CK-12 Trigonometry - Second Edition, Solution Key

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2 CK-1 Trigonometry - Second Edition, Solution Key CK-1 Foundation Say Thanks to the Authors Click (No sign in required)

3 To access a customizable version of this book, as well as other interactive content, visit AUTHOR CK-1 Foundation CK-1 Foundation is a non-profit organization with a mission to reduce the cost of textbook materials for the K-1 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook, CK-1 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform. Copyright 014 CK-1 Foundation, The names CK-1 and CK1 and associated logos and the terms FlexBook and FlexBook Platform (collectively CK-1 Marks ) are trademarks and service marks of CK-1 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-1 Content (including CK-1 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial.0 Unported (CC BY-NC.0) License ( licenses/by-nc/.0/), as amended and updated by Creative Commons from time to time (the CC License ), which is incorporated herein by this reference. Complete terms can be found at Printed: July 1, 014 iii

4 Contents Contents 1 Right Triangles and an Introduction to Trigonometry, Solution Key The Pythagorean Theorem Special Right Triangles Basic Trigonometric Functions Solving Right Triangles Measuring Rotation Applying Trig Functions to Angles of Rotation Trigonometric Functions of Any Angle Relating Trigonometric Functions Graphing Trigonometric Functions, Solution Key 19.1 Radian Measure Applications of Radian Measure Circular Functions of Real Numbers Translating Sine and Cosine Functions Amplitude, Period and Frequency General Sinusoidal Graphs Graphing Tangent, Cotangent, Secant, and Cosecant Trigonometric Identities and Equations, Solution Key 5.1 Fundamental Identities Proving Identities Solving Trigonometric Equations Sum and Difference Identities Double Angle Identities Half-Angle Identities Products, Sums, Linear Combinations, and Applications Inverse Trigonometric Functions, Solution Key Basic Inverse Trigonometric Functions Graphing Inverse Trigonometric Functions Inverse Trigonometric Properties Applications & Models Triangles and Vectors, Solution Key The Law of Cosines Area of a Triangle The Law of Sines The Ambiguous Case General Solutions of Triangles Vectors Component Vectors iv

5 Contents 6 The Polar System, Solution Key Polar Coordinates Graphing Basic Polar Equations Converting Between Systems More with Polar Curves The Trigonometric Form of Complex Numbers The Product & Quotient Theorems De Moivre s and the nth Root Theorems v

6 Chapter 1. Right Triangles and an Introduction to Trigonometry, Solution Key CHAPTER 1 Right Triangles and an Introduction to Trigonometry, Solution Key Chapter Outline 1.1 THE PYTHAGOREAN THEOREM 1. SPECIAL RIGHT TRIANGLES 1. BASIC TRIGONOMETRIC FUNCTIONS 1.4 SOLVING RIGHT TRIANGLES 1.5 MEASURING ROTATION 1.6 APPLYING TRIG FUNCTIONS TO ANGLES OF ROTATION 1.7 TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 1.8 RELATING TRIGONOMETRIC FUNCTIONS 1

7 1.1. The Pythagorean Theorem The Pythagorean Theorem ? ? < 169 The triangle is obtuse ? ? > 11 The triangle is acute ? ? = 1156 This is a right triangle ? ? < 1600 The triangle is obtuse. 5. These lengths cannot make up the sides of a triangle < 9 ( 6. ) ( ) (? ) (4 6) + (6 )?(4 ) ?1 1 = 1 This is a right triangle x = x = 4 x = 75 x = 75 = ( ) = x 5 + (5 ) = x = x 100 = x 10 = x 9. Both legs are = x = x 4 = x 4 = x 11 = x 10. Plug n m,nm,n + m into the Pythagorean Theorem. (n m ) + (nm) = (n + m ) n 4 n m + m 4 + 4n m = n 4 + n m + m 4 n m + 4n m = n m 4n m = 4n m

8 Chapter 1. Right Triangles and an Introduction to Trigonometry, Solution Key 11. (a) (5, -6) and (18, ) ( ) (b), and (, 5 ) d = d = (5 18) + ( 6 ) = ( 1) + ( 9) = = 50 = 5 10 ( ( )) ( + 5 ) ( = ) ( + 6 ) = (9 ) + (6 ) = = 99 = 11

9 1.. Special Right Triangles 1. Special Right Triangles 1. Each leg is 16 = 16 = 16 = 8.. Short leg is 6 6 = = and hypotenuse is.. Short leg is 1 = 1 = 1 = 4 and hypotenuse is The hypotenuse is 4 10 = 4 0 = Each leg is 5 = The short leg is 15 and the long leg is If the diagonal of a square is 6 ft, then each side of the square is 6 or 4.4 ft. 8. These are not dimensions for a special right triangle, so to find the diagonal (both are the same length) do the Pythagorean Theorem: = d = d 500 = d 10 5 = d So, if each diagonal is 10 5, two diagonals would be ft. Pablo needs 45 ft of lights for his yard. 9. : : does not fit into either ratio, so it is not a special right triangle. To see if it is a right triangle, plug these values into the Pythagorean Theorem: ( + = ) = 1 8 < this isnot a right triangle, it is an obtuse triangle. 5 : 15 : 5 is a triangle. The long leg is 5 = 15 and the hypotenuse is 5. 4

10 Chapter 1. Right Triangles and an Introduction to Trigonometry, Solution Key 1. Basic Trigonometric Functions 1. Answers: sina = 9 15 = 5 cosa = 1 15 = 4 5 tana = 9 1 = 4 csca = 15 9 = 5 seca = 15 1 = 5 4 cota = 1 9 = 4. The hypotenuse is 17 sint = cost = 8 17 tant = 15 8 csct = sect = 17 8 cott = Answers: a. The hypotenuse is 1 b. sinx = 1 1,cosX = 5 1 c. sinz = From #, we can conclude that: ( = = ) 89 = 17. ( = = ) 169 = 1.,tanX = 1 5,cosZ = 1 1,tanZ = 5 1,cscZ = 1 5 sinx = cosz cosx = sinz tanx = cotz cotx = tanz cscx = secz secx = cscz Yes, this can be generalized for all complements.,cscx = 1 1,secX = 1 5,cotX = 5 1,secZ = 1 1,cotZ = The hypotenuse is. Each angle is 45, so the sine, cosine, and tangent are the same for both angles. sin45 = = 1 = cos45 = = 1 = tan45 = = 1 6. If the legs are length x, then the hypotenuse is x. For 45, the sine, cosine, and tangent are: sin45 = x = 1 = x cos45 = x = 1 = x tan45 = x x = 1 This tells us that regardless of the length of the sides of an isosceles right triangle, the sine, cosine and tangent of 45 are always the same. 5

11 1.. Basic Trigonometric Functions 7. If the hypotenuse is 10, then the short leg is 5 and the long leg is 5. Recall, that 0 is opposite the short side, or 5, and 60 is opposite the long side, or 5. sin0 = 5 10 = 1 cos0 = 5 10 = tan0 = 5 5 sin60 = 5 = 1 10 = cos60 = 5 10 = 1 tan60 = 5 5 = = 8. If the short leg is x, then the long leg is x and the hypotenuse is x. 0 is opposite the short side, or x, and 60 is opposite the long side, or x. sin0 = x x = 1 cos0 = x x = tan0 = x = 1 x = sin60 = x x = cos60 = x x = 1 tan60 = x x = This tells us that regardless of the length of the sides of a triangle, the sine, cosine and tangent of 0 and 60 are always the same. Also, sin0 = cos60 and cos0 = sin60.}} 9. If sina = 9 41, then the opposite side is 9x (some multiple of 9) and the hypotenuse is 41x. Therefore, working with the Pythagorean Theorem would give us the length of the other leg. Also, we could notice that this is a Pythagorean Triple and the other leg is 40x. 6

12 Chapter 1. Right Triangles and an Introduction to Trigonometry, Solution Key 1.4 Solving Right Triangles 1. A = 50 b 5.8 a 9.. Anna is correct. There is not enough information to solve the triangle. That is, there are infinitely many right triangles with hypotenuse 8. For example: = = = B 7 5. A = sin104 = A = 4 9 sin11 = About 19.9 feet tall 8. About 10. feet 9. The plane has traveled about 0 miles. The two cities are 5 miles apart. 10. About feet 11. About tanθ = opposite ad jacent tanθ = 0.65 θ = 7

13 1.5. Measuring Rotation Measuring Rotation 1. Answers: a. b. c. d. 8. Answers will vary. Examples: 450, 70. Answers: a. Answers will vary. Examples: 40, 480 b. Answers will vary. Examples: 45, 675 c. Answers will vary. Examples: 10, 510, 570 e.

14 Chapter 1. Right Triangles and an Introduction to Trigonometry, Solution Key 4. The front wheel rotates more because it has a smaller diameter. It rotates 00 revolutions versus revolutions for the back wheel, which is a 48,000 difference (( ) 60 ). 9

15 1.6. Applying Trig Functions to Angles of Rotation Applying Trig Functions to Angles of Rotation 1. The radius of the circle is 5. cosθ = 5 sinθ = 4 5 tanθ = 4 secθ = 5 cscθ = 5 4 cotθ = 4. The radius of the circle is 1. cosθ = 5 1 sinθ = 1 1 tanθ = 1 5 = 1 5 secθ = 1 5 cscθ = 1 1 cotθ = 5 1 = 5 1. If tanθ =, it must be in either Quadrant II or IV. Because cosθ > 0, we can eliminate Quadrant II. So, this means that the is negative. (All Students Take Calculus) From the Pythagorean Theorem, we find the hypotenuse: + ( ) = c = c 1 = c 1 = c Because we are in Quadrant IV, the sine is negative. So, sinθ = or (Rationalize the denominator) 4. If cscθ = 4, then sinθ = 1 4, sine is negative, so θ is in either Quadrant III or IV. Because tanθ > 0, we can eliminate Quadrant IV, therefore θ is in Quadrant III. From the Pythagorean Theorem, we can find the other leg: a + ( 1) = 4 a + 1 = 16 a = 15 a = So, cosθ = tanθ = 1 15 or 4,secθ = 4 or ,cotθ = If the terminal side of θ is on (, 6) it means θ is in Quadrant I, so sine, cosine and tangent are all positive. 10

16 Chapter 1. Right Triangles and an Introduction to Trigonometry, Solution Key From the Pythagorean Theorem, the hypotenuse is: + 6 = c = c 40 = c 40 = 10 = c 6. Therefore, sinθ = 6 = = ,cosθ = = 1 = cos70 = and tanθ = 6 = sec70 = unde f ined sin70 = 1 csc70 = 1 1 = 1 7. Answers: tan70 = unde f ined cot70 = 0 a. The triangle is equiangular because all three angles measure 60 degrees. Angle DAB measures 60 degrees because it is the sum of two 0 degree angles. b. BD has length 1 because it is one side of an equiangular, and hence equilateral, triangle. c. BC and CD each have length 1, as they are each half of BD. This is the case because Triangle ABC and ADC are congruent. d. We can use the Pythagorean theorem to show that the length of AC is. If we place angle BAC as an angle in standard position, then AC and BC correspond to the x and y( coordinates ) where the terminal side of the angle intersects the unit circle. Therefore the ordered pair is, 1. e. If we draw the angle 60 in standard position, we will also( obtain a) triangle, but the side lengths will be interchanged. So the ordered pair for 60 1 is,. 11

17 1.6. Applying Trig Functions to Angles of Rotation 8. n + n = 1 n = 1 n = 1 n = ± 1 n = ± 1 n = ± 1 = ± Because the angle is in the first quadrant, the x and y coordinates are positive. 9. An angle in the first quadrant, as the tangent is the ratio of two positive numbers. And, angle in the third quadrant, as the tangent in the ratio of two negative numbers, which will be positive. 10. The terminal side of the angle is a reflection of the terminal side of 0. From this, students should see that the ordered pair is (, 1 ). 11. Students should notice that tangent is the ratio of sin cos, which is y x, which is also slope. 1

18 Chapter 1. Right Triangles and an Introduction to Trigonometry, Solution Key 1.7 Trigonometric Functions of Any Angle 1. Answers: a. 10 b. 60 c. 0 d. 45. Answers: ( ) 1 a., ( ) b., 1 ( ) c.,. Answers: a. 1 b. 0 c. 4. Answers: a. 1 b. c. 5. Answers: a. b. -1 c. 6. Answers: a b c About degrees or about degrees. 8. This is reasonable because tan45 = 1 and the tan60 = 1.7, and the tan50 should fall between these two values. 9. Conjecture: sina + sinb sin(a + b) 10. Answer: TABLE 1.1: a (sina) (cosa)

19 1.7. Trigonometric Functions of Any Angle TABLE 1.1: (continued) a (sina) (cosa) Conjecture: (sina) + (cosa) = 1. 14

20 Chapter 1. Right Triangles and an Introduction to Trigonometry, Solution Key 1.8 Relating Trigonometric Functions 1. Answers: a. 1 4 b. 1 =. (a) TABLE 1.: Angle Sin Csc undefined (b) As the angle gets smaller and smaller, the cosecant values get larger and larger. (c) The range of the cosecant function does not have a maximum, like the sine function. The values get larger and larger. (d) Answers will vary. For example, if we looked at values near 90 degrees, we would see the cosecant values get smaller and smaller, approaching 1.. The values 90, 70, 450, etc, are excluded because they make the function undefined. 4. Answers: a. Quadrant 1; positive b. Quadrant ; negative c. Quadrant 4; negative d. Quadrant ; negative = 4 6. The ratio of sine and cosine will be positive in the third quadrant because sine and cosine are both negative in the third quadrant. 7. cosθ.9 8. cscθ = 5 15

