i i sin 8cos i i Convert to trigonometric notation and then multipl 7. i 8. or divide. 7. i i 8. i i i 5 i i i 0. i i i i Find standard notation, a bi.. i i. i i 6 6i i. cos 0 i sin 0 i i. i.. 6cos 0 i sin 0 i i i i Answers to Eercises,, 5, 8, and 0 can be found on pp. IA-7 and IA-8. Multipl or divide and leave the answer in trigonometric notation. cos 8 i sin 8. cos i sin cos 6 i sin 6

244 Section 7. Comple Numbers: Trigonometric Form 67 Raise the number to the given power and write trigonometric notation for the answer. 78. Eplain wh trigonometric notation for a comple number is not unique, but rectangular, or standard, notation is unique. 5. cos i sin 8cos i sin 6. cos 0 i sin 0 6cos 0 i sin 0 Skill Maintenance 7. i 6 Convert to degree measure. 8. i [5.] [5.] 50 Raise the number to the given power and write Convert to radian measure. standard notation for the answer [5.] 8. 5 [5.] 5 9. cos 0 i sin 0 i cos 0 i sin 0 9 5i 8. Find r. [R.6] 6 5. i 5. 6 i i r i i i 6 Find the square roots of the number. 8. Graph these points in the rectangular coordinate 55. i 56. i sstem:,, 0,, and,. 57. i 58. i Find the function value using coordinates of points on Find the cube roots of the number. the unit circle. 59. i 60. 6i 85. sin [5.5] 86. cos [5.5] 6. i 6. i 6 6. Find and graph the fourth roots of cos [5.5] 88. sin 5 [5.5] 6. Find and graph the fourth roots of i Find and graph the fifth roots of. Snthesis 66. Find and graph the sith roots of. Solve. 67. Find the tenth roots of i i Find the ninth roots of. 90. i i Find the sith roots of. 9. Find polar notation for cos i sin. 70. Find the fourth roots of. 9. Show that for an comple number z, Find all the comple solutions of the equation. z z Show that for an comple number z and its 7. i conjugate z, i 0 z z. (Hint: Let z a bi and z a bi.) Collaborative Discussion and Writing 77. Find and graph the square roots of i. Eplain geometricall wh the are the opposites of each 9. Show that for an comple number z and its conjugate z, zz z. other. (Hint: Let z a bi and z a bi.) Answers to Eercises 7, 8, 55 76, 8, and 89 9 can be found on pp. IA-8 and IA-9.

245 68 Chapter 7 Applications of Trigonometr 95. Show that for an comple number z, z z. 96. Show that for an comple numbers z and w, z w z w. (Hint: Let z r cos i sin and w r cos i sin.) 97. Show that for an comple number z and an nonzero, comple number w, z z. (Use the hint for Eercise 96.) w w 98. On a comple plane, graph z. 99. On a comple plane, graph z z. Answers to Eercises can be found on p. IA-9.

246 68 Chapter 7 Applications of Trigonometr 7. Polar Coordinates and Graphs Origin or pole O P(, ), or P(r, u) r u Polar ais Graph points given their polar coordinates. Convert from rectangular to polar coordinates and from polar to rectangular coordinates. Convert from rectangular to polar equations and from polar to rectangular equations. Graph polar equations. Polar Coordinates All graphing throughout this tet has been done with rectangular coordinates,,, in the Cartesian coordinate sstem. We now introduce the polar coordinate sstem. As shown in the diagram at left, an point P has rectangular coordinates, and polar coordinates r,. On a polar graph, the origin is called the pole and the positive half of the -ais is called the polar ais. The point P can be plotted given the directed angle from the polar ais to the ra OP and the directed distance r from the pole to the point. The angle can be epressed in degrees or radians. To plot points on a polar graph:. Locate the directed angle.. Move a directed distance r from the pole. If r 0, move along ra OP.If r 0,move in the opposite direction of ra OP.

247 Section 7. Polar Coordinates and Graphs 69 Polar graph paper, shown below, facilitates plotting. Points B and G illustrate that ma be in radians. Points E and F illustrate that the polar coordinates of a point are not unique p G (6, ) A (, 60) 0 D (5, 90) 80 0 B (, p) 0 E (, 0) (, 0) 70 5 C (, 0) F (, 85) (, 75) (, 05) EXAMPLE Graph each of the following points. a) A, 60 b) B0, 0 c) C5, 0 d) D, 60 e) E, f ) F, Solution A F 0 B 5 60 D E 0 C 00 70

248 60 Chapter 7 Applications of Trigonometr To convert from rectangular to polar coordinates and from polar to rectangular coordinates, we need to recall the following relationships. u r P(, ), or P(r, u) r cos, r or sin, r or r cos r sin tan u (, ) EXAMPLE Convert each of the following to polar coordinates. a), b), Solution a) We first find r: r 8. Then we determine : tan ; therefore,, or. We know that and not 5 since, is in quadrant I. Thus, r,, 5, or,. Other possibilities for polar coordinates include, 5 and, 5. b) We first find r: 5 r 6. u (, ) Then we determine : 0 tan ; therefore,, or. 6 Thus, r,, 0, or, 6. Other possibilities for polar coordinates for this point include, 6 and, 50. It is easier to convert from polar to rectangular coordinates than from rectangular to polar coordinates.

249 Section 7. Polar Coordinates and Graphs 6 EXAMPLE Convert each of the following to rectangular coordinates. Solution a) The ordered pair 0, gives us r 0 and. We now find and : and a) 0, b) 5, 5 r cos 0 cos r sin 0 sin 0 5. Thus,, 5, 5. b) From the ordered pair 5, 5, we know that r 5 and 5. We now find and : 5 cos and 5 sin Thus,, 5., 5 The conversions above can be easil made with some graphing calculators. Polar and Rectangular Equations Some curves have simpler equations in polar coordinates than in rectangular coordinates. For others, the reverse is true. EXAMPLE Convert each of the following to a polar equation. a) 5 b) 5 Solution a) We have 5 r cos r sin 5 r cos r sin 5 r cos sin 5 r 5 r Substituting for and cos sin This eample illustrates that the polar equation of a circle centered at the origin is much simpler than the rectangular equation.

250 6 Chapter 7 Applications of Trigonometr b) We have 5 r cos r sin 5 r cos sin 5. In this eample, we see that the rectangular equation is simpler than the polar equation. EXAMPLE 5 a) b) c) r r cos 6 r cos sin Solution a) We have Convert each of the following to a rectangular equation. r Substituting for r 6. Squaring In squaring, we must be careful not to introduce solutions of the equation that are not alread present. In this case, we did not, because the graph of either equation is a circle of radius centered at the origin. b) We have r cos 6 6. r cos The graph of r cos 6, or 6, is a vertical line. c) We have r cos sin r r cos r sin Multipling both sides b r. Substituting for r, for r cos, and for r sin Graphing Polar Equations To graph a polar equation, we can make a table of values, choosing values of and calculating corresponding values of r. We plot the points and complete the graph, as we do when graphing a rectangular equation. A difference occurs in the case of a polar equation however, because as increases sufficientl, points ma begin to repeat and the curve will be traced again and again. When this happens, the curve is complete.

