Review of Essential Skills and Knowledge

Review of Essential Skills and Knowledge R Eponent Laws...50 R Epanding and Simplifing Polnomial Epressions...5 R 3 Factoring Polnomial Epressions...5 R Working with Rational Epressions...55 R 5 Slope and Rate of Change of a Linear Function...59 R 6 The Zeros of Linear and Quadratic Functions...55 R 7 Eponential Functions... 553 R 8 Transformations of Functions...55 R 9 Families of Functions...557 R 0 Trigonometric Ratios and Special Angles...559 R Graphing sin and cos...56 R Transformations of Trigonometric Functions...56 R 3 Solving Trigonometric Equations in Degrees...567 R Proving Trigonometric Identities...570 Copright 009 b Nelson Education Ltd. Review of Essential Skills and Knowledge 539

R Eponent Laws Rule Word Description Algebraic Description Eample Multiplication Division Power of a Power Power of a Product Power of a Quotient Zero Eponent Negative Eponents If the bases are the same, add the eponents. If the bases are the same, subtract the eponents. Keep the base, and multipl the eponents. Raise each factor to the eponent. Raise the numerator and the denominator to the eponent separatel. A power with zero as the eponent equals, ecept when zero is also the base. A power with a negative eponent equals the power with the reciprocal base and a positive eponent. a m a n a mn 0 7 0 5 0 a m a n amn, a 0 0 00 0 95 0 5 (a m ) n a mn ( ) (ab) n a n b n ( 5 ) 3 3 ( 5 ) 3 8 5 a a b b n an b 0 b n, a 3 b a 0, if a 0 7 0 Eception: 0 0 is undefined. a n a n, a 0 a b a n a a n a, b 0 b b a b n, a b 9 0 0 0 000 a b a b 6, 0 Rational Eponents with Numerator Rational Eponents with Numerator not The denominator determines the root. The denominator determines the root and the numerator indicates the eponent applied to the root. EXAMPLE a n a m n a m n n a Qn ar m or n a m 000 3 5 3 000 5 0 5 5 3 6 5 Q5 3R 6 6 6 Simplif. Epress our answers using positive eponents. A 3 B a) b) 6 3 A 3 BA B c) A B Copright 009 b Nelson Education Ltd. 50 Advanced Functions: Review of Essential Skills and Knowledge

a) b) 6 3 c) 3 5 6 3 Q 3 6R A 3 BA B A 3 B A B 6 8 8 6 Practising. Simplif. Epress our answers using positive eponents. b) A 3 B 5 f) A B A 3 B j) 5 n) a 5 b 3 c) ( ) ( ) g) 8 0 k) ( ) 3 ( ) a o) 0 b ( ) 5 ( ) d) a 9 h) l) a 3 b a 3 a) 7 e) i) m) Aa 3 bc 0 B 7 A 0 B 0 Copright 009 b Nelson Education Ltd.. Evaluate. Epress our answers in fraction form. a) b) c) d) 6 3 6 3 5 3 7 3 00 0.5 R- Eponent Laws: Review of Essential Skills and Knowledge 5

R Epanding and Simplifing Polnomial Epressions To convert a polnomial epression from factored form to epanded form, use the distributive propert: a(b c) ab ac Some patterns occur frequentl and are worth memorizing. Square of a Sum Square of a Difference Difference of Squares (a b) (a b)(a b) a ab b (a b) (a b)(a b) a ab b (a b)(a b) a ab ab b a b EXAMPLE Epand and simplif ( 5)(3 5). ( 5)(3 5) 3 3 5 5 0 5 3 3 5 5 Use the distributive propert to multipl each term in the binomial b each term in the trinomial. There are 3 6 terms in the epanded form, before it is simplified. Collect like terms to simplif the epanded form. EXAMPLE Epand and simplif ( )( )( 3). ( )( )( 3) 3( ) 3 ( )( 3) ( )( 3 3) ( )( 3) 3 8 6 6 3 0 Since multiplication is associative, ou can multipl the epressions in an order ou like. Use the distributive propert to multipl. Drawing arrows will help ou to keep track of the multiplications. Copright 009 b Nelson Education Ltd. 5 Advanced Functions: Review of Essential Skills and Knowledge