21 1.8. Relating Trigonometric Functions 9. cos θ + sin θ = 1 cos θ + sin θ cos θ 1 + sin θ = 1 cos θ cos θ = 1 cos θ 1 + tan θ = sec θ 10. Using the Pythagorean identities results in a quadratic equation and will have two solutions. Stating that the angle lies in a particular quadrant tells you which solution is the actual value of the expression. In #7, the angle is in the first quadrant, so both sine and cosine must be positive. Chapter Summary 1. Area 1: Area : Add up 4 triangles and inner square. Set the two equal to each other: (a + b) (a + b)(a + b) a + ab + b 4 1 ab + c ab + c a + ab + b = ab + c a + b = c. First, find the diagonal of the base. This is a Pythagorean Triple, so the base diagonal is 5 (you could have also done the Pythagorean Theorem if you didn t see this). Now, do the Pythagorean Theorem with the height and the diagonal to get the three-dimensional diagonal = d = d 74 = d 74 = d C = 90.6 = 66.4 sin.6 = CA cos.6 = AT sin.6 = CA 5 cos.6 = AT CA.9 AT 16

22 Chapter 1. Right Triangles and an Introduction to Trigonometry, Solution Key 4. First do the Pythagorean Theorem to get the third side. 7 + x = x = 4 x = 75 x = 75 = 5 11 Second, use one of the inverse functions to find the two missing angles. 5. sing = 7 ( ) 18 7 sin 1 = G We can subtract G from 90 to get G.89 A = absinc = 16 sin60 = 5 = Make a right triangle with 165 as the opposite leg and w is the hypotenuse. 7. sin85 = 165 w wsin85 = 165 w = 165 sin85 w cos(90 x) = sinx sinx = 7 8. If cos( x) = 4, then cosx = 4. With tanx = 7, we can conclude that sinx = 7 4 and sin( x) = If siny = 1, then we know the opposite side and the hypotenuse. Using the Pythagorean Theorem, we get that the adjacent side is ( 1 + b = b = 9 1 b = 8 = ). Thus, cosy = ± because we don t know if the angle is in the second or third quadrant. 10. sinθ = 1, sine is positive in Quadrants I and II. So, there can be two possible answers for the cosθ. Find the third side, using the Pythagorean Theorem: 1 + b = 1 + b = 9 b = 8 b = 8 = In Quadrant I, cosθ = In Quadrant II, cosθ = 17

23 1.8. Relating Trigonometric Functions cosθ = 5 and is in Quadrant II, so from the Pythagorean Theorem : a + ( ) = 5 a + 4 = 5 a = 1 a = 1 So, sinθ = 1 5 and tanθ = 1 1. If the terminal side of θ is on (, -4) means θ is in Quadrant IV, so cosine is the only positive function. Because the two legs are lengths and 4, we know that the hypotenuse is 5., 4, 5 is a Pythagorean Triple (you can do the Pythagorean Theorem to verify). Therefore, sinθ = 5,cosθ = 4 5,tanθ = 4 1. Reference angle = 15. Possible coterminal angles = 195,

24 Chapter. Graphing Trigonometric Functions, Solution Key CHAPTER Graphing Trigonometric Functions, Solution Key Chapter Outline.1 RADIAN MEASURE. APPLICATIONS OF RADIAN MEASURE. CIRCULAR FUNCTIONS OF REAL NUMBERS.4 TRANSLATING SINE AND COSINE FUNCTIONS.5 AMPLITUDE, PERIOD AND FREQUENCY.6 GENERAL SINUSOIDAL GRAPHS.7 GRAPHING TANGENT, COTANGENT, SECANT, AND COSECANT 19

25 .1. Radian Measure Radian Measure 1. Answers: a. Answer may vary, but 10 seems reasonable. b. Based on the answer in part a., the radian measure would be π c. Again, based on part a., 4π. Answers: a. 4π b. π c. 7π 4 d. 7π 6 e. π f. π 1 g. 5π h. π 5 i. 4π j. 11π 6. Answers: 0 a. 90 b. 96 c. 10 d. 540 e. 60 f. 54 g. 75 h. 10 i j. 48

26 Chapter. Graphing Trigonometric Functions, Solution Key 5. Answers: 4. a b. 57. c d Answers: a. The correct answer is 1 b. Her calculator was is the wrong mode and she calculated the sine of 10 radians. 7. Answer: TABLE.1: x Sin(x) Cos(x) Tan(x) 5π π 6 1 π π undefined 7π 1 0 undefined 1

27 .. Applications of Radian Measure Applications of Radian Measure 1. a. i. π 1 ii. 0. radians iii.15 b. i.0. Answers may vary, anything above 15 and less than 5 is reasonable. ii. π 9 Again, answers may vary. a. π 6 b. 6 cm. a. π 16 b. Let s assume, to simplify, that the chord stretches to the center of each of the dots. We need to find the measure of the central angle of the circle that connects those two dots. Since there are 1 dots, this angle is 1π 16. The length of the chord then is: The chord is approximately.0 m, or 0 cm. = r sin θ = 1. sin( 1 1π 16 ) 4. Each section is π 6 radians. The area of one section of the stands is therefore the area of the outer sector minus the area of the inner sector: A = A outer A inner A = 1 (r outer) π 6 1 (r inner) π 6 The area of each section is approximately 76 ft. A = 1 (110) π 6 1 (55) π 6

28 Chapter. Graphing Trigonometric Functions, Solution Key a. The students have 4 sections or 950 ft b. There are general admission sections or 717 ft c. There is only one press and officials section or 76 ft 5. It is actually easier to calculate the angular velocity first. ω = π 1 = π 6, so the angular velocity is π 6 Because the linear velocity depends on the radius,each girl has her own. rad, or Lois: v = rω = π 6 = π or 1.57 m/sec Doris: v = rω = 10 π 6 = 5π or 5.4 m/sec 6. a. v = d t 10 8 = 7,000 t t = = = or seconds. b. ω = θ t = π ,81 rad/sec c. The proton rotates around once in seconds. So, in one second it will rotate around the LHC = 11, times, or just over 11,111 rotations.

29 .. Circular Functions of Real Numbers Circular Functions of Real Numbers 1. Use similar triangles: So: x 1 = 1 A Ax = 1 A = 1 x cosθ = x 1 cosθ = 1 x 1 cosθ = secθ = 1 x secθ = A 4. Using the Pythagorean theorem, tan θ + 1 = sec θ.

30 Chapter. Graphing Trigonometric Functions, Solution Key 4. b 5. d. 5

31 .4. Translating Sine and Cosine Functions Translating Sine and Cosine Functions 1. B. E. D 4. C 5. A 6. y = + sin(x π) or y = + sin(x + π) ( y = + cos x + π ) ( or y = + cos x π ) Note: this list is not exhaustive, there are other possible answers. 7. C 8. D 9. A 10. B 11. 6

32 Chapter. Graphing Trigonometric Functions, Solution Key.5 Amplitude, Period and Frequency 1.. a. y = secx: period = π, frequency = 1 b. y = cotx: period = π, frequency = c. y = cscx: period = π, frequency = 1 Because these are reciprocal functions, the periods are the same as cosine, tangent, and sine, repectively.... d a. min: -1, max: 1 b. min: -, max: c. min: -1, max: 1 d. there is no minimum or maximum, tangent has a range of all real numbers e. min: 1, max: 1 f. min: -, max: a. period: π, amplitude: 1, frequency: b. period: π, amplitude:, frequency: 1 c. period:, amplitude:, frequency: π d. period: π, amplitude:, frequency: e. period: 4π, amplitude: 1, frequency: 1 f. period: 4π, amplitude:, frequency: 1 a. period: π, amplitude:, frequency:,y = cosx b. period: 4π, amplitude:, frequency: 1,y = sin 1 x c. period:, amplitude:, frequency: π,y = cos π x d. period: π, amplitude: 1, frequency: 6,y = 1 sin6x a. 7

33 .5. Amplitude, Period and Frequency b. c. 8

34 Chapter. Graphing Trigonometric Functions, Solution Key.6 General Sinusoidal Graphs 1. This is a sine wave that has been translated 1 unit to the right and units up. The amplitude is and the frequency is. The period of the graph is π. The function reaches a maximum point of 5 and a minimum of -1.. This is a sine wave that has been translated 1 unit down and π radians to the left. The amplitude is 1 and the period is. The frequency of the graph is π. The function reaches a maximum point of 0 and a minimum of -.. This is a cosine wave that has been translated 5 units up and 10 radians to the right. The amplitude is 1 and the frequency is 40. The period of the graph is π 0. The function reaches a maximum point of 6 and a minimum of This is a cosine wave that has not been translated vertically. It has been translated 5π 4 radians to the left. The amplitude is 1 and the frequency is 1. The period of the graph is 4π. The function reaches a maximum point of 1 and a minimum of -1. The negative in front of the cosine function does not change the amplitude, it simply reflects the graph across the x axis. 5. This is a cosine wave that has been translate up units and has an amplitude of. The frequency is 1 and the period is π. There is no horizontal translation. Putting a negative in front of the x value reflects the function across the y axis. A cosine wave that has not been translated horizontally is symmetric to the y axis so this reflection will have no visible effect on the graph. The function reaches a maximum of 5 and a minimum of 1. other answers are possible given different horizontal translations of sine/cosine 6. y = + cos ( (x π ) 6 ) 7. y = + sinx or y = + cos ( x π ) 8. y = cos(6(x 0)) 9. y = + 4 cos( 1 (x + π) ) 10. y = + 7cos ( 1 (x π ) 4 ) 9

35 .7. Graphing Tangent, Cotangent, Secant, and Cosecant Graphing Tangent, Cotangent, Secant, and Cosecant 1. y = cotx. g(x) = 5csc ( 1 4 (x + π)). f (x) = 4 + tan(0.5(x π)) 0 4. y = + 1 sec(4(x 1))

36 Chapter. Graphing Trigonometric Functions, Solution Key 5. y = tanx 6. h(x) = cot ( 1 x) To make cotangent match up with tangent, it is helpful to graph the two on the same set of axis. First, cotangent needs to be flipped, which would make the amplitude of -1. Once cotangent is flipped, it also needs a phase shift of π. So, tanx = cot( x π ). 8. This is a tangent graph. From the two points we are given, we can determine the phase shift, vertical shift and frequency. There is no phase shift, the vertical shift is and the frequency is 6. y = + tan6x 9. This could be either a secant or cosecant function. We will use a cosecant model. First, the vertical shift is -1. The period is the difference between the two given x values, 7π 4 π 4 = π, so the frequency is π π =. The horizontal shift incorporates the frequency, so in y = cscx the corresponding x value to ( π ( 4,0) is π,1). 1

37 .7. Graphing Tangent, Cotangent, Secant, and Cosecant The difference between the x values is π 4 π = π 4 π 4 π 4 = π. The equation is y = 1 + csc ( (x π ) ). = π 4 and then multiply it by the frequency, Chapter Summary π 180 = 16π 18 = 8π 9 11π π = = 165. cos π 4 = cos15 = 4. For tanθ =, θ must equal 60 or 40. In radians, π or 4π. 5. There are many difference approaches to the problem. Here is one possibility: First, calculate the area of the red ring as if it went completely around the circle: A = A total A gold ( A = π 66 1 ) ( 1 π 66 1 ) A = π π 11 A = 484π 11π = 6π A in Next, calculate the area of the total sector that would form the opening of the c A = 1 r θ A = 1 ( π ) () 4 A in Then, calculate the area of the yellow sector and subtract it from the previous answer. A = 1 r θ A = 1 ( π ) (11) A 47.5 in = 14.6 in Finally, subtract this answer from the first area calculated. The area is approximately 998 in

38 Chapter. Graphing Trigonometric Functions, Solution Key 6. Answers: a. First find the circumference: π 7 = 14π. This will be the distance for the linear velocity. v = dt = 14π 9 = 16π cm/sec b. ω = θ t = π rad/sec 7. Given such a quadrilateral, and given that the two transverse angles are identified as equal (i.e., both are marked as θ in the picture), the orange segment must be parallel to the opposite (pink) radius segment, and this quadrilateral would have to be a square. This means that θ must be equal to 45 degrees, and both the tangent and cotangent of 45 degrees are equal to 1. Also, since the radii of the circle are equal to 1 unit, each of the sides of the quadrilateral (including the cotangent segment) are equal to 1 unit. Therefore, since cot θ = x y, the number of units that is the length of the cotangent segment must be equal to x y. The intersections are 8. ( ) ( ) π 4, 5π and 4,. 9. y = + 4sin5x,A = 4,B = 5, p = π 5,C = 0,D = 10. f (x) = 1 4 cos( 1 (x π ) ),A = 1 4,B = 1, p = 4π,C = π,d = g(x) = 4 + tan ( (x + π ) ),A = 1,B =, p = π π,c =,D = 4

39 .7. Graphing Tangent, Cotangent, Secant, and Cosecant 1. h(x) = 6cos(πx),A = 6,B = π,c = 0,D = 1. y = cosx 14. y = tan6x 4

40 Chapter. Trigonometric Identities and Equations, Solution Key CHAPTER Trigonometric Identities and Equations, Solution Key Chapter Outline.1 FUNDAMENTAL IDENTITIES. PROVING IDENTITIES. SOLVING TRIGONOMETRIC EQUATIONS.4 SUM AND DIFFERENCE IDENTITIES.5 DOUBLE ANGLE IDENTITIES.6 HALF-ANGLE IDENTITIES.7 PRODUCTS, SUMS, LINEAR COMBINATIONS, AND APPLICATIONS 5

41 .1. Fundamental Identities Fundamental Identities 6 1. tan70 = sin70 0, you cannot divide by zero, therefore tan70 is undefined.. If cos ( π x) = 4 5, then, by the cofunction identities, sinx = 4 5. Because sine is odd, sin( x) = If tan( x) = 5 1, then tanx = 5 1. Because sinx = 5 1 1, cosine is also negative, so cosx = Use the reciprocal and cofunction identities to simplify ( π ) secxcos x cos70 = 1 1 cosx sinx sinx cosx tanx 5. (a) Using the sides 5, 1, and 1 and in the first quadrant, it doesn t really matter which is cosine or sine. So, sin θ + cos θ = 1 becomes ( ) 5 ( ) 1 = 1. *Simplifying, we get: = 1, Finally = 1. (b) sin θ + cos θ = 1 becomes ( 1 ( ) ) + = 1. Simplifying we get: = 1 and 4 4 = To prove tan θ + 1 = sec θ, first use sinθ cosθ = tanθ and change sec θ = 1 cos θ. tan θ + 1 = sec θ sin θ cos θ + 1 = 1 cos θ sin θ cos θ + cos θ cos θ = 1 cos θ sin θ + cos θ = 1 7. If cscz = and cosz = 17, then sinz = 17 and tanz = Therefore cotz = (a) Factor sin θ cos θ using the difference of squares. sin θ cos θ = (sinθ + cosθ)(sinθ cosθ) (b) sin θ + 6 sinθ + 8 = (sinθ + 4)(sinθ + ) 9. You will need to factor and use the sin θ + cos θ = 1 identity. sin 4 θ cos 4 θ sin θ cos θ = (sin θ cos θ)(sin θ + cos θ) sin θ cos θ = sin θ + cos θ 10. To rewrite cosx secx 1 so it is only in terms of cosine, start with changing secant to cosine. cosx cosx 1 cosx 1 secx 1 = cosx 1 cosx 1 = cosx cosx 1 cosx = 1 Now, simplify the denominator.