251 Section 7. Polar Coordinates and Graphs 6 EXAMPLE 6 Graph: r sin. Solution We first make a table of values. The TABLE feature on a graphing calculator is the most efficient wa to create this list. Note that the points begin to repeat at. We plot these points and draw the curve, as shown below. 60 r r r sin u Because of its heart shape, this curve is called a cardioid. We plotted points in Eample 6 because we feel that it is important to understand how these curves are developed. We also can graph polar equations using a graphing calculator. The equation usuall must be written first in the form r f. It is necessar to decide on not onl the best window dimensions but also the range of values for. Tpicall, we begin with a range of 0 to for in radians and 0 to 60 for in degrees. Because most polar graphs are curved, it is important to square the window to minimize distortion. EXPLORING WITH TECHNOLOGY Graph r sin. Begin b setting the calculator in POLAR mode, and use either of the following windows: WINDOW WINDOW r sin (Radians) (Degrees) 6 min 0 min 0 ma ma 60 step step 9 9 Xmin 9 Xmin 9 Xma 9 Xma 9 Xscl Xscl 6 Ymin 6 Ymin 6 Yma 6 Yma 6 Yscl Yscl

252 6 Chapter 7 Applications of Trigonometr We observe the same graph in both windows. The calculator allows us to view the curve as it is formed. Now graph each of the following equations and observe the effect of changing the coefficient of sin and the coefficient of : r sin, r 6 sin, r sin, r sin 5, r sin, r sin. Polar equations of the form r a cos n and r a sin n have rose-shaped curves. The number a determines the length of the petals, and the number n determines the number of petals. If n is odd, there are n petals. If n is even, there are n petals. EXAMPLE 7 Graph each of the following polar equations. Tr to visualize the shape of the curve before graphing it. a) b) c) r r 5 sin r csc Solution For each graph, we can begin with a table of values. Then we plot points and complete the graph. a) r For all values of, r is. Thus the graph of r is a circle of radius centered at the origin. r r r 00 We can verif our graph b converting to the equivalent rectangular equation. For r, we substitute for r and square. The resulting equation,, is the equation of a circle with radius centered at the origin.

253 Section 7. Polar Coordinates and Graphs 65 b) r 5 sin r 5 sin θ 6 r r 5 sin θ c) r csc We can rewrite r csc as r sin. r csc θ /sin θ r Not defined Not defined r csc θ We can check our graph in Eample 7(c) b converting the polar equation to the equivalent rectangular equation: r csc r sin r sin. Substituting for r sin The graph of is a horizontal line passing through 0, on a rectangular grid.

254 66 Chapter 7 Applications of Trigonometr EXAMPLE 8 Graph the equation r cos with a graphing calculator. Solution We first solve for r: r cos. We then obtain the following graph. r cos θ 6 6

255 68 Chapter 7 Applications of Trigonometr 7. Eercise Set Graph the point on a polar grid.., 5.,..5, 0., 5 5., 6..75, , 8. 0, 5 9., 5 0..,.., 60., 05 Find polar coordinates of points A, B, C, and D. Give three answers for each point C D A B 00 5p 6 p p B A C p 0 p D Find the polar coordinates of the point. Epress the angle in degrees and then in radians, using the smallest possible positive angle. 5. 0, 6., 7., 8., 9., 0.,.,.,.,. 0, 5. 6.,, 5 Use a graphing calculator to convert from rectangular to polar coordinates. Epress the answer in both degrees and radians, using the smallest possible positive angle. 7., 7 8., 5 7p 6 p p Answers to Eercises,, 5, 6, 8, 0, 7, 8, 5 58, and 6 6 can be found on pp. IA-9 and IA-50. p 5p p 6 p , , 6 Find the rectangular coordinates of the point.. 5, 60. 0, 0, 0., 5. 6, 0, 5., , 7., 5, 8.., 5 9., 0, 0., , 5,., 80 Use a graphing calculator to convert from polar to rectangular coordinates. Round the coordinates to the nearest hundredth..,.9,.05. 5, 5..,.50, , , ,.99 Convert to a polar equation r cos r sin 5. 6 r Convert to a rectangular equation r r sin 6. r sin 6. r r cos 6. r sin, 0

256 Section 7. Polar Coordinates and Graphs r 9 cos 7 sin 66. r 5 sin 7 cos i) j) r 5 sec r cos r cos sin Graph the equation b plotting points. Then check our work using a graphing calculator. 7. r sin 7. r cos 7. r cos 7. r sin 75. r cos 76. r sec 77. r cos 78. r cos In Eercises 79 90, use a graphing calculator to match the equation with one of figures (a) (l), which follow. Tr matching the graphs mentall before using a calculator. a) b) 6 c) 5 e) 6 6 d) 6 f) k) l) r sin (d) 80. r cos (i) 8. r (g) 8. r sin (c) 5 8. r cos (j) 8. r sin (f) 85. r cos (b) 86. r sec (a) 87. r sin (e) 88. r cos 5 (h) 89. r sin (k) 90. r sin 6 (l) Graph the equation using a graphing calculator. 9. r sin tan (Cissoid) 9. r (Spiral of Archimedes) 9. r e/0 (Logarithmic spiral) 9. r 0 (Logarithmic spiral) 95. r cos sec (Strophoid) 96. r cos (Peanut) r tan sec (Semicubical parabola) 98. r sin cos (Twisted sister) g) h) 6 6 Collaborative Discussion and Writing 99. Eplain wh the rectangular coordinates of a point are unique and the polar coordinates of a point are not unique Give an eample of an equation that is easier to graph in polar notation than in rectangular notation and eplain wh. Answers to Eercises 65, 66, 7 78, and 9 98 can be found on p. IA-50.

257 60 Chapter 7 Applications of Trigonometr Skill Maintenance Solve [.] 0. 5 [.] Graph Answers to Eercises 0 06 can be found on p. IA Snthesis 07. Convert to a rectangular equation: r sec. 08. The center of a regular heagon is at the origin, and one verte is the point, 0. Find the coordinates of the other vertices., 0,, 60,, 0,, 80,, 0,, 00

258 60 Chapter 7 Applications of Trigonometr 7.5 Vectors and Applications Determine whether two vectors are equivalent. Find the sum, or resultant, of two vectors. Resolve a vector into its horizontal and vertical components. Solve applied problems involving vectors. We measure some quantities using onl their magnitudes. For eample, we describe time, length, and mass using units like seconds, feet, and kilograms, respectivel. However, to measure quantities like displacement, velocit, or force, we need to describe a magnitude and a direction. Together magnitude and direction describe a vector. The following are some eamples. DISPLACEMENT An object moves a certain distance in a certain direction. A surveor steps 0 d to the northeast. A hiker follows a trail 5 mi to the west. A batter hits a ball 00 m along the left-field line. An object travels at a certain speed in a certain direction. A breeze is blowing 5 mph from the northwest. An airplane is traveling 50 kmh in a direction of. VELOCITY FORCE A push or pull is eerted on an object in a certain direction. A force of 00 lb is required to pull a cart up a 0 incline. A 5-lb force is required to lift a bo upward. A force of 5 newtons is eerted downward on the handle of a jack. (A newton, abbreviated N, is a unit of force used in phsics, and N 0. lb.)