EXAMPLE 3 Epand and simplif ( 3) (3 )(3 ). ( 3) (3 )(3 ) () ()(3) (3) 3 (3) () 9 9 5 3 Use the patterns for (a b) and (a b)(a b). Practising. Epand and simplif. a) 3 A5 3 B b) ( 7) c) 3( ) ( )( ) d) e) f) 6a 3 ba b 5( ) 3( ). Write in simplified epanded form. a) 5( )( )( ) b) ( 3)( 7) c) ( ) 3 d) ( 5)( )( 5)( ) e) (3 ) ( 3) f) ( 3) Copright 009 b Nelson Education Ltd. R Epanding and Simplifing Polnomial Epressions: Review of Essential Skills and Knowledge 53

R 3 Factoring Polnomial Epressions Tpe Eample Comment Common Factoring 0 8 3 6 5 ab ac a(b c) 3 (5 3 ) Factor out the largest common factor of each term. Factoring Trinomials a b c, when a ( 7)( 3) Write the trinomial as the product of two binomials. Determine two numbers whose sum is b and whose product is c. Factoring Trinomials a b c, when a Look for a common factor. If none eists, use decomposition and write the trinomial as the product of two binomials. Check b epanding and simplifing. 3 3 6 (3 ) (6 ) (3 ) (3 ) (3 )( ) Check: (3)() (3)() ()() ()() 3 6 3 Each term has a common factor of 3. () 7(3) 7 (3) and Multipl 3(). Find two numbers whose product is and whose sum is. In this case, the numbers are 6 and. Using these numbers, decompose the -term. Group the terms, and factor out the common factors. Factoring a Difference of 8 Squares ( 9) ( )( ) ( 3)( 3) This is a special case of factoring trinomials, when b 5 0. Practising. Factor. a) 6 5 b) 8 c) 6 d) 3 8 e) 6 f) 8. Factor. a) 6 b) c) 5a 7a 6 d) 8 Common factor first, when possible. 3. Epand to show that ( )( ) is the factored form of 3 3. Copright 009 b Nelson Education Ltd. 5 Advanced Functions: Review of Essential Skills and Knowledge

R Working with Rational Epressions A rational epression is an algebraic epression that can be written as the quotient of two polnomials. A rational epression is undefined if the denominator is zero, so we write restrictions on the variables to avoid this. Simplifing Rational Epressions A rational epression can be simplified b factoring the numerator and the denominator, and then dividing out the common factors. EXAMPLE Simplif and state restrictions. m 3 n a) b) c) 6mn 3 6 9 6 m 3 n a) b) 6mn 3 ( )( ) 3mn (7m ) ( )( 3) 3mn (n ) ( ) ( ) 7m ( ) ( 3) n m 0, n 0 3,3 Factor the numerator and the denominator to find the largest possible common factor to divide out. Write restrictions on the variables to prevent the denominator from equalling zero. Copright 009 b Nelson Education Ltd. c) 6 9 6 (3 ) 3(3 ) 3 3 ( 3) 3(3 ) opposites When factors are opposites, factor out from one of the factors to make the factors identical. R Working with Rational Epressions: Review of Essential Skills and Knowledge 55

Multipling and Dividing Rational Epressions To multipl or divide rational epressions, factor the numerators and the denominators (where possible), and then look for common factors that can be divided out. EXAMPLE Simplif 9 0 6 6 9. 9 0 6 6 9 ( 3)( 3) ( 3) ( 3) ( 3) ( 3) ( 5) 3 3, 3 EXAMPLE 3 Simplif 3 ( 5) ( 3)( 3) ( 5) ( 3) ( 3) 9 5. 5 Factor each polnomial. Divide out the common factors to reduce the epression to lowest terms. Write restrictions to prevent the denominator from equalling zero, which would result in undefined values. 3 3 3 3, 5, 9 5 5 ( )( ) 3 ( )( ) 3 5 9 5 5 ( )( 5) 5 ( ) ( 5) Change the division into multiplication b the reciprocal. Factor. Divide out the identical factors. Write restrictions to avoid undefined values. Copright 009 b Nelson Education Ltd. 56 Advanced Functions: Review of Essential Skills and Knowledge