42 Chapter. Trigonometric Identities and Equations, Solution Key cosx Multiply by the reciprocal 1 cosx = cosx 1 cosx cosx cosx = cosx 1 cosx = cos x 1 cosx cosx 11. The easiest way to prove that tangent is odd to break it down, using the Quotient Identity. tan( x) = sin( x) cos( x) = sinx cosx = tanx from this statement, we need to show that tan( x) = tanx because sin( x) = sinx and cos( x) = cosx 7

43 .. Proving Identities Proving Identities 1. Step 1: Change everything into sine and cosine Step : Give everything a common denominator, cosx. sinxtanx + cosx = secx sinx sinx cosx + cosx = 1 cosx sin x cosx + cos x cosx = 1 cosx Step : Because the denominators are all the same, we can eliminate them. sin x + cos x = 1 We know this is true because it is the Trig Pythagorean Theorem. Step 1: Pull out a cosx cosx cosxsin x = cos x cosx(1 sin x) = cos x Step : We know sin x + cos x = 1, so cos x = 1 sin x is also true, therefore cosx(cos x) = cos x. This, of course, is true, we are done!. Step 1: Change everything in to sine and cosine and find a common denominator for left hand side. sinx 1 + cosx cosx = cscx sinx sinx 1 + cosx cosx = LCD : sinx(1 + cosx) sinx sinx sin x + (1 + cosx) sinx(1 + cosx) Step : Working with the left side, FOIL and simplify. sin x cosx + cos x sinx(1 + cosx) sin x + cos x cosx sinx(1 + cosx) cosx sinx(1 + cosx) + cosx sinx(1 + cosx) (1 + cosx) sinx(1 + cosx) sinx FOIL (1 + cosx) move cos x sin x + cos x = 1 add fator out cancel (1 + cosx) 8

44 Chapter. Trigonometric Identities and Equations, Solution Key 4. Step 1: Cross-multiply sinx 1 + cosx = 1 cosx sinx sin x = (1 + cosx)(1 cosx) Step : Factor and simplify sin x = 1 cos x sin x + cos x = 1 5. Step 1: Work with left hand side, find common denominator, FOIL and simplify, using sin x + cos x = cosx cosx = + cot x 1 cosx cosx (1 + cosx)(1 cosx) 1 cos x sin x Step : Work with the right hand side, to hopefully end up with sin x. = + cot x = + cos x ( sin x ) = 1 + cos x sin x ( sin x + cos ) x = sin x ( ) 1 = sin x = sin x Both sides match up, the identity is true. 6. Step 1: Factor left hand side factor out the common denominator trig pythagorean theorem simply/multiply cos 4 b sin 4 b (cos b + sin b)(cos b sin b) cos b sin b 1 sin b 1 sin b 1 sin b Step : Substitute 1 sin b for cos b because sin x + cos x = 1. (1 sin b) sin b 1 sin b 1 sin b sin b 1 sin b 1 sin b 1 sin b 7. Step 1: Find a common denominator for the left hand side and change right side in terms of sine and cosine. siny + cosy cosy siny = secycscy siny cosy cosy(siny + cosy) siny(cosy siny) 1 = sinycosy sinycosy 9

45 .. Proving Identities Step : Work with left side, simplify and distribute. sinycosy + cos y sinycosy + sin y sinycosy cos y + sin y sinycosy 1 sinycosy 8. Step 1: Work with left side, change everything into terms of sine and cosine. (secx tanx) = 1 sinx 1 + sinx ( 1 cosx sinx ) cosx ( ) 1 sinx cosx (1 sinx) cos x Step : Substitute 1 sin x for cos x because sin x + cos x = 1 (1 sinx) 1 sin be careful, these are NOT the same! x Step : Factor the denominator and cancel out like terms. (1 sinx) (1 + sinx)(1 sinx) 1 sinx 1 + sinx 9. Plug in 5π 6 for x into the formula and simplify. sinxcosx = sinx sin 5π 6 cos 5π 6 = sin 5π ( 6 ) ( 1 ) = sin 5π This is true because sin00 is 10. Change everything into terms of sine and cosine and simplify. secxcotx = cscx 1 cosx cosx sinx = 1 sinx 1 sinx = 1 sinx 40

46 Chapter. Trigonometric Identities and Equations, Solution Key. Solving Trigonometric Equations 1. Answer: Because the problem deals with θ, the domain values must be doubled, making the domain 0 θ < 4π The reference angle is α = sin = θ = 0.645,π 0.645,π ,π θ = 0.645,.4980, 6.966, The values for θ are needed so the above values must be divided by. θ = 0.18, 1.490,.46, The results can readily be checked by graphing the function. The four results are reasonable since they are the only results indicated on the graph that satisfy sinθ = cos x = 1 16 cos x = 1 16 cosx = ± 1 4 Then cosx = 1 4 cos = x x = radians or cosx = 1 4 cos = x x = 1.85 radians. However, cosx is also positive in the fourth quadrant, so the other possible solution for cosx = 1 4 is π = radians cosx is also negative in the third quadrant, so the other possible solution for cosx = 1 4 is π 1.85 = radians tan x = 1 tanx = ± 1 tanx = ±1 41

47 .. Solving Trigonometric Equations So, tanx = 1 or tanx = 1. Therefore, x is all critical values corresponding with π 4 within the interval. x = π 4, π 4, 5π 4, 7π 4 4. Use factoring by grouping. sinx + 1 = 0 or cosx 1 = 0 sinx = 1 cosx = 1 sinx = 1 x = 7π 6, 11π 6 cosx = 1 x = π, 5π 5. You can factor this one like a quadratic. sin x sinx = 0 (sinx )(sinx + 1) = 0 sinx = 0 sinx + 1 = 0 sinx = or sinx = 1 x = sin 1 () x = π 6. For this problem the only solution is π because sine cannot be (it is not in the range). tan x = tanx tan x tanx = 0 tanx(tanx ) = 0 tanx = 0 or tanx = x = 0,π x = 1.5 4

48 Chapter. Trigonometric Identities and Equations, Solution Key 7. sin x 4 cos x 4 = 0 ( 1 cos x ) cos x 4 4 = 0 cos x 4 cos x 4 = 0 cos x 4 + cos x 4 = 0 (cos x 4 1 )(cos x 4 + ) = 0 cos x 4 1 = 0 or cos x 4 + = 0 cos x 4 = 1 cos x 4 = 8. 0π cos x 4 = 1 x 4 = π 5π or x = 4π or 0π is eliminated as a solution because it is outside of the range and cos x 4 = will not generate any solutions because is outside of the range of cosine. Therefore, the only solution is 4π. sin x = 8sinx sin x 8sinx = 0 sin x + 8sinx = 0 (sinx 1)(sinx + ) = 0 sinx 1 = 0 or sinx + = 0 sinx = 1 sinx = 1 sinx = 9. sinxtanx = tanx + secx x = 0.98 radians No solution exists x = π 0.98 =.8018 radians sinx sinx cosx = sinx cosx + 1 cosx sin x cosx = sinx + 1 cosx sin x = sinx + 1 sin x sinx 1 = 0 (sinx + 1)(sinx 1) = 0 sinx + 1 = 0 or sinx 1 = 0 sinx = 1 sinx = 1 sinx = 1 x = 7π 6, 11π 6 4

49 .. Solving Trigonometric Equations One of the solutions is not π, because tanx and secx in the original equation are undefined for this value of x. cos x + sinx = 0 (1 sin x) + sinx = 0 Pythagorean Identity sin x + sinx = 0 sin x + sinx 1 = 0 Multiply by 1 sin x sinx + 1 = 0 (sinx 1)(sinx 1) = 0 sinx 1 = 0 or sinx 1 = 0 sinx = 1 sinx = 1 sinx = 1 x = π 6, 5π 6 x = π 11. tan x + tanx = 0 1 ± 1 4(1)( ) = tanx 1 ± = tanx 1 ± = tanx tanx = or 1 tanx = 1 when x = π 4, in the interval [ π, π ] tanx = when x = rad 1. 5cos θ 6sinθ = 0 over the interval [0,π]. x = sin 1 ( ) ( or sin 1 5 ( 1 sin x ) 6sinx = 0 5sin x 6sinx + 5 = 0 5sin x + 6sinx 5 = 0 6 ± 6 4(5)( 5) (5) = sinx 6 ± = sinx 10 6 ± 16 = sinx 10 6 ± 4 = sinx 10 ± 4 = sinx 5 ) 4 5 x = rad or.598 rad from the first expression, the second expression will not yield any answers because it is out the the range of sine. 44

50 Chapter. Trigonometric Identities and Equations, Solution Key.4 Sum and Difference Identities 1. Answers: a. b. c. d. e. cos 5π ( π 1 = cos = 1 + π ) 1 1 cos 7π ( 4π 1 = cos 1 + π ) 1 = 1 ( π = cos 6 + π ) 4 = 6 4 ( π = cos + π ) 4 = 4 = cos π 6 cos π 4 sin π 6 sin π = 4 = cos π cos π 4 sin π sin π = sin45 = sin( ) = sin00 cos45 + cos00 sin45 = = = 4 tan75 = tan( ) = tan45 + tan0 1 tan45 tan0 = = + = = = = + f. sin 17π cos45 = cos( ) = cos15 cos0 sin15 sin0 = ( ) = 4 ( 9π 1 = sin = 1 + 8π 1 ) ( 1 ) + ( π = sin 4 + π ) = sin π 4 cos π + cos π 4 sin π = 6 4 =. If siny = and in Quadrant II, then by the Pythagorean Theorem cosy = 1 (1 + b = 1 ). And, if sinz = 5 and in Quadrant I, then by the Pythagorean Theorem cosz = 4 5 (a + = 5 ). cos(y z) = cosycosz + sinysinz and = = = If siny = 5 1 and in Quadrant III, then cosine is also negative. By the Pythagorean Theorem, the second leg is 1(5 + b = 1 ), so cosy = 1 1. If the sinz = 4 5 and in Quadrant II, then the cosine is also negative. 45

51 .4. Sum and Difference Identities By the Pythagorean Theorem, the second leg is (4 + b = 5 ), so cos = 5. To find sin(y + z), plug this information into the sine sum formula. 4. Answers: sin(y + z) = sinycosz + cosysinz = = = 65 a. This is the cosine difference formula, so: cos80 cos0 + sin80 0 = cos(80 0 ) = cos60 = 1 b. This is the expanded sine sum formula, so: sin5 cos5 + cos5 sin5 = sin(5 + 5 ) = sin0 = 1 5. Step 1: Expand using the cosine sum formula and change everything into sine and cosine cos(m n) = cotm + tann sinmcosn cosmcosn + sinmsinn = cosm sinmcosn sinm + sinn cosn Step : Find a common denominator for the right hand side. = cosmcosn + sinmsinn sinmcosn The two sides are the same, thus they are equal to each other and the identity is true. 6. cos(π + θ) = cosπcosθ sinπsinθ = cosθ 7. Step 1: Expand sin(a + b) and sin(a b) using the sine sum and difference formulas. sin(a + b) sin(a b) = cos b cos a (sinacosb + cosasinb)(sinacosb cosasinb) Step : FOIL and simplify. sin acos b sinacosasinbcosb + sinasinbcosacosb cos asin bsin acos b cosa sin b Step : Substitute (1 cos a) for sin a and (1 cos b) for sin b, distribute and simplify. (1 cos a)cos b cosa (1 cos b) cos b cos acos b cos a + cos acos b cos b cos a 8. tan(π + θ) = tanπ+tanθ 1 tanπtanθ = tanθ 1 = tanθ 9. sin π = sin( π 4 + π ) 4 = sin π 4 cos π 4 cos π 4 sin π 4 = = 4 4 = 0 This could also be verified by using Step 1: Expand using the cosine and sine sum formulas. cos(x + y)cosy + sin(x + y)siny = (cosxcosy sinxsiny)cosy + (sinxcosy + cosxsiny)siny Step : Distribute cosy and siny and simplify. = cosxcos y sinxsinycosy + sinxsinycosy + cosxsin y = cosxcos y + cosxsin y = cosx(cos y + sin y) } {{ } 1 = cosx 46

