Mathematics 30-1: Co-terminal Angles & Arc Length Exercises

SO 1 Examples Students who achieve the acceptable standard should be able to answer all of the following questions, except for any part of a question labelled SE . Parts labelled SE are appropriate examples for students who achieve the standard of excellence. Note: In the multiple-choice questions that follow, * indicates the correct answer. Please be aware that the worked solutions show possible strategies; there may be other strategies that could be used. 1. An angle, in radians, that is co-terminal with 30° is A. 6 5 r - B. 6 13 r - C. 6 7 r 25 r *D. 6

Alberta Education, Provincial Assessment Sector 61 Mathematics 30–1 SO 1 Use the following information to answer question 2. An angle, i , in standard position, is shown below. 2. The best estimate of the rotation angle i is A. 1.25 radians B. 3.12 radians *C. 4.01 radians D. 5.38 radians

Alberta Education, Provincial Assessment Sector 62 Mathematics 30–1 Use the following information to answer numerical-response question 3. Mary is given the diagram below, showing an angle rotation of 120°. The arc length of the sector is 40 cm. Statement 1 The radius of the circle, to the nearest centimetre, is 19 cm. Statement 2 An equivalent angle rotation is 3 4 r . Statement 3 If the arc length on this circle increases to 80 cm, then the central angle must be 240°. Statement 4 Mary can determine the radius of the circle by dividing the given angle by the arc length. Numerical Response SO 1 3. The two statements above that are correct are numbered and . Solution: 13 or 31 Statement 1: Correct i = r a Statement 2: Incorrect ° ° 1 20 180 # r = 3 2 r r = 40 # r d n ° ° 120 180 r . 19 cm Statement 3: Correct i = r a i = 19 80 # r c 180 i = 240° Statement 4: Incorrect The radius of a circle is determined by dividing the arc length by the given angle. Alberta Education, Provincial Assessment Sector 63 Mathematics 30–1

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Use the following information to answer numerical-response questions 4 and 5. A circle with a radius r , an indicated arc length of 34 r , and two central angles of a 15 r and 17 r is shown below. 15 Numerical Response SO 1 4. For the angle , a 15 r the value of a is . Solution: 13 a 17 + r r = 2 r 15 15 a 17 r r - r = 2 15 15 a 13 r r = 15 15 Therefore, a = 13.

Alberta Education, Provincial Assessment Sector 64 Mathematics 30–1 Numerical Response SO 1 5. The length of the radius, r , of the circle, to the nearest whole number, is . Solution: 30 i = r a r = 34 17 r r 15 Therefore, r = 30. Alberta Education, Provincial Assessment Sector 65 Mathematics 30–1 Use the following information to answer numerical-response question 6.

Point A , 2 3 2 1 d n and Point B , 2 2 2 2 d n - lie on the terminal arm of two different angles in standard position on the circle. The angle, i , where 0 < < i r , can be expressed in the form n m r . Numerical Response SO 1 6. SO 2 The values of m and n are, respectively, and . Solution: In standard position, angle 6 r has a terminal arm that passes through Point A , 2 3 2 1 d n and the angle 4 3 r has a terminal arm that passes through Point B , 2 2 2 2 d n - . 3 r r i = - 4 7 r i = 12 6 m = 7 and n = 12 Alberta Education, Provincial Assessment Sector 66 Mathematics 30–1

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SO 2 SO 2 SO 2 SO 3 Use the following information to answer numerical- response question 7. If the point P (0.2, k ) lies on a circle with a centre at the origin and a radius of 1, then the exact value of k can be expressed as Numerical Response 7. The value of b , to the nearest hundredth, is . Solution: 0.96 x 2 + y 2 = r 2 (0.2) 2 + k 2 = (1) 2 k = ! 0 9. 6 b = 0.96 8. The terminal arm of i , when drawn in standard position, contains the point P ( x , y ) where P is on the unit circle. If sin 10 7 i = and tan 0 i 1 , then what is the value of x ? *A. 10 51 - B. 10 51 C. 10 3 D. 10 3 - 9. On a unit circle, Point P , 13 5 13 12 c m - lies on the terminal arm of angle i in standard position. What are the exact values of the 6 trigonometric ratios for angle i ? Possible solution: Since Point P is on the unit circle, the x- coordinate is the ratio for cosine, the y- coordinate is the ratio for sine, and x y is the ratio for tangent. sin c; ; os tan c; ; sc sec c; ot 13 12 13 5 5 12 12 13 5 13 12 5 ` i i = = - i i = - = = i i - = - Alberta Education, Provincial Assessment Sector 67 Mathematics 30–1