Adding and Subtracting Rational Epressions To add or subtract rational epressions, ou must have a common denominator. To ensure that ou will use the lowest common denominator, factor the numerators and the denominators first. This will keep the epressions as simple as possible. EXAMPLE Simplif 3 6 8 5. 7 Copright 009 b Nelson Education Ltd. 3 6 8 5 7 3 a ba7 7 3( ) ( 3)( 5) ( )( ) 7( 3) 3( ) ( ) ( ) 3 5 7 7( ) 7 0 7( ) ( 7 0) 7( ) 7 0 7( ) 7 7( ),, 3 ( 3) ( 5) 7( 3) 5 b a ba 7 b Factor the numerators and the denominators. If possible, divide out like factors (but onl within each rational epression). The lowest common denominator is 7( ). Multipl the numerator and denominator of each rational epression to create an equivalent epression with the desired common denominator. Simplif the numerator. Write restrictions to avoid undefined values. R Working with Rational Epressions: Review of Essential Skills and Knowledge 57

Practising. State the restrictions (if an) on each rational epression. a) 5 7 b) c) 5 d). Simplif, and state restrictions. Write our answers using positive eponents. a) ab 3h 6h 3 c) e) b h h 5 6 b) 5 b 3 a b t 3 t d) f) 5 b ab a t t 3 3 36 3. Simplif, and state restrictions. 6 a) 8 3 b) c) d) e) f) ab 6ac 5bc 0b ( ) 3 3 7 0 5 6 5 5 3m 7m 6 6m 3m 9m 6 m 5m 3 5 3 9m m m. Simplif, and state restrictions. a) 5 3 b) c) 5 3 5 6 d) 5 6 3 3 e) f) 3 5 5. Show that 3 3, 3. Copright 009 b Nelson Education Ltd. 58 Advanced Functions: Review of Essential Skills and Knowledge

R 5 Slope and Rate of Change of a Linear Function The slope of a line is a ratio that compares the change in the dependent variable,, with the change in the independent variable,. Slope m rise change in run change in The equation of a linear relation can be written in the form m b, where m is the slope and b is the -intercept. Values of the Slope The slope of a line that rises to the right is positive. The slope of a line that drops to the right is negative. The slope of a horizontal line is zero. The equation of the line can be written in the form b. The slope of a vertical line is undefined. The equation of the line can be written in the form a. Equations of Straight Lines point-slope equation of a line: m( ) general form of the equation of a line: A B C 0 slope-intercept equation of a line: m b Parallel and Perpendicular Lines Two lines, with slopes m and m, are parallel if and onl if m m perpendicular if and onl if m m ; that is, if their slopes are negative reciprocals: m m EXAMPLE Copright 009 b Nelson Education Ltd. Find the slope and equation of a line that passes through points (5, 6) and (5, ). Eplain how the slope is a rate of change. The slope is m 6 5 5 0 5. Substituting m 5 and (, ) (5, 6) into m( ), 6 ( 5) 5 6 5 5 8 The slope of the line is 5, and the equation is 5 8. R 5 Slope and Rate of Change of a Linear Function: Review of Essential Skills and Knowledge 59

The slope is a rate of change because will decrease b units for each 5 unit increase in. 0 8 6 run = 5 rise = 0 5 0 Practising. Determine the slope of a line that passes through each pair of points. a) (, 5) and (, 9) b) (, ) and (7, ) c) (5, ) and (5,) d) (3, 5) and (, 9). Describe the graph of a) 3 b) 6 3. Suppose that ou bu a plant. The height of the plant t weeks after ou bu it is h(t) 6.t, where h is the height in centimetres. What is the slope of the height function, and what does the slope mean in the contet of this situation?. Determine the slope and -intercept of each line. a) 3 5 0 0 b) A B C 0 Copright 009 b Nelson Education Ltd. 550 Advanced Functions: Review of Essential Skills and Knowledge

R 6 The Zeros of Linear and Quadratic Functions The Zero of a Linear Function A linear function of the form m b has one zero (-intercept), unless the line is horizontal. (A horizontal line has no -intercepts, unless it lies on the -ais. Then ever point on the line is an -intercept.) Factoring out the slope will give the -intercept. EXAMPLE What is the -intercept of 6? 6 ( 3) Solving for the zero, let 0. 0 ( 3) 3 Factor out the slope. The -intercept is 3, since substituting 3 results in a value of zero. (Hence the name zero of the function for an -intercept.) 6 0 6 6 8 Copright 009 b Nelson Education Ltd. The Zeros of a Quadratic Function A quadratic function can have two zeros, one zero, or no zeros. The zeros of a quadratic function are also found b factoring the equation. When factoring is not possible, the quadratic formula can be used. The discriminant can be used to determine the number of zeros. Quadratic Formula The zeros of the function a b c are b b ac a. Discriminant If b ac 0, there are two zeros. If b ac 0, there is one zero. If b ac 0, there are no zeros. R 6 The Zeros of Linear and Quadratic Functions: Review of Essential Skills and Knowledge 55