52 Chapter. Trigonometric Identities and Equations, Solution Key 11. Step 1: Expand left hand side using the sum and difference formulas cos(c + d) cos(c d) = 1 tanctand 1 + tanctand cosccosd sincsind cosccosd + sincsind = 1 tanctand 1 + tanctand Step : Divide each term on the left side by cosccosd and simplify cosccosd cosccosd sincsind cosccosd cosccosd cosccosd + sincsind cosccosd = 1 tanctand 1 + tanctand 1 tanctand 1 + tanctand = 1 tanctand 1 + tanctand 1. To find all the solutions, between [0,π), we need ( to expand using the sum formula and isolate the cosx. cos x + π ) = 1 ( cos x + π ) = 1 ( cos x + π ) = ± cosxcos π sinxsin π = ± cosx 0 sinx 1 = ± sinx = ± sinx = ± 1 = ± This is true when x = π 4, π 4, 5π 4, or 7π 4 1. First, solve for tan(x + π 6 ). ( tan x + π ) + 1 = 7 ( 6 tan x + π ) = 6 ( 6 tan x + π ) = ( 6 tan x + π ) = ± 6 Now, use the tangent sum formula to expand for when tan(x + π 6 ) =. tanx + tan π 6 1 tanxtan π = 6 tanx + tan π 6 = ( 1 tanxtan π ) 6 tanx + tanx + = tanx = tanx tanx = tanx = 47

53 .4. Sum and Difference Identities This is true when x = π 6 or 7π 6. If the tangent sum formula to expand for when tan(x + π 6 ) =, we get no solution as shown. tanx + tan π 6 1 tanxtan π = 6 tanx + tan π 6 = ( 1 tanxtan π ) 6 tanx + tanx + = + tanx = + tanx = Therefore, the tangent sum formula cannot be used in this case. However, since we know that tan(x + π 6 ) = when x + π 6 = 5π 11π 6 or 6, we can solve for x as follows. x + π 6 = 5π 6 x = 4π 6 x = π x + π 6 = 11π 6 x = 10π 6 x = 5π Therefore, all of the solutions are x = π 6, π, 7π 6, 5π 14. To solve, expand each side: 48 ( sin x + π ) = sinxcos π cosxsin π 6 = sinx + 1 cosx ( sin x π ) = sinxcos π 4 4 cosxsin π 4 = sinx cosx

54 Chapter. Trigonometric Identities and Equations, Solution Key Set the two sides equal to each other: sinx + 1 cosx = sinx cosx sinx + cosx = sinx cosx sinx sinx = cosx cosx ( ) ( sinx = cosx 1 ) sinx cosx = 1 tanx = = + 6 = + 6 As a decimal, this is.69677, so tan 1 (.69677) = x,x = 90.5 and

55 .5. Double Angle Identities Double Angle Identities 1. If sinx = 4 5 and in Quadrant II, then cosine and tangent are negative. Also, by the Pythagorean Theorem, the third side is (b = 5 4 ). So, cosx = 5 and tanx = 4. Using this, we can find sinx,cosx, and tan x. sinx = sinxcosx = = = 4 5. This is one of the forms for cosx. cosx = 1 sin x tanx = tanx 1 tan x ( ) 4 = 1 = ( 4 = 1 5 = 7 5 = = = 4 7 ) = Step 1: Use the cosine sum formula cos 15 sin 15 = cos(15 ) = cos0 = cosθ = 4cos θ cosθ cos(θ + θ) = cosθcosθ sinθsinθ Step : Use double angle formulas for cosθ and sinθ = (cos θ 1)cosθ (sinθcosθ)sinθ Step : Distribute and simplify. = cos θ cosθ sin θcosθ = cosθ( cos θ + sin θ + 1) = cosθ[ cos θ + (1 cos θ) + 1] Substitute 1 cos θ for sin θ = cosθ[ cos θ + cos θ + 1] = cosθ( 4cos θ + ) = 4cos θ cosθ 4. Step 1: Expand sint using the double angle formula. sint tant = tant cost sint cost tant = tant cost 50

56 Chapter. Trigonometric Identities and Equations, Solution Key Step : change tant and find a common denominator. sint cost sint cost sint cos t sint cost sint(cos t 1) cost sint cost (cos t 1) tant cost 5. If sinx = 9 41 So, 6. Step 1: Expand sin x and in Quadrant III, then cosx = and tanx = 9 40 (Pythagorean Theorem, b = 41 ( 9) ). cosx = cos x 1 ( sinx = sinxcosx = 40 ) 1 tanx = sinx 41 cosx = = Step : Separate and solve each for x. = = sinx + sinx = 0 sinxcosx + sinx = 0 sinx(cosx + 1) = 0 cosx + 1 = 0 = = sinx = 0 cosx = 1 x = 0,π or x = π, 4π 7. Expand cosx and simplify cos x cosx = 0 cos x (cos x 1) = 0 cos x + 1 = 0 cos x = 1 cosx = ±1 cosx = 1 when x = 0, and cosx = 1 when x = π. Therefore, the solutions are x = 0,π. 8. a..49 b c

57 .5. Double Angle Identities 9. a. cscx x = sinx cscx x = sinxcosx 1 cscx x = ( sinxcosx )( ) sinx 1 cscx x = sinx sinxcosx cscx x = sinx sin xcosx cscx x = 1 sin x sinx cosx cscx x = csc xtanx b. cos 4 θ sin 4 θ = (cos θ + sin θ)(cos θ sin θ) cos 4 sin 4 θ = 1(cos θ sin θ) cosθ = cos θ sin θ cos 4 θ sin 4 θ = cosθ c. 10. cosx 1 = sin x sinx 1 + cosx = sinxcosx 1 + (1 sin x) sinx 1 + cosx = sinxcosx sin x sinx 1 + cosx = sinxcosx (1 sin x) sinx 1 + cosx = sinxcosx cos x sinx 1 + cosx = sinx cosx sinx 1 + cosx = tanx (1 sin x) 1 = sin x sin x = sin x 0 = sin x 0 = sin x 0 = sinx x = 0,π 5

58 Chapter. Trigonometric Identities and Equations, Solution Key 11. cosx = cosx cos x 1 = cosx cos x cosx 1 = 0 (cosx + 1)(cosx 1) = 0 cosx + 1 = 0 or cosx 1 = 0 cosx = 1 cosx = 1 cosx = 1 1. cosx = 1 when x = 0 and cosx = 1 when x = π. 1. sinx cosx = 1 cscxtanx = sec x sinx sinx cosx = 1 cos x sinxcosx sinx cosx = 1 cos x 1 cos x = 1 cos x sinxcosx (1 sin x) = 1 sinxcosx 1 + sin x = 1 sinxcosx + sin x = sinxcosx + sin x = 1 sinxcosx = 1 sin x sinxcosx = cos x ( ) ± 1 cos x cosx = cos x ( 1 cos x ) cos x = cos 4 x cos x cos 4 x = cos 4 x cos x cos 4 x = 0 cos x(1 cos x) = 0 1 cos x = 0 cos x = 0 cos x = 1 cosx = 0 or cos x = 1 x = π, π cosx = ± x = π 4, 5π 4 Note: If we go back to the equation sinxcosx = cos x, we can see that sinxcosx must be positive or zero, since cos x is always positive or zero. For this reason, sinx and cosx must have the same sign (or one of them 5

59 .5. Double Angle Identities must be zero), which means that x cannot be in the second or fourth quadrants. This is why π 4 and 7π 4 valid solutions. 14. Use the double angle identity for cosx. are not sin x = cosx sin x = cosx sin x = 1 sin x sin x = sin x = 1 sinx = ±1 x = π, π 54

60 Chapter. Trigonometric Identities and Equations, Solution Key.6 Half-Angle Identities 1. Answers: a. b. c. cos11.5 = cos cos5 = 1 = = = = sin105 = sin 10 1 cos10 = 1 = = = 4 = 4 d. tan π 8 = tan 1 π 4 = 1 cos π 4 sin π 4 e. sin67.5 = sin 15 = tan 7π 8 = tan 1 7π 4 = 1 cos 7π 4 sin 7π 4 = 1 = 1 f. tan165 = tan 0 = 1 cos0 = = 1 cos15 = sin0 = 1 1 = = + = + 1 = = = = = = 1 ( = + = 4 ) = + +. If sinθ = 7 5, then by the Pythagorean Theorem the third side is 4. Because θ is in the second quadrant, cosθ = 4 5. sin θ 1 cosθ = = 49 = 50 = 7 5 = 7 10 cos θ 1 + cosθ = = 1 = 50 = 1 = 5 10 = tan θ 1 cosθ = 1 + cosθ = = 49 = 7 55

61 .6. Half-Angle Identities Step 1: Change right side into sine and cosine. tan b = secb secbcscb + cscb = 1 cosb = 1 cosb 1 cscb(secb + 1) ( ) 1 sinb cosb + 1 ( ) 1 + cosb = 1 cosb 1 sinb = 1 cosb 1 + cosb sinbcosb = 1 cosb sinbcosb 1 + cosb = sinb 1 + cosb Step : At the last step above, we have simplified the right side as much as possible, now we simplify the left side, using the half angle formula. 1 cosb 1 + cosb = sinb 1 + cosb 1 cosb 1 + cosb = sin b (1 + cosb) cosb (1 cosb)(1 + cosb) = sin b(1 + cosb) (1 cosb)(1 + cosb) = sin b 1 cos b = sin b 4. Step 1: change cotangent to cosine over sine, then cross-multiply. cot c = sinc 1 cosc = cos c sin c = 1 + cosc 1 cosc = sinc 1 cosc 1 + cosc 1 cosc = sin c (1 cosc) 1 + cosc 1 cosc (1 + cosc)(1 cosc) = sin c(1 cosc) (1 + cosc)(1 cosc) = sin c 1 cos c = sin c 5. sinxtan x ( ) 1 cosx + cosx = sinx + cosx sinx sinxtan x + cosx = 1 cosx + cosx sinxtan x + cosx = 1 + cosx sinxtan x + cosx = cos x 56

62 Chapter. Trigonometric Identities and Equations, Solution Key 6. First, we need to find the third side. Using the Pythagorean Theorem, we find that the final side is ( 105 b = 1 ( 8) ). Using this information, we find that cosu = ± Plugging this into the half angle formula, we get: cos u = 1 ± = = ± ± To solve cos x = 1, first we need to isolate cosine, then use the half angle formula. cos x = 1 cos x = cosx = cosx = 1 cosx = 0 cosx = 0 when x = π, π 8. To solve tan a = 4, first isolate tangent, then use the half angle formula. tan a = 4 1 cosa 1 + cosa = 4 1 cosa 1 + cosa = cosa = 1 cosa 17cosa = 15 cosa = Using your graphing calculator, cosa = when a = 15,08 57

63 .6. Half-Angle Identities 9. ( ± cos x = 1 + cosx 1 + cosx ± = 1 + cos x Half angle identity ) 1 + cosx = (1 + cosx) square both sides 1 + cosx = 1 + cosx + cos x ( ) 1 + cosx = (1 + cosx + cos x) 1 + cosx = + 4cosx + cos x cos x + cosx + 1 = 0 (cosx + 1)(cosx + 1) = 0 Then cosx + 1 = 0 cosx = 1 x = π, 4π Or cosx + 1 = 0 cosx = 1 x = π sinx cosx = 1 cosx sinx This is the two formulas for tan x. Cross-multiply. sinx 1 + cosx = 1 cosx sinx (1 cosx)(1 + cosx) = sin x 1 + cosx cosx cos x = sin x 1 cos x = sin x 1 = sin x + cos x 58

64 Chapter. Trigonometric Identities and Equations, Solution Key.7 Products, Sums, Linear Combinations, and Applications 1. Using the sum-to-product formula:. Using the difference-to-product formula:. Using the difference-to-product formulas: 4. Using the product-to-sum formula: 5. (a) If 5cosx 5sinx, then A = 5 and B = 5. sin9x + sin5x ( ( ) ( )) 1 9x + 5x 9x 5x sin cos 1 sin7xcosx cos4y cosy ( ) ( ) 4y + y 4y y sin sin sin 7y sin y cosa cos5a sina sin5a = tan4a sin ( ) ( a+5a sin a 5a ) sin ( ) ( a 5a cos a+5a ) sin4a cos4a tan4a sin6θsin4θ 1 (cos(6θ 4θ) cos(6θ + 4θ)) 1 (cosθ cos10θ) By the Pythagorean Theorem, C = 5 and cosd = 5 = 1 = So, because B is negative, D is in Quadrant IV. Therefore, D = 7π 4. Our final answer is 5 cos ( x 7π ) 4. (b) If 15cosx 8sinx, then A = 15 and B = 8. By the Pythagorean Theorem, C = 17. Because A and B are both negative, D is in Quadrant III, which means D = cos 1 ( 15 17) = 0.49+π =.6 rad. Our final answer is 17 cos (x.6)

65 .7. Products, Sums, Linear Combinations, and Applications 6. Using the sum-to-product formula: sin11x sin5x = 0 11x 5x 11x + 5x sin cos = 0 sinxcos8x = 0 sinxcos8x = 0 sinx = 0 or cos8x = 0 x = 0,π,π,π,4π,5π x = 0, π, π,π, 4π, 5π 8x = π, π, 5π, 7π, 9π, 11π, 1π, 15π x = π 16, π 16, 5π 16, 7π 16, 9π 16, 11π 16, 1π 16, 15π Using the sum-to-product formula: cos4x + cosx = 0 4x + x 4x x cos cos = 0 cosxcosx = 0 cosxcosx = 0 So, either cosx = 0 or cosx = 0 x = π, π, 5π, 7π, 9π, 11π x = π 6, π, 5π 6, 7π 6, π, 11π 6 8. Move sinx over to the other side and use the sum-to-product formula: So sinx = 0 ( 5x + x cos ) sin x = 0,π or cos4x = 1 sin5x + sinx = sinx sin5x sinx + sinx = 0 ( ) 5x x + sinx = 0 cos4xsinx + sinx = 0 sinx(cos4x + 1) = 0 cos4x = 1 4x = π, 4π, 8π, 10π, 14π, 16π, 0π, π = π 6, π, π, 5π 6, 7π 6, 4π, 5π, 11π 6 x = 0, = π 6, π, π, 5π 6,π, 7π 6, 4π, 5π, 11π 6 60