259 Section 7.5 Vectors and Applications 6 Vectors W N S E Wind vector Vectors can be graphicall represented b directed line segments. The length is chosen, according to some scale, to represent the magnitude of the vector, and the direction of the directed line segment represents the direction of the vector. For eample, if we let cm represent 5 kmh, then a 5-kmh wind from the northwest would be represented b a directed line segment cm long, as shown in the figure at left. Vector A vector in the plane is a directed line segment. Two vectors are equivalent if the have the same magnitude and direction. A v O C B P D Consider a vector drawn from point A to point B. Point A is called the initial point of the vector, and point B is called the terminal point. 0 Smbolic notation for this vector is AB (read vector AB ). Vectors are also denoted b boldface letters such as u, v, and w. The four vectors in the figure at left have the same length and direction. Thus the represent equivalent vectors; that is, AB CD OP v. In the contet of vectors, we use to mean equivalent. 0 0 The length, or magnitude, of AB is epressed as AB. In order to determine whether vectors are equivalent, we find their magnitudes and directions. 0 EXAMPLE The vectors u, OR, and w are shown in the figure below. 0 Show that u OR w. (, ) u (, ) R (, ) O(0, 0) w (, ) (, )

260 6 Chapter 7 Applications of Trigonometr v u u v Same magnitude; same direction u u v (not equivalent) Different magnitudes; different directions u v Same magnitude; different directions u u u v Different magnitudes; same direction v Vector sum 5 v v Solution We first find the length of each vector using the distance formula: u 9 0, 0 OR , w 9 0. Thus 0 u OR w. 0 The vectors u, OR, and w appear to go in the same direction so we check their slopes. If the lines that the are on all have the same slope, the vectors have the same direction. We calculate the slopes: 0 u OR w Slope Since u, OR, and w have the same magnitude and the same direction, 0 u OR w. Keep in mind that the equivalence of vectors requires onl the same magnitude and the same direction not the same location. In the illustrations at left, each of the first three pairs of vectors are not equivalent. The fourth set of vectors is an eample of equivalence. Vector Addition Suppose a person takes steps east and then steps north. He or she will then be 5 steps from the starting point in the direction shown at left. A vector units long and pointing to the right represents steps east and a vector units long and pointing up represents steps north. The sum of the two vectors is the vector 5 steps in magnitude and in the direction shown. The sum is also called the resultant of the two vectors. In general, two nonzero vectors u and v can be added geometricall b placing the initial point of v at the terminal point of u and then finding the vector that has the same initial point as u and the same terminal point as v, as shown in the following figure. C u v v A u B The sum u v is the vector represented b the directed line segment from the initial point A of u to the terminal point C of v.that is, if 0 u AB and v BC 0,

261 Section 7.5 Vectors and Applications 6 then u v AB BC AC. We can also describe vector addition b placing the initial points of the vectors together, completing a parallelogram, and finding the diagonal of the parallelogram. (See the figure on the left below.) This description of addition is sometimes called the parallelogram law of vector addition. Vector addition is commutative. As shown in the figure on the right below, both u v and v u are represented b the same directed line segment. D C u v u v v v u u v v A u B u Applications If two forces F and F act on an object, the combined effect is the sum, or resultant, F F of the separate forces. F F F F C 5 O u v 5 B 5 A EXAMPLE Forces of 5 newtons and 5 newtons act on an object at right angles to each other. Find their sum, or resultant, giving the magnitude of the resultant and the angle that it makes with the larger force. 0 Solution We make a drawing this time, a rectangle using v or OB to represent the resultant. To find the magnitude, we use the Pthagorean theorem: v 5 5 v 5 5 v 9.. Here v denotes the length, or magnitude, of v. To find the direction, we note that since OAB is a right triangle, tan(.6) tan Using a calculator, we find, the angle that the resultant makes with the larger force:. 0 The resultant OB has a magnitude of 9. and makes an angle of with the larger force. tan 0.6

262 6 Chapter 7 Applications of Trigonometr aerial bearings review section 5.. W W N 0 S Windspeed 00 E 90 km/h S Airplane airspeed N 8 km/h E Pilots must adjust the direction of their flight when there is a crosswind. Both the wind and the aircraft velocities can be described b vectors. EXAMPLE Airplane Speed and Direction. An airplane travels on a bearing of 00 at an airspeed of 90 kmh while a wind is blowing 8 kmh from 0. Find the ground speed of the airplane and the direction of its track, or course, over the ground. Solution We first make a drawing. The wind is represented b OC 0 and 0 the velocit vector of the airplane b OA. The resultant velocit vector is 0 v, the sum of the two vectors. The angle between v and OA is called a drift angle O 0 N C 0 60 u v A 8 B sin(.907) Note that the measure of COA Thus the measure of CBA is also 60 (opposite angles of a parallelogram are equal). Since the sum of all the angles of the parallelogram is 60 and OCB and OAB have the same measure, each must be 0. B the law of cosines in OAB, we have v cos 0 v 7,5 v 8. Thus, v is 8 kmh. B the law of sines in the same triangle, 8 8, sin sin 0 or 8 sin 0 sin Thus,, to the nearest degree. The ground speed of the airplane is 8 kmh, and its track is in the direction of 00, or 89.

263 Section 7.5 Vectors and Applications 65 Components A w u v u C v B Given a vector w, we ma want to find two other vectors u and v whose sum is w. The vectors u and v are called components of w and the process of finding them is called resolving, or representing, a vector into its vector components. When we resolve a vector, we generall look for perpendicular components. Most often, one component will be parallel to the -ais and the other will be parallel to the -ais. For this reason, the are often called the horizontal and vertical components of a vector. In the figure at left, 0 the vector w AC 0 is resolved as the sum of u AB and v BC 0. The horizontal component of w is u and the vertical component is v. EXAMPLE A vector w has a magnitude of 0 and is inclined 0 with the horizontal. Resolve the vector into horizontal and vertical components. w 0 C v Solution We first make a drawing showing horizontal and vertical vectors u and v whose sum is w. From ABC, we find u and v using the definitions of the cosine and sine functions: A 0 u B cos 0 u, 0 or u 0 cos 0 00, sin 0 v, 0 or v 0 sin 0 8. Thus the horizontal component of w is 00 right, and the vertical component of w is 8 up. EXAMPLE 5 Shipping Crate. A wooden shipping crate that weighs 86 lb is placed on a loading ramp that makes an angle of 5 with the horizontal. To keep the crate from sliding, a chain is hooked to the crate and to a pole at the top of the ramp. Find the magnitude of the components of the crate s weight (disregarding friction) perpendicular and parallel to the incline. Solution We first make a drawing illustrating the forces with a rectangle. We let 5 D C A 86 B 5 R 0 CB the weight of the crate 86 lb (force of gravit), CD 0 the magnitude of the component of the crate s weight perpendicular to the incline (force against the ramp), and CA 0 the magnitude of the component of the crate s weight parallel to the incline (force that pulls the crate down the ramp).