EXAMPLE Determine the zeros of each function. a) f () 6 b) g() a) Factoring gives f () 6 f () ( 3)( ) Solving for the zeros, let f () 0. 0 ( 3)( ) 0 3 or 0 3 or 6 6 0 6 6 b) To solve for the zeros, let g() 0. 0 Using the quadratic formula gives b b ac a () () ()() () 7 Since the square root of 7 is not a real number, the function g has no zeros. 8 6 6 0 6 The graph of g is entirel above the -ais. Therefore, the function has no zeros. Practising. What are the zeros of each function? a) 3( ) b) ( 3)( 7) c) ( 9) d) 5. Determine the zero of each linear function. a) 3 b) line with -intercept 3 and slope 3. Determine the zeros of each quadratic function. a) f () c) g() 3 b) 6 8 d) 3 5. The zeros of a quadratic function are and, and the -intercept is 8. Write the equation of the function. 5. Use the discriminant to determine the number of zeros for each quadratic function. a) 6 0 5 b) 6 0 3 Copright 009 b Nelson Education Ltd. 55 Advanced Functions: Review of Essential Skills and Knowledge

R 7 Eponential Functions The eponential function f () b has the following characteristics: The base is restricted to 0 b or b. The domain is {R}, and the range is { R 0}. The -ais is a horizontal asmptote. The -intercept is. If b, the graph increases (is a growth function). If 0 b, the graph decreases (is a deca function). EXAMPLE Sketch the graph of each eponential function. a) b) f () a 3 b a) b) 3 0 8 0 3 8 a 3 b 8 9 6 3 0 3 0 3 3 3 0 3 Note that is a growth curve and A 3B is a deca curve. 9 0 8 6 Copright 009 b Nelson Education Ltd. Practising. Sketch the graph of each eponential function. a) f () 3 c) f () a b b) f () 0 d) A 3B f () (.5). Compare the graphs of and 3. How are the related? 3. For the function f (), state the domain, range, intercepts, and equation of the asmptote.. The function T 0 76(0.9) t models the temperature, in C, of a cup of coffee t minutes after it is poured. a) What is the initial temperature of the coffee? b) What is the temperature after 0 min? c) What is the temperature after 60 min? d) Determine the equation of the horizontal asmptote. What does it represent? e) What is the significance of the number 76 in the equation? R 7 Eponential Functions: Review of Essential Skills and Knowledge 553

R 8 Transformations of Functions You can graph functions of the form af (k( d )) c b appling the appropriate transformations to ke points on the parent function f (). Stretches/compressions and reflections (based on a and k) must be applied before translations (based on c and d ). The value of a determines whether there is a vertical stretch or compression and whether there is a reflection in the -ais. The -coordinate of each point is multiplied b a. If a, the graph of f () is stretched verticall b the factor ZaZ. If 0 a, the graph is compressed verticall b the factor a. If a is negative, the graph is also reflected in the -ais. The value of k determines whether there is a horizontal stretch or compression and whether there is a reflection in the -ais. The -coordinate of each point is multiplied b k. If k, the graph of f () is compressed horizontall b the factor k. If 0 k, the graph is stretched horizontall b the factor k. If k is negative, the graph is also reflected in the -ais. The value of c determines the vertical translation. This value is added to the -coordinate of each point. If If the graph is translated c units up. the graph is translated c units down. The value of d determines the horizontal translation. This value is added to the -coordinate of each point. If If c 0, c 0, d 0, d 0, the graph is translated d units to the right. the graph is translated d units to the left. EXAMPLE What transformations to the parent function f () would ou perform to create the graph of f (3( )) 5? What happens to the coordinates of each point on the parent function? Comparing the transformed function with the general form af (k( d)) c, we have a, k 3, d, and c 5. Copright 009 b Nelson Education Ltd. 55 Advanced Functions: Review of Essential Skills and Knowledge