66 Chapter. Trigonometric Identities and Equations, Solution Key 9. Using the sum-to-product formula: 10. Derive a formula for tan4x. 11. Let y = 0. f (t) = sin(00t + π) + sin(00t π) ( ) ( ) (00t + π) + (00t π) (00t + π) (00t π) = sin cos ( ) ( ) 400t π = sin cos = sin00t cosπ = sin00t( 1) = sin00t tan4x = tan(x + x) tanx + tanx = 1 tanxtanx = tanx 1 tan x = tanx 1 tan x ( ) 1 tanx 1 tan x.50sint + 1.0sint = 0 = 4tanx 1 tan x (1 tan x) 4tan x (1 tan x) = 4tanx 1 tan x 1 tan x + tan 4 x 4tan x (1 tan x) = 4tanx 1 tan x (1 tan x) 1 6tan x + tan 4 x = 4tanx 4tan x 1 6tan x + tan 4 x.50 sint +.40 sint cost = 0, Double-Angle Identity sint( cost) = 0 sint = 0 or cost = 0.40cost =.50 cost = 1.46 no solution because 1 cost 1. t = 0,π Chapter Summary 1. If the terminal side is on ( 8,15), then the hypotenuse of this triangle would be 17 (by the Pythagorean Theorem, c = ( 8) + 15 ). Therefore, sinx = 15 17,cosx = , and tanx = 8.. If sina = 5 and tana < 0, then a is in Quadrant II. Therefore seca is negative. To find the third side, we 61

67 .7. Products, Sums, Linear Combinations, and Applications need to do the Pythagorean Theorem. ( 5 ) + b = 5 + b = 9 b = 4 b = So seca =.. Factor top, cancel like terms, and use the Pythagorean Theorem Identity. Note that this simplification doesn t hold true for values of x that are π 4 + nπ, where n is a positive integer since the original expression is undefined for these values of x. cos 4 x sin 4 x cos x sin x (cos x + sin x)(cos x sin x) cos x sin x cos x + sin x Change secant and cosecant into terms of sine and cosine, then find a common denominator sinx = secx(cscx + 1) cosxsinx = 1 ( ) 1 cosx sinx + 1 = 1 ( ) 1 + sinx cosx sinx = 1 + sinx cosxsinx ( sec x + π ) + = 0 ( sec x + π ) = ( cos x + π ) = 1 x + π = π, 4π x = π π, 4π π x = π 6, 5π 6 6

68 Chapter. Trigonometric Identities and Equations, Solution Key 6. ( x 8sin 8 = 0 ) 8sin x = 8 sin x = 1 x = x x = π 7. sin x + sinx = 0 sin x + sinxcosx = 0 sinx(sinx + cosx) = 0 So, sinx = 0 or sinx + cosx = 0 sinx = 0 sinx + cosx = 0 sinx = 0 sinx = cosx x = 0,π x = π 4, 7π 4 8. tan x = 1 9. tan x = 1 tanx = ± x = π 6, 5π 6, 7π 6, 11π 6, 1π 6, 17π 6, 19π 6, π 6 x = π 1, 5π 1, 7π 1, 11π 1, 1π 1, 17π 1, 19π 1, π 1 sin 1 ( 1+ 1 sinx = sinx 1 = sinx + sinx ( 1 = sinx 1 + ) = sinx ) 1 = x or x =.747 radians and x =.7669 radians 10. Because this is cosx, you will need to divide by at the very end to get the final answer. This is why we 6

69 .7. Products, Sums, Linear Combinations, and Applications went beyond the limit of π when finding x. cosx 1 = 0 cosx = 1 cosx = 1 ( ) 1 x = cos 1 = π, 5π, 7π, 11π, 1π, 17π x = π 9, 5π 9, 7π 9, 11π 9, 1π 9, 17π Rewrite the equation in terms of tan by using the Pythagorean identity, 1 + tan θ = sec θ. 64 sec x tan 4 x = (1 + tan x) tan 4 x = + tan x tan 4 x = tan 4 x tan x + 1 = 0 (tan x 1)(tan x 1) = 0 Because these factors are the same, we only need to solve one for x. Where k is any integer. 1. Use the half angle formula with Use the sine sum formula. tan x 1 = 0 tan x = 1 tanx = ±1 x = π 4 + πk and π 4 + πk cos157.5 = cos cos15 = 1 + = = = sin 1π ( 10π 1 = sin 1 + π ) 1 ( 5π = sin 6 + π ) 4 = sin 5π 6 cos π 4 + cos 5π 6 sin π 4 = 1 + ( ) 6 = 4

70 Chapter. Trigonometric Identities and Equations, Solution Key 14. [ ( ) ( )] 5x + 9x 5x 9x 4(cos5x + cos9x) = 4 cos cos = 8cos7xcos( x) = 8cos7xcosx 15. cos(x y)cosy sin(x y)siny cosy(cosxcosy + sinxsiny) siny (sinxcosy cosxsiny) cosxcos y + sinxsinycosy sinxsinycosy + cosxsin y cosxcos y + cosxsin y cosx(cos y + sin y) cosx 16. Use the sine and cosine sum formulas. ( ) ( 4π sin x + cos x + 5π ) 6 sin 4π cosx cos 4π sinx + cosxcos 5π 6 sinxsin 5π 6 cosx + 1 sinx cosx 1 sinx cosx 17. Use the sine sum formula as well as the double angle formula. sin6x = sin(4x + x) = sin4xcosx + cos4xsinx = sin(x + x)cosx + cos(x + x)sinx = cosx (sinxcosx + cosxsinx) + sinx(cosxcosx sinx sinx) = sinxcos x + sinxcos x sin x = sinx cos x sin x = sinx(cos x sin x) = sinxcosx[(cos x sin x) (sinxcosx) = sinxcosx[(cos 4 x sin xcos x + sin 4 x) 4 sin xcos x] = sinxcosx[cos 4 x 6sin xcos x + sin 4 x 4 sin xcos x] = sinxcosx[cos 4 x + sin 4 x 10sin xcos x] = 6sinxcos 5 x + 6sin 5 xcosx 0sin xcos x 18. Using our new formula, cos 4 x = [ 1 (cosx + 1)] Now, our final answer needs to be in the first power of cosine, so we need to find a formula for cos x. For this, we substitute x everywhere there is an x and the formula translates to cos x = 1 (cos4x + 1). Now we can write cos4 x in terms of the first power of cosine as 65

71 .7. Products, Sums, Linear Combinations, and Applications follows. cos 4 x = [ 1 (cosx + 1)] = 1 4 (cos x + cosx + 1) = 1 4 (1 (cos4x + 1) + cosx + 1) = 1 8 cos4x cosx = 1 8 cos4x + 1 cosx Using our new formula, sin 4 x = [ 1 (1 cosx)] Now, our final answer needs to be in the first power of cosine, so we need to find a formula for cos x. For this, we substitute x everywhere there is an x and the formula translates to cos x = 1 (cos4x + 1). Now we can write sin4 x in terms of the first power of cosine as follows. sin 4 x = [ 1 (1 cosx)] = 1 4 (1 cosx + cos x) = 1 4 (1 cosx + 1 (cos4x + 1)) = cosx cos4x = 1 8 cos4x 1 cosx Answers: (a) First, we use both of our new formulas, then simplify: sin xcos x = 1 (1 cosx)1 (cos4x + 1) ( 1 = 1 )( 1 cosx cos4x + 1 ) = 1 4 cos4x cosxcos4x 1 4 cosx = 1 (1 cosx + cos4x cosxcos4x) 4 (b) For tangent, we use the identity tanx = sinx cosx and then substitute in our new formulas. tan4 x = sin4 x Now, use the formulas we derived in #18 and #19. tan 4 x = sin4 x cos 4 x = = 1 8 cos8x 1 cos4x cos8x + 1 cos4x + 8 cos8x 4cos4x + cos8x + 4cos4x + cos 4 x 66

72 Chapter 4. Inverse Trigonometric Functions, Solution Key CHAPTER 4 Inverse Trigonometric Functions, Solution Key Chapter Outline 4.1 BASIC INVERSE TRIGONOMETRIC FUNCTIONS 4. GRAPHING INVERSE TRIGONOMETRIC FUNCTIONS 4. INVERSE TRIGONOMETRIC PROPERTIES 4.4 APPLICATIONS & MODELS 67

73 4.1. Basic Inverse Trigonometric Functions Basic Inverse Trigonometric Functions 1. Answers: a. 1 b. c.. Answers: π a., π π b., 5π 11π c. 6, 7π 6. cosθ = cos 1 17 = sinθ = sin 1 6 = This problem uses tangent inverse. tanx = 14 x = tan 1 14 = (value graphing calculator will produce). *However, this is the reference angle. Our angle is in the third quadrant because both the x and y values are negative. The angle is = cosa = 4 9 cos = A A = f (x) = x 5 y = x 5 x = y 5 x + 5 = y x + 5 = y x + 5 = y 68

74 Chapter 4. Inverse Trigonometric Functions, Solution Key 8. y = 1 tan 1 ( 4 x 5 ) x = 1 tan 1 ( 4 y 5 ) x = tan 1 ( 4 y 5 ) tan(x) = 4 y 5 9. tan(x) + 5 = 4 y 4(tan(x) + 5) = y sin 1 ( x sin 1 ( x 4 g(x) = sin(x 1) + 4 y = sin(x 1) + 4 x = sin(y 1) + 4 x 4 = sin(y 1) x 4 = sin(y 1) ) = y 1 ) = y 10. h(x) = 5 cos 1 (x + ) y = 5 cos 1 (x + ) x = 5 cos 1 (y + ) x 5 = cos 1 (y + ) 5 x = cos 1 (y + ) cos(5 x) = y + cos(5 x) = y cos(5 x) = y 69

75 4.. Graphing Inverse Trigonometric Functions 4. Graphing Inverse Trigonometric Functions 1. The graph represents a one-to-one function. It passes both a vertical and a horizontal line test. The inverse would be a function.. The graph represents a function, but is not one-to-one because it does not pass the horizontal line test. Therefore, it does not have an inverse that is a function.. The graph does not represent a one-to-one function. It fails a vertical line test. However, its inverse would be a function. 4. By selecting 4-5 points and switching the x and y values, you will get the red graph below. 5. By selecting 4-5 points and switching the x and y values, you will get the red graph below y = 1 cos 1 (x+1) is in blue and y = cos 1 (x) is in red. Notice that y = 1 cos 1 (x+1) has half the amplitude and is shifted over -1. The seems to narrow the graph.

76 Chapter 4. Inverse Trigonometric Functions, Solution Key 7. y = tan 1 (x ) is in blue and y = tan 1 x is in red. y = tan 1 (x ) is shifted up and to the right (as indicated by point C, the center ) and is flipped because of the tan

77 4.. Graphing Inverse Trigonometric Functions ( y = cos x π ) ( x = cos y π ) cos 1 x = y π π + cos 1 x = y sin 1 x π + cos 1 x, graphing the two equations will illustrate that the two are not the same. This is because of the restricted domain on the inverses. Since the functions are periodic, there is a phase shift of cosine that, when the inverse is found, is equal to sine inverse. 7

78 Chapter 4. Inverse Trigonometric Functions, Solution Key 4. Inverse Trigonometric Properties 1. Answers: a. π 6 b. π 4 c. π 4 d. π e. π 4 f. π 4 <. Answers: a..747 b c Answers: ( ) a. csc cos 1 = csc π 6 = b. sec 1 (tan(cot 1 1)) = sec 1 ( tan π ) 4 = sec 1 1 = 0 c. tan 1 ( cos π ) ( = tan 1 0 = 0 ) ( ) d. cot sec 1 = cot cos 1 = cot π 6 = 1 tan π 6 = 1 = 4. y = sec 1 x when plugged into your graphing calculator is y = cos 1 1 x. The domain is all reals, excluding the interval (-1, 1). The range is all reals in the interval [0,π],y π. There are no y intercepts and the only x intercept is at y = csc 1 x when plugged into your graphing calculator is y = sin 1 1 x. 7

79 4.. Inverse Trigonometric Properties The domain is all reals, excluding the interval (-1, 1). The range is all reals in the interval [ π, π ],y 0. There are no x or y intercepts. 6. The domain is all real numbers and the range is from (0,π). There is an x intercept at π. 7. Answers: a. cosθ = 5 ( ( )) 1 5 sin cos 1 = sinθ 1 74 sinθ = 1 1 b. tan ( sin 1 ( 11)) 6 sinθ = The third side is b = 11 6 = 85. tanθ = 6 = c. cos ( csc 1 ( )) 5 7 cscθ = 5 7 sinθ = 7 5. This two lengths of a Pythagorean Triple, with the third side being 4. cosθ = Answers: a. 1 x +1 b. 1 x 9. The adjacent side to θ is 1 x, so the three trig functions are:

80 Chapter 4. Inverse Trigonometric Functions, Solution Key a. b. sin(sin 1 x) = sinθ = x cos(sin 1 x) = cosθ = 1 x tan(sin 1 x) = tanθ = x 1 x tan(sin 1 (x )) = x 1 (x ) = x 1 4x The opposite side to θ is a. b. 1 x, so the three trig functions are: sin(cos 1 x) = sinθ = 1 x cos(cos 1 x) = cosθ = x 1 x tan(cos 1 x) = tanθ = x ( ( )) ( ) 1 sin cos 1 x = 1 1 x = x 75