264 66 Chapter 7 Applications of Trigonometr The angle at R is given to be 5 and BCD R 5 because the sides of these angles are, respectivel, perpendicular. Using the cosine and sine functions, we find that cos 5 CD0, 86 or CD 0 86 cos 5 70 lb, and sin 5 CA0, 86 or CA 0 86 sin 5 5 lb.

265 66 Chapter 7 Applications of Trigonometr 7.5 Eercise Set Sketch the pair of vectors and determine whether the are equivalent. Use the following ordered pairs for the initial and terminal points. A, E, I6, B, F, J, C, 5 G, K, D, H, O0, 0 0. GE, 0 BJ Yes 0. DJ, 0 OF No 0. DJ, 0 AB No. CG 0, FO 0 Yes 5. DK, BH 0 Yes 6. BA 0, 0 DI Yes 7. EG 0, 0 BJ No 8. GC 0, 0 FO No 9. GA 0, BH 0 No 0 0. JD, 0 CG No. AB 0, 0 ID Yes 0. OF, 0 HB No. Two forces of N (newtons) and 5 N act on an object at right angles. Find the magnitude of the resultant and the angle that it makes with the smaller force. 55 N, 55. Two forces of 50 N and 60 N act on an object at right angles. Find the magnitude of the resultant and the angle that it makes with the larger force. 78 N, 0 5. Two forces of 0 N and 600 N act on an object. The angle between the forces is 7. Find the magnitude of the resultant and the angle that it makes with the larger force. 99 N, 9 6. Two forces of 55 N and 5 N act on an object. The angle between the forces is 6. Find the magnitude of the resultant and the angle that it makes with the smaller force. 9 N, 6 In Eercises 7, magnitudes of vectors u and v and the angle between the vectors are given. Find the sum of u v.give the magnitude to the nearest tenth and give the direction b specifing to the nearest degree the angle that the resultant makes with u. 7. u 5, v 5, 57.0, u 5, v, 7., u 0, v, 8., u 5, v 0,.7, 7. u 0, v 0, 0.9, 58. u 0, v 0, 8.6, 6 7. u, v 7, 68., 8 7. u, v 7, 89., 5

266 Section 7.5 Vectors and Applications Hot-air Balloon. A hot-air balloon is rising verticall 0 ftsec while the wind is blowing horizontall 5 ftsec. Find the speed v of the balloon and the angle that it makes with the horizontal. ftsec, 6 6. Boat. A boat heads 5,propelled b a force of 750 lb. A wind from 0 eerts a force of 50 lb on the boat. How large is the resultant force F, and in what direction is the boat moving? 76 lb, 7 N v 5 ft/sec Wind 0 ft/sec Balloon Wind. A wind has an easterl component (from the east) of 0 kmh and a southerl component (from the south) of 6 kmh. Find the magnitude and the direction of the wind. 8.9 kmh, from 8. A vector w has magnitude 00 and points southeast. Resolve the vector into easterl and southerl components east; 70.7 south. A vector u with a magnitude of 50 lb is inclined to the right and upward 5 from the horizontal. Resolve the vector into components. Horizontal: 9 lb; vertical: 8 lb. Airplane. An airplane takes off at a speed S of 5 mph at an angle of 7 with the horizontal. Resolve the vector S into components. Horizontal: 5.7 mph forward; vertical: mph up. Wheelbarrow. A wheelbarrow is pushed b appling a 97-lb force F that makes a 8 angle with the horizontal. Resolve F into its horizontal and vertical components. (The horizontal component is the effective force in the direction of motion and the vertical component adds weight to the wheelbarrow.) Horizontal: 76. lb forward; vertical: 59.7 lb down 750 F F = Ship. A ship sails first N80E for 0 nautical mi, and then S0W for 00 nautical mi. How far is the ship, then, from the starting point, and in what direction? 7 nautical mi, S5 E 8. Airplane. An airplane flies 0 for 0 km, and then 80 for 70 km. How far is the airplane, then, from the starting point, and in what direction? 5 km, 5 9. Airplane. An airplane has an airspeed of 50 kmh. It is to make a flight in a direction of 070 while there is a 5-kmh wind from 0.What will the airplane s actual heading be? Luggage Wagon. A luggage wagon is being pulled with vector force V,which has a magnitude of 780 lb at an angle of elevation of 60.Resolve the vector V into components. Horizontal: 90 lb forward; vertical: lb up V =

267 68 Chapter 7 Applications of Trigonometr 8. Horizontal: 50.9 mph forward; vertical: 50.9 mph up 9. Perpendicular: 90.6 lb; parallel:. lb 6. Hot-air Balloon. A hot-air balloon eerts a Collaborative Discussion and Writing 00-lb pull on a tether line at a 5 angle with the horizontal. Resolve the vector B into components.. Describe the concept of a vector as though ou were eplaining it to a classmate. Use the concept of an arrow shot from a bow in the eplanation. 0. Eplain wh vectors QR and RQ 0 are not equivalent. 5 B = 00 Horizontal: 88.5 lb toward the balloon; vertical: 88.5 lb up 7. Airplane. An airplane is fling at 00 kmh in a direction of 05. Find the westerl and northerl components of its velocit. Northerl: 5 kmh; westerl: 6 kmh 8. Baseball. A baseball plaer throws a baseball with a speed S of 7 mph at an angle of 5 with the horizontal. Resolve the vector S into components. 9. A block weighing 00 lb rests on a 5 incline. Find the magnitude of the components of the block s weight perpendicular and parallel to the incline. a 0. A shipping crate that weighs 50 kg is placed on a loading ramp that makes an angle of 0 with the horizontal. Find the magnitude of the components of the crate s weight perpendicular and parallel to the incline. Perpendicular: 89.7 kg; parallel: 5 kg. An 80-lb block of ice rests on a 7 incline. What force parallel to the incline is necessar in order to keep the ice from sliding down? 8. lb. What force is necessar to pull a 500-lb truck up a 9 incline? 57.5 lb b Skill Maintenance In each of Eercises 5 5, fill in the blank with the correct term. Some of the given choices will not be used. angular speed linear speed acute obtuse secant of cotangent of identit inverse absolute value sines cosine common natural horizontal line vertical line double-angle half-angle coterminal reference angle 5. Logarithms, base e,are called logarithms. [.] natural 6. identities give trigonometric function values in terms of function values of. [6.] half-angle 7. is distance traveled per unit of time. [5.] linear speed 8. The sine of an angle is also the of the angle s complement. [5.] cosine 9. A(n) is an equation that is true for all possible replacements of the variables. [6.] identit 50. The is the length of the side adjacent to divided b the length of the side opposite. [5.] cotangent of 5. If two or more angles have the same terminal side, the angles are said to be. [5.] coterminal

268 Section 7.5 Vectors and Applications In an triangle, the sides are proportional to the of the opposite angles. [7.] sines 5. If it is possible for a to intersect the graph of a function more than once, then the function is not one-to-one and its is not a function. [.] horizontal line; inverse 5. The for an angle is the angle formed b the terminal side of the angle and the -ais. [5.] reference angle; acute Snthesis 55. Eagle s Flight. An eagle flies from its nest 7 mi in the direction northeast, where it stops to rest on a cliff. It then flies 8 mi in the direction S0W to land on top of a tree. Place an -coordinate sstem so that the origin is the bird s nest, the -ais points east, and the -ais points north. a) At what point is the cliff located?.950,.950 b) At what point is the tree located? 0.950,.978