Since a, there is a vertical stretch b a factor of and also a reflection in the -ais. The -coordinate of each point is multiplied b. Since k 3, there is a horizontal compression b a factor of 3. The -coordinate of each point is multiplied b 3. Since c 5, there is a vertical translation 5 units down. The value 5 is added to the -coordinate of each point. Since d, there is a horizontal translation units to the right. The value is added to the -coordinate of each point. EXAMPLE Graph the function 3 (3) b appling of the appropriate transformations to the parent function. Table of values for Graph of We start with points on the parent function. 6 0 0 6 Copright 009 b Nelson Education Ltd. Table of values for 3 Graph of 3 Appl an stretches/compressions and reflections net. 6 Since a 3, there is a vertical 3 stretch. Each -coordinate is 3 8 multiplied b. 3 0 Since k there is a horizontal 6 stretch and also a reflection in the 3 0 -ais. Each -coordinate is multiplied b. 3 6 R 8 Transformations of Functions: Review of Essential Skills and Knowledge 555

Table of values for 3 (3) 7 3 8 5 3 3 7 Graph of 3 (3) 6 0 6 Appl an translations last. Since c, there is a translation up. The value is added to each -coordinate. Since d 3, there is a translation to the right. The value 3 is added to each -coordinate. Notice that the horizontal asmptote is shifted up to. Practising. Describe the transformations that ou would appl to the graph of f () to graph each of the following functions. a) 3f () b) c) d) e) f a ( 3)b f () 7 3f (( )) f () f) f () 3 5. The point (, 5) is on the graph of f (). State the coordinates of the image of this point under each of the following transformations. a) f (3) c) f ( ) b) f () d) f () 7 3. Given the function f (), state the equation of the transformed function under a vertical stretch of factor 3, a reflection in the -ais, a horizontal translation 3 units to the right, and a vertical translation units up.. Consider the function f () 3. a) Make a table of values for f using {,, 0,, }. b) Describe the transformations to f that result in the function g() ( ) 3 5. c) Determine the five points on the graph of g that are the images of the five points in our table of values for f in part a). 5. Consider the functions Y and Y. What transformations to Y result in Y? Copright 009 b Nelson Education Ltd. 556 Advanced Functions: Review of Essential Skills and Knowledge

R 9 Families of Functions Families of Linear Functions Consider the equation m 3. It represents a straight line with -intercept 3 and slope m. Different values of the parameter m will result in lines with different slopes. Together, these lines make up a famil of lines with the same -intercept. 6 0 Families of Quadratic Functions The equation a( )( 3) represents a famil of quadratic functions. Each member of the famil has zeros at and 3. Their differences are determined b the value of the vertical stretch factor a. 6 0 6 6 EXAMPLE What member of the famil of quadratic functions with the verte (3, ) passes through point (5, 5)? Copright 009 b Nelson Education Ltd. The graph shows several members of the famil of quadratic functions with the verte (3, ). This famil has an equation of the form f () a( 3). Substitute point (5, 5) into the equation, and solve for a. 5 a(5 3) 5 a() 6 a a 6 3 The equation is f () 3 ( 3). 6 0 6 6 R 9 Families of Functions: Review of Essential Skills and Knowledge 557

Practising. a) Determine the general equation of the famil of straight lines with slope 3, but varing -intercepts. b) Find the equation of the member of this famil that passes through point (, 7).. a) Determine the equation of the famil of quadratic functions with zeros at and. b) What is the equation of the member of this famil with -intercept? 3. A famil of eponential functions has equation (k). a) At what point do all the members of this famil meet? b) Wh does the parameter k var in the graphs of this famil? c) Show that k 3 results in a curve that passes through point (, 8).. Determine the equation of the quadratic function that has verte (, 5) and passes through (, 8). 5. Determine the equation of the quadratic function that has -intercepts 5 and, and passes through (7, 0). 6. Determine the equation of the quadratic function f () a 6 7 if f () 3. Copright 009 b Nelson Education Ltd. 558 Advanced Functions: Review of Essential Skills and Knowledge