81 4.4. Applications & Models Applications & Models 1. I = I 0 sinθcosθ I = I 0 sinθcosθ I 0 I 0 I = sinθcosθ I 0 I = sinθcosθ I 0 I = sin4θ I 0 1 I sin = 4θ I 0 1 I sin 1 = θ 4 I 0. The volume is 10 feet times the area of the end. The end consists of two congruent right triangles and one rectangle. The area of each right triangle is 1 (sinθ)(cosθ) and that of the rectangle is (1)(cosθ). This means that the volume can be determined by the function V (θ) = 10(cosθ + sinθcosθ), and this function can be graphed as follows to find the maximum volume and the angle θ where it occurs. Therefore, the maximum volume is approximately 1 cubic feet and occurs when θ is about 0.. See the figure below. cosx = 7 0 x = 7 cos 1 0 x =

82 Chapter 4. Inverse Trigonometric Functions, Solution Key 4. i = I m [sin(wt + α)cosϕ + cos(wt + α)sinϕ] i = sin(wt + α)cosϕ + cos(wt + α)sinϕ I m } {{ } sin(wt+α+ϕ) i = sin(wt + α + ϕ) I m sin 1 i = wt + α + ϕ I m sin 1 i α ϕ = wt I ( m 1 sin 1 i ) α ϕ = t w I m 5. Answers: a. 64 on the 16 th of November = cos [ (0 + 10) 60 ] 65 = 6.64 b. 15 on the 8 th of August = cos [ (0 + 10) 60 ] 65 = We need to find y and z before we can find x. sin70 = y 40 y = 40sin70 = 7.59 cos70 = z 40 z = 40cos70 = 1.68 Using as the adjacent side for x, we can now find the missing angle. tanx = = 8.48 x = tan 1 (8.48) = Therefore, the bearing from the ship back to the point of departure is W87.99 S. 7. The maximum displacement for this equation is simply the amplitude, 4. 77

83 4.4. Applications & Models 8. You can use the same picture from Example 5 for this problem. tan18 = x 10, 000 x = 10,000tan18 tan(18 + y) = x = , 000 tan(18 + y) = , y = tan 10, y = 0.55 y =.55 So, the towers are.6 apart. Chapter Summary 1. Answers: 78 a. π 6 π b. 6 c. π π d. 4 e. 0 π f. 4. Answers: a b c d e f Answers: a. b. 1 c. 5 d. 1 5 e. or π f. 4. Answers: 1

84 Chapter 4. Inverse Trigonometric Functions, Solution Key a. b. 5. Answers: f (x) = 5 + cos(x 1) y = 5 + cos(x 1) x = 5 + cos(y 1) x 5 = cos(y 1) cos 1 (x 5) = y cos 1 (x 5) = y 1 + cos 1 (x 5) sin ( sin ( x 4 = y g(x) = 4sin 1 (x + ) y = 4sin 1 (x + ) x = 4sin 1 (y + ) x 4 = sin 1 (y + ) ) = y + x ) 4 = y a. b. 79

85 4.4. Applications & Models c. d. 6. Answers: a. sin(cos 1 x ) = 1 (x ) = 1 x 6 ) x 1 x = ( ) 1 x ( c. cos 4 (arctan(x) ) = cos 4 (tan 1 4x ) = b. tan ( sin = ( x4 9 ) 1 x 4 = ( x x4 9 ) = x4 9 x 4 7. x can help us find our final answer, but we need to find y and z first = (4x ) = + 1) 16x (16x 4 +1) 80

86 Chapter 4. Inverse Trigonometric Functions, Solution Key sin40 = y 65 y = 65sin40 = cos40 = 0 + z 0 + z = 65cos z = z = 9.79 tanx = x = tan x = Now, the bearing from the ship to the point of departure is north, and then ( ) west, which is written as N5.49 W on the 1 th of May = cos [ (1 + 10) 60 ] 65 = sin(x ± y) = sinxcosy ± cosxsiny,a = sinx and b = siny x = sin 1 a and y = sin 1 b sin(x ± y) = a 1 sin y ± b 1 sin x sin(x ± y) = a 1 b ± b 1 a ( ) x ± y = sin 1 a 1 b ± b 1 a ( ) sin 1 a ± sin 1 b = sin 1 a 1 b ± b 1 a 10. cos(x ± y) = cosxcosy sinxsiny,a = cosx and b = cosy x = cos 1 a and y = cos 1 b cos(x ± y) = ab (1 cos x)(1 cos y) cos(x ± y) = ab (1 a )(1 b ) ) x ± y = cos (ab 1 (1 a )(1 b ) ) cos 1 a ± cos 1 b = cos (ab 1 (1 a )(1 b ) 81

87 CHAPTER 5 Triangles and Vectors, Solution Key Chapter Outline 5.1 THE LAW OF COSINES 5. AREA OF A TRIANGLE 5. THE LAW OF SINES 5.4 THE AMBIGUOUS CASE 5.5 GENERAL SOLUTIONS OF TRIANGLES 5.6 VECTORS 5.7 COMPONENT VECTORS 8

88 Chapter 5. Triangles and Vectors, Solution Key 5.1 The Law of Cosines 1. Answers: a. side a b. T, R, and I c. side l d. R and D e. side b f. C, D, M. Answers: a. a = cos50,a 8.5 b. 11 = cosi, I c. l = cos79.5,l.8 d. 1.8 = cosd, D 41.8 e. b = cos67.,b 45.5 f. 11 = cosd, D Answer: 4. Answer: 6 = cosc 5 = cosi = 41.4 C 8.5, I 55.1, R 41.4 First, find AB. AB = cos7.4 AB = 9.4 sin. = AD 9.4 AD = Answer: HJI = = 8.7 (these two angles are a linear pair). 6.7 = HJ HJ 1.9 cos8.7. This simplifies to the quadratic equation HJ 0.417HJ Using the quadratic formula, we can determine that HJ So, since HJ + JK = HK,HK Answer: To determine this, use the Law of Cosines and solve for d to determine if the picture is accurate. d = cos0,d = 14.9, which means d in the picture is off by Answers: 8. Answers: (a) First, find x: x = cos47,x =.187 miles. Dividing the miles by his speed will tell us how long he will have service = 0.5 hr or 0.9 min. (b) = 0.66 hr or 9.7 min, so he will have service for 8.8 minutes longer. a = cosa. The angle formed, a, is 7. b. 07 = cosb. The new angle, b, will need to be 4.8 rather than 7 or. less. 8

89 5.1. The Law of Cosines 9. x = cos9, making the ball 10.6 yards away from the flag. 10. Students answers will vary. The goal is to have each student create their own word problem. 11. Draw a figure as shown below. We need to find the height in order to get the area. 1. Draw a figure as shown below. 7 = cosx,x = 9.6 sin9.6 = h 15 h = 7.4 A = = 144. Recall that the area of a trapezoid is A = 1 h(b 1 + b ). We need to find the angle x, in order to find y and then h. 100 = cosx,x = = 5.9 = y.cos5.9 = h 00 h = A = ( ) = 4,811,670 sq. ft. or acres. 84

90 Chapter 5. Triangles and Vectors, Solution Key 5. Area of a Triangle 1. Answers: a. A = 1 bh b. Heron s formula c. K = 1 bcsina d. A = 1 bh. Answers: a. A = b. A = 14. c. A = 514. d. A = Answers: a. A = 1 bh b. K = 1 bcsina c. A = 1 bh 4. Answers: a. h = 89. b. A = 67 c. Area of ABC = Answers: (a) Use Heron s Formula, then multiply your answer by 4, for the 4 sides. s = 1 ( ) = 670 A = 670(670 75)(670 75)( ) = 68, ,97.4 multiplied by 4 = 7,189.8 total square feet. (b) 7, ,98 gallons of paint are needed. 6. Using Heron s Formula, s and calculate the area: s = 1 ( ) = A = 19.55( )( )( ) = sq. ft. He will need bundles ( = 1.8). The shingles will cost him $15.45 = $ square f eet will go to waste ( = 6.9). 7. Answers: a. Use K = 1 bcsina,k = 1 (186)(05)sin148. So, the area that needs to be replaced is square yards. b. K = 1 (186)(88)sin148 = , the area has increased by yards. 8. You need to use the K = 1 bcsina formula to find DE and GF = 1 (1.6)DE sin9 DE = = 1 (1.6)EF sin60 EF = 14.4 Second, you need to find sides DG and GF using the Law of Cosines. DG = cos9 DG = 8.95 GF = cos60 GF =

91 5.. Area of a Triangle The perimeter of the quadrilateral is Answer: 10. Answer: = 7.4. First, find BD by using the Pythagorean Theorem. BD = Then, using the area and formula (A = 1 bh), you can find AC..96 = 1 (7.4)AC AC = DC = = d = e + f e f cosd e = d + f d f cose f = d + e decosf All three versions of the Law of Cosines Add the three formulas together, we get: d + e + f = e + f e f cosd + d + f d f cose + d + e decosf d + e + f = (d + e + f ) (e f cosd + d f cose + decosf) (d + e + f ) = (e f cosd + d f cose + decosf) d + e + f = (e f cosd + d f cose + decosf) 86

92 Chapter 5. Triangles and Vectors, Solution Key 5. The Law of Sines 1. Answers: a. ASA b. AAS c. neither d. ASA e. AAS f. AAS. Student answers will vary but they should notice that in both cases you know or can find an angle and the side across from it.. Answers: sin11.7 a. a = sin ,a = 5.6 sin41. b = sin9.7 d,d = 08.0 c. not enough information sin40. d. l = sin1.5 6.,l = 4.9 sin9 e. o = sin1 15,o = 4.6 sin17 f. q = sin1.8.6,q = G = = Answers: sin96.6 g = sin1.,g = sina = sinb a b a(sin B) = b(sin A) a b = sina sinb sin6.1 h = sin1.,h = Law of Sines Cross multiply a. tan54 = h 7.15 h = 9.8,cos67 = 9.8 x x = 5. b. The angle we are finding is the one at the far left side of the triangle. Divide by b(sinb) 8.9 = cosA A = First we need to find the other two sides in the triangle. sin64 18 = sin11 x = sin105 y sin4.4 x = sin1 11. x = 14.9.,x = 46.,y = 4., where y is the length of the original fight plan. The modified flight plan is = 64.. Dividing both by 495 mi/hr, we get min (modified) and 8.4 min (original). Therefore, the modified flight plan is.6 minutes longer. 8. First, we need to find the distance between Stop B (B) and Stop C (C). 87

93 5.. The Law of Sines sin6 1. = sin41 B = sin10 C B = 1.7,C = 0.4. The total length of her route is = 48.6 miles. Dividing this by 45 mi/hr, we get that it will take her 1.08 hours, or 64.8 minutes, of actual driving time. In addition to the driving time, it will take her 6 minutes (three stops at minutes per stop) to deliver the three packages, for a total roundtrip time of 70.8 minutes. Subtracting this 70.8 minutes from 10:00 am, she will need to leave by 8:49 am. 88

94 Chapter 5. Triangles and Vectors, Solution Key 5.4 The Ambiguous Case TABLE 5.1: a >, =, or < b sin A Diagram Number of solutions a. 6 > 19.9 b. 16 < c = d..4 >.. Answers:. sin.5 a. 6 = sinb 7 B = 49.9 or = 10.1 b. no solution sin47.8 c = sinb 18. B = 90 d. sin = sinb 4. B = 75. or = sina = sinc a c csina = asinc csina csinc = asinc csinc c(sina sinc) = sinc(a c) sina sinc sinc = a c c 4. Student answers will vary. Student should mention using a > bsina in their explanation. 5. The answers can be determined as follows: a. a < bsina a b < sina 1 < sina A >

95 5.4. The Ambiguous Case b. a = bsina a b = sina 1 = sina A = 45. c. a > bsina a b > sina 1 > sina A < This problem can be done entirely with right triangle trig, but there are several different ways to solve this particular problem. 7. Answers: BD = = 9.6 tan4.6 = 9.8 DC = 10.7 DC sin4.6 = 9.8 AC = 14.5 AC BC = = 0. sinb = B = 45.7 A = = 91.7 a. EBD 7.6 = cosEBD 44.4 b. BDE sinbde 9.9 = sin c. DEB d. BDC e. BEA f. DBC g. ABE h. BAE i. BC sin114. BC = sin j. AB sin110.1 AB = sin k. AE sin1. AE = sin l. DC sin4.9 DC = sin m. AC = = We need to find the distance between sensors and. If it is less than 6000 ft, then the sensor will be able to detect all motion between the two. First, 4500 > 4000, so there is going to be one solution. S = S = 180 S1 = 76.5 sin76.5 x = sin x = 579 sins 4000 = sin Since x < 6000 ft., Sensor will be able to pick up all movement from its location to the location of Sensor. 9. The length from S to S4 is x and from S4 to S is y = 95, which is the angle at S4. sin = sin6 = sin49 x y x = 114.8,y =

96 Chapter 5. Triangles and Vectors, Solution Key 5.5 General Solutions of Triangles 1. Answers: a. AAS, Law of Sines, one solution b. SAS, Law of Cosines, one solution c. SSS, Law of Cosines, one solution d. SSA, Law of Sines, no solution (15 < 5sin58 ) e. SSA, Law of Sines, two solutions (45 > 60sin47 ). Answers: sin69 a.. = sin1 b,b = 4.97 b. c = (1.4)(.)cos58,c =.0 c.. = (6.1)(4.8)cosA,A =.6 d. No solution e. sin47 45 = sinb 60,B = 77. or = (a) need angle C and side c. (b) need angle A and angle B. (c) need angle B and angle C. C = = 99 sin99 = sin69,c =.6 c. sin58 = sina,a = ,B = 85.6 sin.6 = sinb,b = ,C = 6.6 (d) No solution (e) Both cases need angle C and side c. Case 1 : C = = 55.8, sin55.8 c Case : C = = 0., sin0. c = sin47,c = = sin47,c = To find the area of the rhombus, use the formula K = 1 bcsina and then multiply that by. We first need to find one of the angles that are opposite the given diagonal (they are both the same measurement). We will call it angle A. 1.5 = (1)(1)sinA,A = 17., which means the other two angles are both 5.8 ( and then divide by ). ( ) 1 K = (1)(1)sin17. =