269 Section 7.6 Vector Operations Vector Operations b (a, b) v a, b a Perform calculations with vectors in component form. Epress a vector as a linear combination of unit vectors. Epress a vector in terms of its magnitude and its direction. Find the angle between two vectors using the dot product. Solve applied problems involving forces in equilibrium. Position Vectors Let s consider a vector v whose initial point is the origin in an coordinate sstem and whose terminal point is a, b. We sa that the vector is in standard position and refer to it as a position vector. Note that the ordered pair a, b defines the vector uniquel. Thus we can use a, b to denote the vector. To emphasize that we are thinking of a vector and to avoid the confusion of notation with ordered-pair and interval notation, we generall write v a, b. The coordinate a is the scalar horizontal component of the vector, and the coordinate b is the scalar vertical component of the vector. B scalar, we mean a numerical quantit rather than a vector quantit. Thus, a, b is considered to be the component form of v. Note that a and b are not vectors and should not be confused with the vector component definition given in Section 7.5.

270 650 Chapter 7 Applications of Trigonometr Now consider AC 0 with A, and C,. Let s see how to find the position vector equivalent to AC 0. As ou can see in the figure below, the initial point A is relocated to the origin 0, 0. The coordinates of P are found b subtracting the coordinates of A from the coordinates 0 of C. Thus, P, and the position vector is OP. C (, ) A(, ) P(, ) O 0 It can be shown that OP and AC 0 have the same magnitude and direction and are therefore equivalent. Thus, AC 0 0 OP,. Component Form of a Vector The component form of AC 0 with A, and C, is AC 0,. P(5, 8) F(, 5) v 5, 8 O C (, ) v CF OP EXAMPLE Find the component form of CF if C, and F, 5. Solution We have 0 CF, 5 5, Note that vector CF is equivalent to position vector OP with P 5, 8 as shown in the figure at left. Now that we know how to write vectors in component form, let s restate some definitions that we first considered in Section 7.5. The length of a vector v is eas to determine when the components of the vector are known. For v v, v, we have v v v v v v. 0 Using the Pthagorean theorem v (v, v ) v v

271 Section 7.6 Vector Operations 65 Length of a Vector The length,or magnitude,ofa vector v v, v is given b v v v. Two vectors are equivalent if the have the same magnitude and the same direction. Equivalent Vectors Let u u, u and v v, v. Then u, u v, v if and onl if u v and u v. Operations on Vectors To multipl a vector v b a positive real number, we multipl its length b the number. Its direction stas the same. When a vector v is multiplied b for instance, its length is doubled and its direction is not changed. When a vector is multiplied b.6, its length is increased b 60% and its direction stas the same. To multipl a vector v b a negative real number, we multipl its length b the number and reverse its direction. When a vector is multiplied b, its length is doubled and its direction is reversed. Since real numbers work like scaling factors in vector multiplication, we call them scalars and the products kv are called scalar multiples of v..6v v v v Scalar multiples of v Scalar Multiplication For a real number k and a vector v v, v, the scalar product of k and v is kv kv, v kv, kv. The vector kv is a scalar multiple of the vector v.

272 65 Chapter 7 Applications of Trigonometr EXAMPLE Let u 5, and w,. Find 7w, u, and w. Solution 7w 7, 7, 7, u 5, 5,, w,, In Section 7.5, we used the parallelogram law to add two vectors, but now we can add two vectors using components. To add two vectors given in component form, we add the corresponding components. Let u u, u and v v, v. Then u v u v, u v. u v u (u, u ) u u v (u v, u v ) v (v, v ) v u v u v For eample, if v, and w 5, 9, then v w 5, 9, 7. Vector Addition If u u, u and v v, v, then u v u v, u v. v (v, v ) v (v, v ) Before we define vector subtraction, we need to define v. The opposite of v v, v, shown at left, is v v v, v v, v. Vector subtraction such as u v involves subtracting corresponding components. We show this b rewriting u v as u v. If u u, u and v v, v, then u v u v u, u v, v u v, u v u v, u v.

273 Section 7.6 Vector Operations 65 We can illustrate vector subtraction with parallelograms, just as we did vector addition. u v u v u (v) u v u v u u v v v v Sketch u and v. Sketch v. Sketch u (v), or u v, using the parallelogram law. u v is the vector from the terminal point of v to the terminal point of u. Vector Subtraction If u u, u and v v, v, then u v u v, u v. It is interesting to compare the sum of two vectors with the difference of the same two vectors in the same parallelogram. The vectors u v and u v are the diagonals of the parallelogram. u u v u v v EXAMPLE Do the following calculations, where u 7, and v, 5. a) u v b) u 6v c) u v d) 5v u Solution a) b) c) u v 7,, 5 7, 5, 7 u 6v 7, 6, 5 7, 8, 0 5, 8 u v 7,, 5, 6, 0 9, 6

274 65 Chapter 7 Applications of Trigonometr d) 5v u 5, 5 7, 5, 5, 9, Before we state the properties of vector addition and scalar multiplication, we need to define another special vector the zero vector. The vector whose initial and terminal points are both 0, 0 is the zero vector, denoted b O, or 0, 0. Its magnitude is 0. In vector addition, the zero vector is the additive identit vector: v O v. v, v 0, 0 v, v Operations on vectors share man of the same properties as operations on real numbers. Properties of Vector Addition and Scalar Multiplication For all vectors u, v, and w, and for all scalars b and c:. u v v u.. u v w u v w.. v O v.. v v; 0v O. 5. v v O. 6. bcv bcv. 7. b cv bv cv. 8. bu v bu bv. E, R v Unit Vectors A vector of magnitude, or length, is called a unit vector. The vector is a unit vector because v 5, 5 v 5, EXAMPLE Find a unit vector that has the same direction as the vector w, 5. Solution We first find the length of w: w 5.

275 Section 7.6 Vector Operations 655 Thus we want a vector whose length is of w and whose direction is the same as vector w. That vector is u. w, 5, 5 The vector u is a unit vector because u w Unit Vector If v is a vector and v O, then v, or, v v v is a unit vector in the direction of v. j i i, 0 j 0, Although unit vectors can have an direction, the unit vectors parallel to the - and -aes are particularl useful. The are defined as i, 0 and j 0,. An vector can be epressed as a linear combination of unit vectors i and j. For eample, let v v, v. Then v v, v v,0 0, v v, 0 v 0, v i v j. EXAMPLE 5 Epress the vector r, 6 as a linear combination of i and j. Solution r, 6 i 6j i 6j EXAMPLE 6 Write the vector q i 7j in component form. Solution q i 7j i 7j, 7