R 0 Trigonometric Ratios and Special Angles Right-Triangle Definitions of Trigonometric Ratios The trigonometric ratios for an acute angle can be defined using a right triangle, as shown below. sin u opposite hpotenuse cos u adjacent hpotenuse csc u hpotenuse opposite sec u hpotenuse opposite hpotenuse u adjacent opposite tan u opposite adjacent cot u adjacent opposite The Pthagorean theorem is often useful for solving problems that involve right triangles: (adjacent) (opposite) (hpotenuse) The right-triangle definitions given above cannot be used for an angle that is not acute, so we need to broaden the definitions. Definitions of Trigonometric Ratios for An Angle An angle in standard position has its verte at the origin and rotates counterclockwise from the positive -ais to its terminal arm. If point (, ) is on the terminal arm of angle u, at a distance r units from the origin, we define the trigonometric ratios of u as follows: sin u r csc u r (, ) Copright 009 b Nelson Education Ltd. cos u r tan u sec u r cot u In the diagram above, b is the acute angle related to u. The related acute angle alwas has one arm on the -ais. The trigonometric ratios for angle b are equal in magnitude to those for angle u, but the are alwas positive. The CAST rule is an eas wa to remember which trigonometric ratios are positive in each quadrant. Since r is alwas positive, the sign of each ratio depends on the signs of the coordinates of the point. In quadrant, all (A) ratios are positive because both and are positive. In quadrant, onl sine (S) and its reciprocal cosecant are positive, since is negative and is positive. In quadrant 3, onl tangent (T) and its reciprocal cotangent are positive, because both and are negative. In quadrant, onl cosine (C) and its reciprocal secant are positive, since is positive and is negative. r b u 3 S T 0 A C R 0 Trigonometric Ratios and Special Angles: Review of Essential Skills and Knowledge 559

The eact values of the primar trigonometric ratios for 30, 5, and 60 angles can be found b using an isosceles right triangle and half of an equilateral triangle, as shown below. These triangles are often referred to as special triangles. 5 30 5 3 60 u 30 5 60 sin u cos u tan u 0.5 3 0.8660 3 0.577 3 3 0.707 0.707 3 0.8660 0.5 3.73 EXAMPLE Determine the eact value of sin 0. The following diagram shows that a 0 angle is related to a 60 acute angle. 0 60 Copright 009 b Nelson Education Ltd. 560 Advanced Functions: Review of Essential Skills and Knowledge

sin 0 sin 60 3 sin 0 3 EXAMPLE To determine the sign of sin 0, consider that sine is the ratio. Since 0 is in quadrant III, is negative. Therefore, sine is negative. If tan u, find an eact value for sin u. Confirm the value with a calculator. There are two possible terminal arms, as shown in the following diagram. (, ) Since tan u, we know that an point on the terminal arm of the angle must satisf. 5 u 0 u A point on the terminal arm in quadrant II is (, ). A point on the terminal arm in quadrant IV is (, ). 5 (, ) Using r gives r 5. Therefore, sin u r. 5 Copright 009 b Nelson Education Ltd. Practising. Using eact values, show that sin u cos u for each angle. a) u 60 b) u 5. Determine the acute angle that each line makes with the -ais. a) 5 b) 53 3. Determine the angle(s) u between 0 and 360 if a) cos u sin u b) sin u. Determine an eact value for a) sin 35 b) cos 0 c) tan 5 d) csc 300 R 0 Trigonometric Ratios and Special Angles: Review of Essential Skills and Knowledge 56

R Graphing = sin and = cos The functions sin and cos are periodic functions since their graphs consist of a regularl repeating shape. The period of both of these functions is 360. The minimum value of these functions is, and the maimum value is. The amplitude of a function is defined as, () of these functions is. maimum minimum, so the amplitude The ais (or midline) of a function is the horizontal line halfwa between the maimum value and the minimum value. For both of these functions, the equation maimum minimum () of the ais is 0. = sin = cos 80 0 80 360 EXAMPLE At what values of, in the interval 360 to 360, does sin? Draw the sine function from 360 to 360 along with the line. 360 80 0 80 360 We can read the intersection points directl from the graph. The values of in the interval 3360, 360 are 330, 0, 30, and 50. Copright 009 b Nelson Education Ltd. 56 Advanced Functions: Review of Essential Skills and Knowledge