97 5.5. General Solutions of Triangles 5. Divide the pentagon into three triangles, drawing segments from to 5, called x below, and to 4, called y below. With these three triangles, only the middle triangle needs us to find two sides and the angle between them (called Z below) to use K = 1 bcsina (the outer two triangles already have two sides and an angle that fit this criteria). x = (19)(190.5)cos81 x = 48.4 y = (146)(17.8)cos7 x = = (48.4)(191.5)cosZ Z = 7.4 Areas: K = 1 (190.5)(19)sin81 = k = 1 (48.4)(191.5)sin7.4 = K = 1 (17.8)(146)sin7 = 11.0 Total Area : altitude, x: sin56.8 = x 8 x = 1.8 GT = = 0.8 GI = = 67. A small = 1 (0.8)(1.8) = 0.8 A big = 1 (67.)(1.8) = RI = = 74. I sini = The headings would be as follows: a. W8. N b. S8.8 W c. N90 E 8. Answers: a. a = (187)(18)cos115, he would need to hit the ball 4.0 yards. sin115 b. 4 = sinb 187, he would have to hit the ball at a 9.7 angle. 9. Answers: a =.6 sin14. b. b = sin16. 0,56.8 yards sin16. c.,65.7 yards 0 = sin.6 c 9

98 Chapter 5. Triangles and Vectors, Solution Key 5.6 Vectors 1. For each problem below, use the Pythagorean Theorem to find the magnitude and tanθ = n m a. magnitude = 48.1, direction = 51.7 b. magnitude = 6.1, direction = 6.6 c. magnitude = 15., direction = 8.. Answers: a. b. c. 9

99 5.6. Vectors d. e. 94 f.. When two vectors are summed, the magnitude of the resulting vector is almost always different than the sum of the magnitudes of the two initial vectors. The only times that a + b = a + b would be true is when 1) the magnitude of at least one of the two vectors to be added is zero, or ) both of the vectors to be added have the same direction. 4. Speed (magnitude): = 5.7 and its direction is tanθ = 18 5 = N4.6 W. 5. The magnitude is = 56. Newtons and the direction is tan 1 ( 410 0) = E51. N. 6. Answers: a. a = = 1.6, direction = tan 1 ( 18 1) = 56.. b. a = ( ) + 6 = 6.7, direction = tan 1 ( 6 ) = Answers: a. a = ( 8) + (4 6) = 6., direction = tan 1 ( 4 6 8) = b. a = (5 ) + ( 1) =.6, direction = tan 1 ( 1 5 ) = 1.7. Note that when you use your

100 Chapter 5. Triangles and Vectors, Solution Key calculator to solve for tan 1 ( 1 5 ), you will get 56.. The calculator produces this answer because the range of the calculator s y = tan 1 x function is limited to 90 < y < 90. You need to sketch a draft of the vector to see that its direction when placed in standard position is into the second quadrant (and not the fourth quadrant), and so the correct angle is calculated by moving the angle into the second quadrant through the equation = In both a and b, we have the SAS case, so you can do the Law of Cosines, followed by the Law of Sines. a. ( a + b) = (1)(1)cos1, a + b = 56.6, sin1 b. ( a + b) = (9)(44)cos6, a + b =, sinx 56.6 = sinx 1 44 = sin6,x = 4,x =

101 5.7. Component Vectors Component Vectors 1. Answers: a. b = 5,4 = 10,8 = 10î + 8 ĵ b. 1 c = 1,7 = 1.5,.5 = 1.5î.5 ĵ c. 0.6 b = 0.6 5,4 =,.4 = î +.4 ĵ d. b = 5,4 = 15, 1 = 15î 1 ĵ. All of these need to be translated to (0,0). Also, recall that magnitudes are always positive. (a) (,8) + (, 8) = (0,0) horizontal = 5, vertical = 9 (, 1) + (, 8) = (5, 9) (b) (7,1) + ( 7, 1) = (0,0) (11,19) + ( 7, 1) = (4,6) horizontal = 4, vertical = 6 (c) (4., 6.8) + ( 4.,6.8) = (0,0) horizontal = 5.5, vertical =.6. Answers: a. cos5 = x 75,sin5 = y 75,x = 61.4,y = 4 b. cos16 = x.4,sin16 = y.4,x =.,y = 1.1 c. cos1 = x 15.9,sin1 = y 15.9,x = 15.6,y =. 4. magnitude =., direction = 75. from the smaller force 5. magnitude = 04,18. between resultant and larger force 6. y = 1sin8. = 5.7,x = 1cos8. = Answer: ( 1., 9.4) + ( 4.,6.8) = ( 5.5,.6) Recall that headings and angles in triangles are complementary. So, an 8 heading translates to 7 from the horizontal. Adding that to 5 (70 from 05 ) we get 4 for two of the angles in the parallelogram. So, the other angles in the parallelogram measure 18 each, 60 (4). Using 18 in the Law of Cosines, we can find the diagonal or resultant, x = (4)(155)cos18, so x = We then need to find the angle between the resultant and the speed using the Law of Sines. sina so a = 8.6. To find the actual heading, this number needs to be added to 8, getting = sin , 8. The heading is just tanθ = 10, or 11. against the current. 9. BA is the same vector as AB, but because it starts with B it is in the opposite direction. Therefore, when you add the two together, you will get (0,0). Chapter Summary Use Law of Cosines to solve for angle B = (15)(0.9)cosB,B = 55.8, which means the picture was drawn accurately.

102 Chapter 5. Triangles and Vectors, Solution Key. CD = (.6)(51.4)cos7,CD = 6.8 C, C =.5 AB = (64.1)(51.4)cos.5,AB = 5.4 This means that the artist s drawing is off by 1.1 foot.. The area of a trapezoid is A = 1 h(b 1 + b ). In the problem, we were given both bases, a side and a diagonal. So, we need to find the height. In order to do this, we first need to find the angle between 00 and 400(x), use that to find it s complement, then we can take the cosine to find the height. 100 = (400)(00)cosx x = = 5.9 cos5.9 = h 00 h = A = 1 (178.1)( ) = 4,811,670 sq. ft. 4. AB = (4)(1)cos10,AB =. minus 5, new AB is 18. new E is 18. = (4)(1)cosE, E = Answers: a. x = (1.)(17.6)cos1,x = 8. b. cos5 = x 19.,x = 11.9 sin4 c = sinx 15.9,x = 4.9 sin46 d. 9.7 = sin1 x,x = Answers: a. We will call the distance across the canyon h. sin 1 = sin87 x,x =.6.sin61 = h.6,19.8 km. b. We will call the distance between the two boulders y. y = a + b, where a is the distance from boulder 1 to the first stop (that the surveyor took) and b is the distance from that spot to boulder. tan = a 19.8 So, a + b = 19.8(tan + tan7 ) = 16.0 km. tan7 = 7. Company A coverage: x = sinx (8)(47)cos7.8,x = 51 Company B coverage: 58 = sin1 59,x = 11.8 sin1 59 = sin156. b,b = Overlap: sina 8 = sin7.8 51,a = 45.4 sin1 15 = sin1.6 o,o = Answers: a. magnitude = = 67.8, direction = tanθ = b. magnitude = = 5.5, direction = tanθ = Answers: a. magnitude = (19 1) + ( 4 1) = 4.1, direction = tanθ = b. magnitude = ( ) + (11 1) = 14.1, direction = tanθ = This is the SSA or ambiguous case. Because 17 > 45sin16, we will have two solutions or two different sin16 lengths that the golfer could have hit the ball. 17 = sinx 45,x = 58.8 or 11. (this is the angle at the ball) = sin16 17, distance is yards. Case : Use 11., d 17, distance is 7.7 yards. It is safe to say that the golfer did not hit the ball yards, considering that Tiger Woods longest recorded drive is 45 yards. So, we can logically rule out Case 1 and our answer is that the golfer s drive was 7.7 yards. Case 1: Use 58.8, the angle at the pin is 105. sin105. d the angle at the pin is 4.8 sin4.8 = sin16 b

103 5.7. Component Vectors cos6 = x 0.5,x = 0.4 miles 1. The third angle in the triangle is 5, from the Triangle Sum Theorem. sin5 17. = sin17,x = 177. x Therefore, the ball was hit 177. feet. The distance between second base and the ball can be calculated as follows: sin5 17. = sin18,x = 68.6 x 1. Therefore, the distance from second base to the ball is 68.6 feet. 14. The distance between the tower and target is: x = (7)(18)cos67.,4. miles. This means that the second target is out of range by 4. miles. 15. Answer: We need to find area, use Heron s Formula. s = 1 ( ) = 615 A = 615( )(615 47)(615 96) = 7,5.1, times = bacteria. 98

104 Chapter 6. The Polar System, Solution Key CHAPTER 6 The Polar System, Solution Key Chapter Outline 6.1 POLAR COORDINATES 6. GRAPHING BASIC POLAR EQUATIONS 6. CONVERTING BETWEEN SYSTEMS 6.4 MORE WITH POLAR CURVES 6.5 THE TRIGONOMETRIC FORM OF COMPLEX NUMBERS 6.6 THE PRODUCT & QUOTIENT THEOREMS 6.7 DE MOIVRE S AND THE NTH ROOT THEOREMS 99

105 6.1. Polar Coordinates Polar Coordinates 1. Answers: a. b. c. 100 d.

106 Chapter 6. The Polar System, Solution Key. ( 4, π ) 4 all positive both negative negative angle ( 4, 5π ) 4 ( 4, 7π ) 4 ( 4, π ) Use P 1 P = a. b. c. d. (,10 ) negative angle only (, 40 ) r 1 + r r 1r cos(θ θ 1 ). P 1 P = negative radius only (,00 ) both negative (, 60 ) (1)(6)cos(15 0 ) P 1 P 6. units P 1 P = P 1 P = P 1 P = + 9 ()(9)cos150 = ()(10)cos( 7) = (5)(16)cos(00 5) =

107 6.. Graphing Basic Polar Equations 6. Graphing Basic Polar Equations 1. Answers: a. a limaçon with an innerloop. r = sinθ b. a cardioid r = + sinθ c. a dimpled limaçon r = sinθ. Answers: a. r = cosθ b. r = + 4sinθ 10 c. r = 4

108 Chapter 6. The Polar System, Solution Key d. θ = π e. r = 5 + cosθ f. r = 6 5sinθ 10

109 6.. Graphing Basic Polar Equations To determine the equation of these curves, first notice that cosine curves are along the horizontal axis and sine curves are along the vertical axis. Second, where the curve passes through the axis on the non-dimpled side is a in r = a + bsinθ. b is a little harder to see, but it is the average of the two intercepts where the curve crosses the axis on the dimpled axis. If there is an inner loop, use the innermost value of the loop.. Answer: TABLE 6.1: θ cos θ

110 Chapter 6. The Polar System, Solution Key TABLE 6.: θ cos θ In the graph of r = 4cosθ, the rose has four petals on it but the graph of r = 4cosθ has only three petals. It appears, that if n is an even positive integer, the rose will have an even number of petals and if n is an odd positive integer, the rose will have an odd number of petals. 4. Answers: a. r = sin4θ b. r = sin5θ 105

111 6.. Graphing Basic Polar Equations c. r = cosθ d. r = 4sinθ 106 e. r = 5cos4θ

112 Chapter 6. The Polar System, Solution Key f. r = cos6θ For roses, the general equation is r = asinnθ or r = acosnθ. a indicates how long each petal is, and depending on if n is even or odd indicates the number of pedals. If n is ood, there are n pedals and if n is even there are n pedals. 107

113 6.. Converting Between Systems 6. Converting Between Systems 1. For A,r = 4 and θ = 5π 4 x = r cosθ y = r sinθ For B,r = and θ = 15 x = 4cos 5π ( 4 ) x = 4 y = 4sin 5π ( 4 ) y = 4 x = y = x = r cosθ y = r sinθ x = cos15 y = sin15 x = y = x = y = For C, r = 5 and θ = ( ) π x = r cosθ y = r sinθ. Answers: a. x = 5cos π ( x = 5 1 ) y = 5sin π ( ) y = 5 x =.5 y = 5 r = 6cosθ r = 6r cosθ x + y = 6x x 6x + y = 0 x 6x y = 9 (x ) + y = 9 b. r sinθ = y = 108

114 Chapter 6. The Polar System, Solution Key c. d.. Answers: r = sinθ r = r sinθ x + y = y y y = x y y + 1 = x + 1 (y 1) = x + 1 x + (y 1) = 1 r sin θ = cosθ r sin θ = r cosθ y = x a. For A(,5)x = and y = 5. The point is located in the second quadrant and x < 0. r = ( ) + (5) = 9 5.9, θ = Arc tan 5 + π = The polar coordinates for the rectangular coordinates A(,5) are A(5.9,1.95) b. For B(5, 4)x = 5 and y = 4. The point is located in the fourth quadrant and x > 0. r = (5) + ( 4) = ( ) , θ = tan The polar coordinates for the rectangular coordinates B(5, 4) are A(6.40, 0.67) c. C(1,9) is located in the first quadrant. r = = , θ = tan d. D( 1, 5) is located in the third quadrant and x < 0. r = ( 1) + ( 5) = 169 = 1, θ = tan Answers: a. (x 4) + (y ) = 5 x 8x y 6y + 9 = 5 x 8x + y 6y + 5 = 5 x 8x + y 6y = 0 x + y 8x 6y = 0 From graphing r 8cosθ 6sinθ = 0, we see that the r 8(r cosθ) 6(r sinθ) = 0 additional solutions are 0 and 8. r 8r cosθ 6r sinθ = 0 r(r 8cosθ 6sinθ) = 0 r = 0 or r 8cosθ 6sinθ = 0 r = 0 or r = 8cosθ + 6sinθ 109