276 656 Chapter 7 Applications of Trigonometr Vector operations can also be performed when vectors are written as linear combinations of i and j. EXAMPLE 7 If a 5i j and b i 8j, find a b. Solution a b 5i j i 8j 5i 6j i 8j 6i j Unit circle O (0, ) u u (sin u)j (cos u)i (, 0) u cos u, sin u (cos u)i (sin u)j P(cos u, sin u) Direction Angles The terminal point P of a unit vector in standard position is a point on the unit circle denoted b cos,sin. Thus the unit vector can be epressed in component form, u cos,sin, or as a linear combination of the unit vectors i and j, u cos i sin j, where the components of u are functions of the direction angle measured counterclockwise from the -ais to the vector. As varies from 0 to, the point P traces the circle. This takes in all possible directions for unit vectors so the equation u cos i sin j describes ever possible unit vector in the plane. unit circle review section 5.5. q, u u i (, 0) EXAMPLE 8 sin j for Solution Calculate and sketch the unit vector u cos i. Include the unit circle in our sketch. u cos i sin j i j Let v v, v with direction angle. Using the definition of the tangent function, we can determine the direction angle from the components of v: v v v, v v i v j v (v, v ) tan v v tan v v. u v

277 Section 7.6 Vector Operations 657 u w (, ) w, i j EXAMPLE 9 Determine the direction angle of the vector w i j. Solution We know that w i j,. Thus we have tan and. Since w is in the third quadrant, we know that is a third-quadrant angle. The reference angle is tan tan , and, or 7. It is convenient for work with applied problems and in subsequent courses, such as calculus, to have a wa to epress a vector so that both its magnitude and its direction can be determined, or read, easil. Let v be a vector. Then vv is a unit vector in the same direction as v. Thus we have v v cos i sin j v vcos i sin j Multipling b v v vcos i vsin j. Let s revisit the applied problem in Eample of Section 7.5 and use this new notation. EXAMPLE 0 Airplane Speed and Direction. An airplane travels on a bearing of 00 at an airspeed of 90 kmh while a wind is blowing 8 kmh from 0. Find the ground speed of the airplane and the direction of its track, or course, over the ground. Solution We first make a drawing. The wind is represented b OC 0 and the velocit vector of the airplane b OA 0. The resultant velocit vector is v, the sum of the two vectors: v OC 0 OA 0. N 8 C O 0 50 v A

278 658 Chapter 7 Applications of Trigonometr The bearing (measured from north) of the airspeed vector OA 0 is 00. Its direction angle (measured counterclockwise from the positive -ais) is 50. The bearing (measured from north) of the wind vector OC 0 is 0. Its direction angle (measured counterclockwise from the positive -ais) is 50. The magnitudes of OA 0 and OC 0 are 90 and 8, respectivel. We have OA 0 90cos 50i 90sin 50j, and OC 0 8cos 50i 8sin 50j. Thus, v OA 0 OC 0 90cos 50i 90sin 50j 8cos 50i 8sin 50j 90cos 50 8cos 50i 90sin 50 8sin 50j 7.97i.78j. From this form, we can determine the ground speed and the course: We let Ground speed kmh. be the direction angle of v.then tan Thus the course of the airplane (the direction from north) is 90, or 89. tan Angle Between Vectors When a vector is multiplied b a scalar, the result is a vector. When two vectors are added, the result is also a vector. Thus we might epect the product of two vectors to be a vector as well, but it is not. The dot product of two vectors is a real number, or scalar. This product is useful in finding the angle between two vectors and in determining whether two vectors are perpendicular. Dot Product The dot product of two vectors u u, u and v v, v is u v u v u v. (Note that u v u v is a scalar, not a vector.)

279 Section 7.6 Vector Operations 659 EXAMPLE Find the indicated dot product when u, 5, v 0,, and w,. a) b) u w w v Solution a) b) u w w v 0 0 The dot product can be used to find the angle between two vectors. The angle between two vectors is the smallest positive angle formed b the two directed line segments. Thus the angle between u and v is the same angle as between v and u, and 0. Angle Between Two Vectors If is the angle between two nonzero vectors u and v, then cos u v. uv EXAMPLE Find the angle between u, 7 and v,. Solution We begin b finding u v, u, and v: v, a u, 7 Then u v 7, u 7 58, and v 0. cos u v uv 58 0 cos Forces in Equilibrium When several forces act through the same point on an object, their vector sum must be O in order for a balance to occur. When a balance occurs, then the object is either stationar or moving in a straight line without acceleration. The fact that the vector sum must be O for a balance, and vice versa, allows us to solve man applied problems involving forces.

280 660 Chapter 7 Applications of Trigonometr EXAMPLE Suspended Block. A 50-lb block is suspended b two cables, as shown at left. At point A, there are three forces acting: W, the block pulling down, and R and S, the two cables pulling upward and outward. Find the tension in each cable. 55 A 7 Solution We draw a force diagram with the initial points of each vector at the origin. For there to be a balance, the vector sum must be the vector O: R S W O. 50 lb We can epress each vector in terms of its magnitude and its direction angle: R 5 55 S 7 W R Rcos 5i sin 5j, S Scos 7i sin 7j, and W Wcos 70i sin 70j 50cos 70i 50sin 70j 50j. cos 70 0; sin 70 Substituting for R, S, and W in R S W O, we have Rcos 5 Scos 7i Rsin 5 Ssin 7 50j 0i 0j. This gives us a sstem of equations: Rcos 5 Scos 7 0, Rsin 5 Ssin Solving this sstem, we get R 80 and S 0. The tensions in the cables are 80 lb and 0 lb.

281 Section 7.6 Vector Operations Eercise Set Find the component form of the vector given the initial and terminal points. Then find the length of the vector.. MN 0 ; M6, 7, N, 9, 5; 06. CD 0 ; C, 5, D5, 7, ; 5 0. FE; E8,, F,, 6; 5. BA 0 ; A9, 0, B9, 7 0, 7; KL; K,, L8,, 0; 6. GH 0 ; G6, 0, H,, 8; 7 7. Find the magnitude of vector u if u, Find the magnitude of vector ST 0 if 7 0 ST, 5. Do the indicated calculations in Eercises 9 6 for the vectors u 5,, v, 7, and w,. 9. u w, 5 0. w u, 5. w v 57. 6v 5u,. v u 9, 9. w u v, v 0, 5 7. u v v u v u w, 0. w u v 0, 9. v O 8,. 07w u u w. w u 5. u v 6. v w 7 The vectors u, v, and w are drawn below. Cop them on a sheet of paper. Then sketch each of the vectors in Eercises Vectors u, v, and w are determined b the sides of ABC below. (a) w u v; (b) v w u B a) Find an epression for w in terms of u and v. b) Find an epression for v in terms of u and w.. In ABC, vectors u and w are determined b the sides shown, where P is the midpoint of side BC. Find an epression for v in terms of u and w. v u w B A A u v u Find a unit vector that has the same direction as the given vector.. v 5,. u,, 5 5. w, 0 6. a 6, 7 7. r, 8 8. t, Epress the vector as a linear combination of the unit vectors i and j. 9. w, 6 0. r 5, 9 w i 6j r 5i 9j. s, 5 s i 5j. u, u i j Epress the vector as a linear combination of i and j.. 7i 5j. i 8j w w P v C C Q(, ) Q(, ) u w v 7. u v 8. v u 9. u v w 0. u w P(, ) P(, ) Answers to Eercises 7 0,, and 5 8 can be found on p. IA-50.