EXAMPLE At what values of does the function cos have a minimum value? Eamine the graph. 360 80 0 80 360 50 70 We can see that the minimum values occur at 80, 80, 50, and so on. There are an infinite number of values of, so we cannot list all of them. (This would take forever!) Notice, however, that the occur at regular intervals due to the periodic nature of the function. Each minimum value is a multiple of 360 that is either less than or more than 80. Therefore, we can write all the values of as follows: 80 k(360), where ki Copright 009 b Nelson Education Ltd. Practising. At what values of does the function sin have a maimum value?. At what values of do the functions sin and cos meet? 3. Consider the graph of sin from 0 to 360. 0 90 80 70 360 a) Determine the values of that correspond to an angle in i) quadrant I ii) quadrant II iii) quadrant III iv) quadrant IV b) Describe the behaviour of the sin function in each quadrant. R Graphing = sin and = cos : Review of Essential Skills and Knowledge 563

R Transformations of Trigonometric Functions We can transform the sine and cosine functions in the same wa that we transform other functions. The general forms of the transformed functions are a sin(k( d)) c and a cos(k( d)) c. Transformation The value of a determines whether there is a vertical stretch/compression and whether there is a reflection in the -ais. Characteristics of the Transformed Function Since the vertical stretch/compression factor is a, the amplitude of the transformed function is a. The value of k determines whether there is horizontal stretch/compression and whether there is a reflection in the -ais. Since the horizontal stretch/compression factor is the period of the transformed function is 360 k. The value of c determines the vertical translation. The ais of the transformed function is c. The value of d determines the horizontal translation. The horizontal shift of the transformed function is d. k, EXAMPLE Describe the transformations that have been applied to a parent function to get each of the following transformed functions. State the amplitude, period, and ais of the transformed function, and determine whether there is a horizontal shift. Then sketch the graph of the transformed function. a) cos b) g() 3 sina 5b a) To obtain the transformed function cos, the parent function cos undergoes a vertical stretch of factor, a reflection in the -ais, a horizontal compression of factor, and a vertical translation unit down. 360 The amplitude of the transformed function is, the period is 80, and the ais is. There is no horizontal shift. 90 0 3 90 80 70 360 50 Start with the parent function The amplitude is. The period is 360. cos. Copright 009 b Nelson Education Ltd. 56 Advanced Functions: Review of Essential Skills and Knowledge

90 0 3 90 80 70 360 50 Appl the stretch, compression, and reflection. cos The function is reflected in the -ais. The amplitude is. The period is 80. 90 0 3 90 80 70 360 50 Appl the translation. cos The function is reflected in -ais. The amplitude is. The period is 80. The ais is. There is no horizontal shift. b) The transformed function g() 3 sin A 5B is not in the general form we work with, so we must factor the argument of the function. g() 3 sin a 5b g() 3 sin a ( 30)b The argument of a function is the input to the function. For the function g() 3 sin A 5B, the argument is A 5B. Copright 009 b Nelson Education Ltd. The parent function f () sin undergoes a vertical compression of factor 3, a horizontal stretch of factor, and a horizontal translation 30 to the right. 360 The amplitude of the curve is 3, the period is 70, and the ais of the curve is 0. There is a horizontal shift 30 to the right. 80 0 80 360 50 70 Start with the parent function f () sin. The amplitude is. The period is 360. R Transformations of Trigonometric Functions: Review of Essential Skills and Knowledge 565

80 0 80 360 50 70 Appl the compression and stretch. There is no reflection. 3 sin a b The amplitude is 3. The period is 360. 80 0 80 360 50 70 Appl the translation. g() 3 sin a ( 30)b The amplitude is 3. The period is 360. The ais is 0. There is a horizontal shift of 30 to the right. Practising. For each of the following transformed functions, identif the parent function. Describe the transformations that have been applied to create the transformed function. State the amplitude, period, and ais, and determine whether there is a horizontal shift. Then sketch a graph of the transformed function. a) b) f () sin() f () cos(3 90) c) sin c ( 60)d 3 Copright 009 b Nelson Education Ltd. 566 Advanced Functions: Review of Essential Skills and Knowledge

R 3 Solving Trigonometric Equations in Degrees Trigonometric functions can have man solutions, due to their periodic nature. The number of solutions for a function depends on the domain of the function. The solutions can be found using various approaches, as shown in the following eamples. EXAMPLE Solve USING SPECIAL TRIANGLES AND THE CAST RULE sin u 0.5 where u30, 360. Since sin u r opposite hpotenuse, we can see that u 30 is a solution. The terminal arm for 30 is in quadrant I. S A ( 3, ) ( 3, ) 50 30 3 3 The value of 0.5 is recognizable as a special value from the special 306090 triangle. The CAST rule states that sine values are positive for first quadrant and second quadrant angles. The terminal arms of these angles are mirror images in the -ais. T C Copright 009 b Nelson Education Ltd. The angle in quadrant II is 80 30 50. There are no other values of u in the domain 30, 360. Therefore, the solutions are u 30 or 50. R 3 Solving Trigonometric Equations in Degrees: Review of Essential Skills and Knowledge 567