115 6.. Converting Between Systems b. c. x y = 1 r cosθ r sinθ = 1 r(cosθ sinθ) = 1 1 r = cosθ sinθ x + y 4x + y = 0 r cos θ + r sin θ 4r cosθ + r sinθ = 0 r (sin θ + cos θ) 4r cosθ + r sinθ = 0 r(r 4cosθ + sinθ) = 0 r = 0 or r 4cosθ + sinθ = 0 r = 0 or r = 4cosθ sinθ d. x = 4y (r cosθ) = 4(r sinθ) r cos θ = 4r sin θ 4r sin θ r cos θ = 1 4tan θsecθ = 1 r 4tan θsecθ = r 110

116 Chapter 6. The Polar System, Solution Key 6.4 More with Polar Curves 1. Answer: There appears to be one point of intersection. r = sin θ r = sinθ 0 = sin θ 0 = sinθ 0 = θ 0 = sinθ 0 = θ The point of intersection is (0, 0). Answer: 111

117 6.4. More with Polar Curves r = + sinθ ( ) π r = + sin 4 r = + sinθ r = + sin 7π 4 r.4 r 0.59 r = + sinθ r = cosθ 0 = + sinθ 0 = cosθ 1 = sinθ 1 = cosθ θ = π θ = 0 Since both equations have a solution at r = 0, that is (0, π ) and (0, 0), respectively, and these two points are equivalent, the two equations will intersect at (0, 0). r = + sinθ r = cosθ + sinθ = cosθ sinθ = cosθ sinθ cosθ = cosθ cosθ sinθ cosθ = 1 tanθ = 1 θ = π 4 and θ = 7π 4 The points of intersection are (.4, π 4. Answer: ) ( ), 0.59, 7π 4 and (0,0). 11

118 Chapter 6. The Polar System, Solution Key x + y = 6x r = 6(r cosθ) r = x + y and x = ycosθ r = 6cosθ divide by r Both equations produced a circle with center (,0) and a radius of. 4. r = + sinθ and r = sinθ are not equivalent because the sine has the opposite sign. r = + sinθ will be primarily above the horizontal axis and r = sinθ will be mostly below. However, the two do have the same pole axis intercepts. 5. r = + 4cos( π) and r = + 4cosπ are equivalent because the sign of a does not matter, nor does the sign of θ. 6. Yes, the equations produced the same graph so they are equivalent. 7. Students answers will vary, but they need to include that b must be the same sign. They should also mention that the sign of a does not matter, nor does the sign of θ. 8. There are several answers here. The most obvious are any two pairs of circles, for example r = and r = 9. 11

119 6.5. The Trigonometric Form of Complex Numbers The Trigonometric Form of Complex Numbers 1. Answers: a. 5 cis π 6 = 5 π 6 = 5( cos π 6 + isin π ) 6 b. 15 = cis15 = (cos15 + isin15 ) c. ( cos π + isin π ) = cis π = π. 6 8i 6 8i x = 6 and y = 8 tanθ = y x r = x + y tanθ = 8 6 r = (6) + ( 8) θ = 5.1 r = 10 Since θ is in the fourth quadrant then θ = = 06.9 Expressed in polar form 6 8i is 10(cos isin06.9 ) or Answers: a. 4 + i x = 4,y = r = b. + 9i x =,y = 9 r = 4 + = 5,tanθ = 4 θ = (cos isin6.87 ) ( ) + 9 = 85 9.,tanθ = 9 θ = (cos isin10.5 ) 114

120 Chapter 6. The Polar System, Solution Key c. 7 i x = 7,y = 1 r = = ,tanθ = 1 7 θ = (cos isin51.87 ) d. 5 i x = 5,y = r = ( 5) + ( ) = 9 5.9,tanθ = 5 θ = (cos isin01.8 ) 4. Answer: Note: The range of a graphing calculator s tan 1 function is limited to Quadrants I and IV, and for points located in the other quadrants, such as + 9i in part b (in Quadrant II), you must add 180 to get the correct angle θ for numbers given in polar form. ( cos π 4 + isin π ) 4 r = x = cos π 4 = y = sin π 4 = The standard form of the polar complex number ( cos π 4 + isin π ) 4 is + i. 5. Answers: a. cis π x = cos π = 0,y = sin π = 1 (0) + (1i) = i 5π b. 4 6 x = cos 5π 6 =,y = sin 5π 6 = 1 ( 4 c. 8 ( cos ( π ) + isin ( π 4i )) ( ) x = cos π = 1,y = sin( π ) + 4 ( i 1 ) = + i ) = 8 ( ( ) i ) = 4 115

121 6.6. The Product & Quotient Theorems The Product & Quotient Theorems 1. Answers: a. 56,7 11 = ()(7) ( ) = b. (cosπ + isinπ),10 ( cos 5π + isin 5π c. + i, i change to polar ) = ()(10)cis ( π + 5π ) = 0cis 8π = 0cis π x =,y = x = 5,y = 11 r = + = 1.61 r = ( 5) + 11 = tanθ = θ = 56.1 tanθ = 11 5 θ = d. 6 i, 0i change to polar (.61)(1.08) ( ) = x = 6,y = 1 r = 6 + ( 1) = r = tanθ = 1 6 θ = x = 0,y = ( 0) = 40 = 0 tanθ = 0 0 = und θ = (6.08)(0) ( ) = = Without changing complex numbers to polar form, you mulitply by FOIL-ing. ( + i)( i) = 10 + i 15i + i = i = 4 + 7i The answer is student opinion, but they seem about equal in the degree of difficulty.. Answer: ( 4 cos π 4 + isin π ) ( = 4 cos π isin π ) ( 4 cos π isin π ) ( ( 4 π = 16 cos 4 + π ) ( π + isin π )) ( 4 = 16 cos π isin π ) 4. Answer: 5. Answers: a. 56 b = 7 (56 11 ) = (cos 5π +isin 5π ) 5(cosπ+isinπ) P = (6.80)(7.05) ( ),P = watts = ( cos ( 5π π) + isin ( 5π π)) = ( cos π + isin π )

122 Chapter 6. The Polar System, Solution Key +i c. 5+11i change each to polar. x =,y = x = 5,y = 11 r = + = 1.61 r = ( 5) + 11 = tanθ = θ = 56.1 tanθ = 11 5 θ = ( ) = change both to polar d. 6 i 1 0i x = 6,y = 1 r = 6 + ( 1) = r = tanθ = 1 6 θ = ( ) = Answer: x = 1,y = ( 0) = 401 = 0.0 tanθ = 0 1 θ = i i = + i 5 11i 10 i 15i + 7i = = i 5 11i Again, this is opinion, but in general, using the polar form is easier. 7. Answer: ( [4 cos π 4 + isin π ) ( ] = [4 cos π isin π ) ( ] 4 cos π isin π ). 4 From #, [4 ( cos π 4 isin π 4 ) ] = 16 ( cos π isin π ). So, ( [4 cos π 4 isin π ) 4 ( ] = 16 cos π isin π ) ( = 64 cos π 4 isin π 4 ( 4 ) cos π 4 isin π 4 8. Even though 1 is not a complex number, we can still change it to polar form. 1 x = 1,y = 0 ) r = = 1 tanθ = 0 1 = 0 θ = 0 1 So, 4cis π 6 = 1cis0 4cis π 6 = 1 ( 4 cis 0 π ) = 1 ( 6 4 cis π )

123 6.7. De Moivre s and the nth Root Theorems De Moivre s and the nth Root Theorems 1. Express z in polar form: The polar form is z = 1(cos10 + isin10 ). Answers: a. b. [ ( cos π 4 + isin π ) ] 8 4 = r = x + y r = ( 1 ) ( ) + 1 r = = 1 ) θ = tan ( 1 = 10 1 z n = [r(cosθ + isinθ) n = r(cosnθ + isinnθ) z = 1 [cos(10 ) + isin(10 )] z = 1(cos60 + isin60 ) z = 1(1 + 0i) z = 1 ( ) 8 ( cos 8π 4 + isin 8π ) 4 = 1 16 cosπ + i 16 sinπ = 1 16 [ ( i ] 4 ) = ( i ) 4 r = ( ) + ( ) = 6,tanθ = = 1 45 ( = ( 6 cos π 4 + isin π )) ( 4 = ( 6) 4 cos 4π isin 4π 4 = 81(6)[ 1 + i(0)] = 96 ) c. ( 5 i) 7 r = ( 5) + ( 1) = 6,tanθ = 1 5 θ = 5.9 6(cos5.9 + isin5.9 ) 7 = ( 6) 7 (cos(7 5.9 ) + isin(7 5.9 )) = 16 6(cos51. + isin51. ) = 16 6( i) = i d. [ ( cos π 6 + isin π 6 )] 1 = 1 (cosπ + isinπ) = 51,

124 Chapter 6. The Polar System, Solution Key. Answer: r = and θ = 15 or 7π 4. z n = [r(cosθ + isinθ)] n = r n (cosnθ + isinnθ) ( ) ( )] 7π 7π z = [(cos + isin 4 4 ( z = 8 cos 1π ) 1π + isin 4 4 ( ) z = 8 i z = 4 4i 1π 4 is in the third quadrant so both are negative. 4. Answer: a = 0 and b = 7 7i = (0 + 7i) 1 x = 0 and y = 7 5. Answer: Polar Form r = x + y θ = π r = (0) + (7) r = 7 [ ( 7i = 7 cos( π )] 1 + πk) + isin(π + πk) for k = 0,1, [ ( ) 1 (π ) ( ) 1 (π 7i = 7 cos + πk + isin ( 7i = cos π 6 + isin π ) for k = 0 ( 6 7i = cos 5π 6 + isin 5π ) for k = 1 6 ( 7i = cos 9π 6 + isin 9π ) for k = 6 ( ) ( 7i = + 1 i, + ), i 1 i r = ( 1 x + y θ = tan 1 1 r = (1) + (1) r = ) = [ ( (1 + i) 1 5 = cos π 4 + isin π )] 1 5 ( ) 4 (1 + i) = 5 1 (π ) ( ) 1 (π ) [cos ] + isin (1 + i) ( = cos π 0 + isin π ) 0 In standard form (1 + i) 1 5 = ( i) and this is the principal root of (1 + i) πk )] for k = 0,1, Polar Form = cis π 4 119

125 6.7. De Moivre s and the nth Root Theorems i in polar form is: r = = 81,tanθ = 81 0 = und θ = π ( 81 cos π + isin π ) [ ( ( π ) ( π ))] 1 81 cos + πk + isin + πk 4 ( ( π cos + πk ) ( π + isin + πk )) 4 4 ( ( π cos 8 + πk ) ( π + isin 8 + πk )) ( ( π z 1 = cos 8 + 0π ) ( π + isin 8 + 0π )) = cos π 8 + isin π = i 8 ( ( π z = cos 8 + π ) ( π + isin 8 + π )) = cos 5π 8 + isin 5π = i ( ( 8 π z = cos 8 + π ) ( π + isin 8 + π )) = cos 9π 8 + isin 9π = i 8 z 4 = ( cos ( π 8 + π ) + isin ( π 8 + π )) = cos 1π 1π + isin = i Answer: x = 0 r = x + y x 4 = 1 r = ( 1) + (0) x 4 = 1 + 0i r = 1 ( ) 0 θ = tan 1 + π = π 1 Write an expression for determining the fourth roots of x 4 = 1 + 0i ( 1 + 0i) = [1(cos(π + πk) + isin(π + πk))] 4 ( ( 1 + 0i) = 1 4 cos π + πk + isin π + πk ) 4 4 ( x 1 = 1 cos π 4 + isin π ) = 4 + i ( x = 1 cos π 4 + isin π ) = 4 + i ( x = 1 cos 5π 4 + isin 5π ) = 4 i ( x 4 = 1 cos 7π 4 + isin 7π ) = 4 i for k = 0 for k = 1 for k = for k = If a line segment is drawn from each root on the polar plane to its adjacent roots, the four roots will form the corners of a square. 10

126 Chapter 6. The Polar System, Solution Key 8. Answer: x 64 = 0 x = i i = 64(cos(0 + πk) + isin(0 + πk)) x = (x ) ( ( ) ( )) πk 0 + πk = (64 + 0i) = 64 cos + isin ( ( ) ( )) 0 + π0 0 + π0 z 1 = 4 cos + isin = 4cos0 + 4isin0 = 4 for k = 0 ( ( ) ( )) 0 + π 0 + π z = 4 cos + isin = 4cos π + 4isin π = + i for k = 1 ( ( ) ( )) 0 + 4π 0 + 4π z = 4 cos + isin = 4cos 4π + 4isin 4π = i for k = If a line segment is drawn from each root on the polar plane to its adjacent roots, the three roots will form the vertices of an equilateral triangle. Chapter Summary 1. A (, π 4 ) three equivalent coordinates (, π ) ( ) ( ) 4,, 7π 4,, 5π 4.. (,94 ) and (7, 7 ) d = + 7 ()(7)cos(94 ( 7 )) = cos167 = Answers: a. r = sin5θ 11

127 6.7. De Moivre s and the nth Root Theorems b. r = 6 cosθ c. r = + 5cos9θ 1 4. Answers: a. r = 6cosθ b. r = 7 + sinθ 5. Answers: a. A( 6,11) r = ,tanθ = 11 6,θ = (1.59,118.6 ) b. B(15, 8) r = = 17,tanθ = 8 15,θ = 8.1 (17, 8.1 ) c. C(9,40) r = = 41,tanθ = 40 9,θ = 77. (41,77. )