282 66 Chapter 7 Applications of Trigonometr For Eercises 5 8, use the vectors u i j, v i 0j, and w i 5j. Perform the indicated vector operations and state the answer in two forms: (a) as a linear combination of i and j and (b) in component form. 5. u 5w 6. v w 7. u v w 8. u v w Sketch (include the unit circle) and calculate the unit vector u cos i sin j for the given direction angle. 9. j,or 0, j,or 0, Determine the direction angle of the vector, to the nearest degree. 5. u, w, 55. q i j w 5i j t 5, b 8, 07 Find the magnitude and the direction angle of the vector. 59. u cos 5i sin 5j u ; 60. w 6cos 50i sin 50j 6. v ; 0, w 6; u i j u ; Find the angle between the given vectors, to the nearest tenth of a degree. 6. u, 5, v,. 6. a,, b 5, w, 5, r 5, v,, t, a i j, b i j u i j, v i j 70. Epress each vector in Eercises 69 7 in the form ai bj and sketch each in the coordinate plane. 69. The unit vectors u cos i sin j for and. Include the unit circle in our sketch The unit vectors u cos i sin j for and. Include the unit circle in our sketch. 7. The unit vector obtained b rotating j counterclockwise radians about the origin 7. The unit vector obtained b rotating j clockwise radians about the origin For the vectors in Eercises 7 and 7, find the unit vectors u cos i sin j in the same direction. 7. i j 7. 6i 8j u 5 i 5 j For the vectors in Eercises 75 and 76, epress each vector in terms of its magnitude and its direction i j 5i j i 77. Use a sketch to show that v i 6j and u i j have opposite directions. 78. Use a sketch to show that v i 6j and u i j have the same direction. Answers to Eercises 5 8, 50, 5, and 69 7, 75, 77, and 78 can be found on pp. IA-50 and IA-5. Eercises 79 8 appeared first in Eercise Set 7.5, where we used the law of cosines and the law of sines to solve the applied problems. For this eercise set, solve the problem using the vector form v vcos i sin j. 79. Ship. A ship sails first N80E for 0 nautical mi, and then S0W for 00 nautical mi. How far is the ship, then, from the starting point, and in what direction? 7 nautical mi, S7 E 80. Boat. A boat heads 5,propelled b a force of 750 lb. A wind from 0 eerts a force of 50 lb on the boat. How large is the resultant force, and in what direction is the boat moving? 76 lb, 7 8. Airplane. An airplane has an airspeed of 50 kmh. It is to make a flight in a direction of 070 while there is a 5-kmh wind from 0. What will the airplane s actual heading be? Airplane. An airplane flies 0 for 0 mi, and then 80 for 70 mi. How far is the airplane, then, from the starting point, and in what direction? 5 mi, 5 j

283 Section 7.6 Vector Operations Two cables support a 000-lb weight, as shown. Find the tension in each cable. 500 lb on left, 866 lb on right 86. A weight of 00 lb is supported b a frame made of two rods and hinged at points A, B, and C. Find the forces eerted b the two rods. Horizontal rod: 68-lb tension; other rod: 6-lb compression 60 0 B C 000 lb 8. A 500-kg block is suspended b two ropes, as shown. Find the tension in each rope. kg on left, 80 kg on right A 0 00 lb 5 60 Let u u, u and v v, v. Prove each of the following properties. 87. u v v u 500 kg 88. u v v u 85. A 50-lb sign is hanging from the end of a hinged boom, supported b a cable inclined with the horizontal. Find the tension in the cable and the compression in the boom. Cable: -lb tension; boom: 67-lb compression Collaborative Discussion and Writing 89. Eplain how unit vectors are related to the unit circle. 90. Write a vector sum problem for a classmate for which the answer is v 5i 8j. 50 lb Skill Maintenance Find the slope and the -intercept of the line with the given equation. [.] 9. 5 ; 0, [.] 0; 0, 7 Find the zeros of the function [.] 0, [.], Snthesis If the dot product of two nonzero vectors u and v is 0, then the vectors are perpendicular (orthogonal). Let u u, u and v v, v. a) Prove that if u v 0, then u and v are perpendicular. b) Give an eample of two perpendicular vectors and show that their dot product is 0. Answers to Eercises 87, 88, and 95 can be found on p. IA-5.

284 66 Chapter 7 Applications of Trigonometr 96. If is an vector, what is PQ 0 0 PQ 0 QP? 99. Given the vector AB 0 i j and A is the point, 9, find the point B. 5, Find all the unit vectors that are parallel to the vector,., 5 i 5 i 5 j 5 j 00. Find vector v from point A to the origin, where AB 0 i j and B is the point, 5. 6i 7j 98. Find a vector of length whose direction is the opposite of the direction of the vector v i j.how man such vectors are there? 5 5 i 5 5 j Answer to Eercise 96 can be found on p. IA-5.

285 66 Chapter 7 Applications of Trigonometr Chapter 7 Summar and Review Important Properties and Formulas The Law of Sines a sin A b sin B c sin C The Area of a Triangle K bc sin A ab sin C ac sin B The Law of Cosines a b c bc cos A, b a c ac cos B, c a b ab cos C Comple Numbers Absolute Value: Trigonometric Notation: Multiplication: Division: a bi a b a bi rcos i sin r cos i sin r cos i sin r r cos i sin r cos i sin r cos i sin r cos i sin r, r 0 (continued)

286 Chapter 7 Summar and Review 665 DeMoivre s Theorem rcos i sin n r n cos n i sin n Roots of Comple Numbers The nth roots of rcos i sin are r /ncos n k n i sin k n n, r 0, k 0,,,..., n. Vectors If u u, u and v v, v and k is a scalar, then: Length: Addition: Subtraction: Scalar Multiplication: Dot Product: v v v u v u v, u v u v u v, u v kv kv, kv u v u v u v Angle Between Two Vectors: cos u v uv

287 Chapter 7 Review Eercises 665 Review Eercises Solve ABC, if possible.. a. ft, b 5.7 ft, c 8. ft [7.] A 5, B 8, C 9. B 7, C 5, b 9 in. [7.] A 8, a 7 in., c in.. A 8, C, b 890 m [7.] B 50, a 5 m, c 87 m. B 7, b d, c 8 d [7.] No solution 5. Find the area of ABC if b 9.8 m, c 7. m, and A 67.. [7.] m 6. A parallelogram has sides of lengths. ft and 7.85 ft. One of its angles measures 7. Find the area of the parallelogram. [7.].7 ft