EXAMPLE USING A CALCULATOR AND THE CAST RULE Solve cos u 0.8 for u, to the nearest degree, where u30, 360. The cosine ratio is negative in quadrants II and III. Since 0.8 is not from a special triangle, we use a calculator to determine the related acute angle. The related acute angle is about 37. S A 3 T C The angle u in quadrant II is 80 37 3. The angle u in quadrant III is 80 37 7. Therefore, the two solutions in the required domain are 3 and 7. EXAMPLE 3 Solve USING THE --r DEFINITIONS AND A CALCULATOR tan u for u, to the nearest degree, where u30, 360, so we can use (, ) (,) and (, ) as the points on the terminal arm. B definition, tan u. Copright 009 b Nelson Education Ltd. 568 Advanced Functions: Review of Essential Skills and Knowledge

S (, ) A Determine the inverse cosine of to find the related acute angle. 7 97 63 T (, ) C The related acute angle is about 63. The diagram shows that the angles in the required domain are u 360 63 97 and u 80 63 7 Therefore, the two solutions in the required domain are 7 and 97. Copright 009 b Nelson Education Ltd. Practising. Solve each trigonometric function, to the nearest degree, where 0 u 360 0. a) cos u 3 c) tan u d) sin u 0.55 b) tan u 5 d) sin u f) cos u5.5. Solve, where 0 u 360 0. a) cos u 0 c) sin u b) 3 tan u 3 d) 3 cos u R 3 Solving Trigonometric Equations in Degrees: Review of Essential Skills and Knowledge 569

R Proving Trigonometric Identities An identit is an equation that is true for all possible values of its variable. To disprove an identit (in other words, to prove that an equation is not an identit), we need to find onl one value of the variable that does not satisf the equation. To prove that an equation is an identit, we need to show that the two sides of the equation are equal for all possible values of the variable. To do this, we need to rewrite one or both sides of the equation b substituting known identities and/or using algebraic techniques. Algebraic Techniques factoring Reciprocal Identities Quotient Identities csc u sin u, tan u sin u cos u, sin u 0 cos u 0 Pthagorean Identities cos u sin u epanding and simplifing adding or subtracting rational epressions using a common denominator sec u cos u, cot u cos u sin u, cos u 0 sin u 0 cot u tan u, tan u 0 tan u sec u cot u sec u EXAMPLE Show that cos u sin u is not an identit. To prove that this equation is not an identit, we need to find a value of u that does not satisf the equation. Tr u 5. Left side cos 5 sin 5 Right side Copright 009 b Nelson Education Ltd. For u 5, the equation is not satisfied. Therefore, the equation cos u sin u is not an identit. 570 Advanced Functions: Review of Essential Skills and Knowledge

EXAMPLE Prove each identit. tan u a) b) cos u sin u sin u tan u tan u sin u cos u Copright 009 b Nelson Education Ltd. a) To prove the identit, we must work with each side independentl. sin u LS tan u RS cos u sin u Since we have shown that the left side and right side are equal, the equation is an identit ever possible value of u will satisf the equation. There are values of u that will result in each side being undefined, however, so a restriction is needed: cos u 0. b) Work with each side independentl. LS tan u tan u sin u cos u Since the left side and right side are equal, the equation is an identit. Again, restrictions are needed to avoid undefined values: sin u 0, cos u 0. Practising sin u cos u cos u sin u cos u sin u cos u cos u sin u sin u sin u cos u cos u sin u cos u RS sin u cos u sin u cos u. Prove each of the following identities. a) sin u tan u cos u cos u c) cos u sin u cos u tan u b) sin u cos u5 sin 3 u cos u d) tan u tan u sin u. Prove that sin a b, where sin 0. tan 3. Prove that ( cos )( cos ) sin sin. A Proving Trigonometric Identities: Review of Essential Skills and Knowledge 57