288 666 Chapter 7 Applications of Trigonometr 7. Sandbo. A child-care center has a triangularshaped sandbo. Two of the three sides measure 5 ft and.5 ft and form an included angle of. To determine the amount of sand that is needed to fill the bo, the director must determine the area of the floor of the bo. Find the area of the floor of the bo to the nearest square foot. [7.] 6 ft 8. Flower Garden. A triangular flower garden has sides of lengths m, 9 m, and 6 m. Find the angles of the garden to the nearest degree. [7.] 9,, In an isosceles triangle, the base angles each measure 5. and the base is 5 ft long. Find the lengths of the other two sides to the nearest foot. [7.] 9 ft 0. Airplanes. Two airplanes leave an airport at the same time. The first flies 75 kmh in a direction of 05.6.The second flies 0 kmh in a direction of 95.5.After hr, how far apart are the planes? [7.] About 650 km Graph the comple number and find its absolute value.. 5i.. i. i Find trigonometric notation. 5. i 6. i i 8. Find standard notation, a bi. 9. cos 60 i sin 60 [7.] i 0. 7cos 0 i sin 0 [7.] 7. 5 cos i sin [7.] 5. cos sin [7.] i i Answers to Eercises 8,, and 0 can be found on p. IA-5..5 ft 5 ft 5 i Convert to trigonometric notation and then multipl or divide, epressing the answer in standard notation. i. i i. [7.] i i i 5. [7.] i 6. i i i [7.] i Raise the number to the given power and write trigonometric notation for the answer. 7. cos 60 i sin 60 [7.] 8cos 80 i sin i [7.] cos 7 i sin 7 Raise the number to the given power and write standard notation for the answer. 9. [7.] 8i 0. 0 i 6 i. Find the square roots of i.. Find the cube roots of i.. Find and graph the fourth roots of 8.. Find and graph the fifth roots of. Find all the comple solutions of the equation. 5. i Find the polar coordinates of each of these points. Give three answers for each point. 0 A B D 0 Find the polar coordinates of the point. Epress the answer in degrees and then in radians. 8., 9. 0, Use a graphing calculator to convert from rectangular to polar coordinates. Epress the answer in degrees and then in radians. 0., 5.., 7 60 C 00 0

289 Chapter 7 Review Eercises 667 Find the rectangular coordinates of the point..,. 6, 0 [7.], Use a graphing calculator to convert from polar to rectangular coordinates. Round the coordinates to the nearest hundredth. [7.].86,.5 [7.].9, 0.5., 5 5.., 5 Convert to a polar equation [7.] r5 cos sin 6 Magnitudes of vectors u and v and the angle between the vectors are given. Find the magnitude of the sum, u v,to the nearest tenth and give the direction b specifing to the nearest degree the angle that it makes with the vector u. 58. u, v 5, [7.5].7, u, v 60, [7.5] 98.7, 5 The vectors u, v, and w are drawn below. Cop them on a sheet of paper. Then sketch each of the vectors in Eercises 60 and [7.] r sin 8. 9 [7.] r v w u [7.] r sin r cos 6 0 Convert to a rectangular equation. 50. r 6 5. r r sin 5. r 5. r cos sin cos In Eercises 5 57, match the equation with one of figures (a) (d), which follow. a) 80 c) b) 80 d) r sin [7.] (b) 55. r cos [7.] (d) 56. r cos [7.] (a) 57. r sin [7.] (c) 60. u v 6. u w 6. Forces of 0 N and 500 N act on an object. The angle between the forces is 5. Find the resultant, giving the angle that it makes with the smaller force. [7.5] 666.7N, 6 6. Wind. A wind has an easterl component of 5 kmh and a southerl component of 5 kmh. Find the magnitude and the direction of the wind. [7.5] 9 kmh, 9 6. Ship. A ship sails N75E for 90 nautical mi, and then S0W for 00 nautical mi. How far is the ship, then, from the starting point, and in what direction? [7.5] 0. nautical mi, S E Find the component form of the vector given the initial and terminal points. 65. AB 0 ; A, 8, B, 5 [7.6], 66. TR 0 ; R0, 7, T, [7.6], Find the magnitude of vector u if u 5, 6. [7.6] 6 Do the calculations in Eercises 68 7 for the vectors u,, v, 9 and w, u w [7.6] 0, 69. w 6v [7.6], u w 7. u w [7.6] [7.6] Find a unit vector that has the same direction as v 6,. 7. Epress the vector t 9, as a linear combination of the unit vectors i and j. [7.6] 9i j Answers to Eercises, 50 5, 60, 6, and 7 can be found on p. IA-5.

290 668 Chapter 7 Applications of Trigonometr 7. Determine the direction angle of the vector w, to the nearest degree. [7.6] Find the magnitude and the direction angle of u 5i j. [7.6] ; 76. Find the angle between u, 7 and v, to the nearest tenth of a degree. [7.6] Airplane. An airplane has an airspeed of 60 mph. It is to make a flight in a direction of 080 while there is a 0-mph wind from 0.What will the airplane s actual heading be? [7.6] 85. Do the calculations in Eercises 78 8 for the vectors u i 5j, v i 0j, and w i 7j u 8v [7.6] i 55j 79. u v w 80. u v [7.6] 5 8. w v 8. Epress the vector PQ 0 in the form ai bj, if P is the point, and Q is the point,. [7.6] 5i 5j Epress each vector in Eercises 8 and 8 in the form ai bj and sketch each in the coordinate plane. 8. The unit vectors u cos i sin j for and. Include the unit circle in our sketch The unit vector obtained b rotating j counterclockwise radians about the origin. 85. Epress the vector i j as a product of its magnitude and its direction..0 Collaborative Discussion and Writing 86. Eplain wh these statements are not contradictor: The number has one real cube root. The number has three comple cube roots. 87. Summarize how ou can tell algebraicall when solving triangles whether there is no solution, one solution, or two solutions. 88. Golf: Distance versus Accurac. It is often argued in golf that the farther ou hit the ball, the more accurate it must be to sta safe. (Safe means not in the woods, water, or some other hazard.) In his book Golf and the Spirit (p. 5), M. Scott Peck asserts Deviate 5 from our aiming point on a 50-d shot, and our ball will land approimatel 0 d to the side of where ou wanted it to be. Do the same on a 00-d shot, and it will be 0 d off target. Twent ards ma well be in the range of safet; 0 ards probabl won t. This principle not infrequentl allows a mediocre, short-hitting golfer like mself to score better than the long hitter. Check the accurac of the mathematics in this statement, and comment on Peck s assertion. Snthesis 89. Let u i 5j. Find a vector that has the same 6 5 direction as u but has length. [7.6] i j 90. A parallelogram has sides of lengths. and Its area is 8.. Find the sizes of its angles. [7.] 50.5, 9.8 Answers to Eercises 79, 8, and 8 88 can be found on pp. IA-5 and IA-5.

291 Answers Chapter R Eercise Set R.. 8,0,9, 5. 7, 5...,, 5 5, 5., 5., 7, 8, 0,.96, 9,, 5, , 7,.96,, 7 9., 0., ;., ; 5., ; 7..8, ; 9. 7, ;. 0, 5. 9, 5., h 7. p, 9. True. False. True 5. False 7. False 9. True. True. True 5. False 7. Commutative propert of multiplication 9. Multiplicative identit propert 5. Associative propert of multiplication 5. Commutative propert of multiplication 55. Commutative propert of addition 57. Multiplicative inverse propert Discussion and Writing Answers ma var; Answers ma var; Eercise Set R... 5.,or ,or 9. 7,or a,or 7 5b a b 5,or 9.,or a n 5 5. b 7.,or 8 9.,or. 8 5,or. 8a b c d m 8,or 5 m z b 5.,or. 5 a 0 b 5 c 0,or z 9 a 0 c ,700, ,960,000, ,900, mi m 77. $ a A-

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