SO 3 Use the following information to answer question 10. 10. Given that csc 5 8 i = , where < < 2 r i r , determine the exact value of tan i . Possible solution: x 2 + (5) 2 = 8 2 x 2 = 39 x = ! 39 Since i is a second quadrant angle, x = - 39. 5 39 i i = - - or = . 5 SO 3 Therefore, tan tan 39 39 11. The exact value of sin cos 6 4 7 + r r d- n d n is *A. 2 2 1 - B. 2 2 1 + 3 2 + C. 2 3 2 - D. 2

Alberta Education, Provincial Assessment Sector 68 Mathematics 30–1 SO 3 SO 5 SO 3 12. If tan 2 5 i = , where 0 2 # i < r , then the largest positive value of i , to the nearest tenth, is rad. 1 5 i = - d n Possible solution: tan 2 i . 1 1. 9 Tangent is positive in quadrants I and III; therefore, the largest positive value for the angle in the given domain is i = r + 1 1. 9 i . 4 3. Use the following information to answer numerical- response question 13. Each of the trigonometric ratios listed below results undefined. tan 2 r b l cot 2 3 r b l sin r csc 2 ` j r Numerical Response 13. Use the following code to indicate that the value of the ratio is zero or that the ratio is undefined. 1 = The value of the ratio is zero. 2 = The ratio is undefined. b l cot 2 3 r b l sin r csc 2 ` j r Ratio: tan 2 r Solution: 2112

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Alberta Education, Provincial Assessment Sector 69 Mathematics 30–1 SO 1 SO 3 Use the following information to answer numerical- response question 14. For the angles 6 r , 6 5 r , 6 7 r , and 6 11 r , the following statements are given. Statement 1 These angles in degrees are, respectively, 30°, 150°, 210°, and 300°. Statement 2 They are all part of the general solution Statement 3 The values of sin 6 7 r b l and Numerical Response 14. The statement that is true from the list above is number . Solution: 3 Statement 1: False ° ° 6 11 180 # 330 r r = Statement 3: True Sine is negative in Quadrant III and cosine is negative in Quadrant II. Numerical Response Statement 2: False The general solution n 6 + 2 r i = r , n I ! does not include 6 5 r , 6 7 r , and 6 11 r . SO 4 15. For the function y = a d cos i + , where the range is [–4, 10], the values of a and d are, respectively, _____ and _____.

Solution: a = 7 and d = 3 Since the maximum value is 10 and the minimum value is –4, the amplitude is 10 - - 4 . ( ) 2 The vertical displacement is maximum – amplitude = 10 – 7. Alberta Education, Provincial Assessment Sector 70 Mathematics 30–1 SE SO 4 16. A function is represented by the equation y = sin( ) 3 7 x + + r . The numerical value of the phase shift, when compared to y = sin x, is i and the period of the corresponding graph is ii . The statement above is completed by the information in row Row i ii A. r 3 B. r 2 r 3 *C. r 3 2 r 3 D. r 3 3 Note : This item is SE since the b value must be factored out in order to identify the phase shift. Numerical Response SO 4 17. Given that f ( ) i i = cos( ) n has the same period as the graph of g ( ) i i = tan , the value of n is . Solution: 2 Period of g ( ) i i = tan is r . Period of f ( ) i i = cos( ) n is n 2 r . In order for the two functions to have the same period of r , n = 2.

Alberta Education, Provincial Assessment Sector 71 Mathematics 30–1 Use the following information to answer numerical-response question 18. The partial graph of the cosine function below has a minimum point at , 2 2 r b - l and a maximum point at ( , r 8) as shown below. The equation of the function can be expressed in the form y = - a b cos _ i ( ) x c + d , a , , b c , d W d . Numerical Response SE 18. With a minimum possible phase shift, the values of a , b , c , and d are, respectively, SO 4 , , , and . Solution: ( ) a 2 8 2 = 5 - - = period b 2 2 2 r r r = = = c = 0 d = - 8 5 = 3 ` solution is 5203 Note : This item is SE since all 4 parameters must be determined. Alberta Education, Provincial Assessment Sector 72 Mathematics 30–1

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SO 4 Use the following information to answer question 19. For the graph of the function f ( ) x d = - 3 sin were made. Statement 1 The amplitude is 3. Statement 2 The maximum value is ( d – 3). Statement 3 The period is 2 r . Statement 4 When compared to the graph of y = f x ( ) has been horizontally translated 5 units to the right. Statement 5 If d > 3, then the graph of y 19. The true statements are A. 1, 5 only B. 2, 3 only *C. 1, 4, 5 only D. 2, 4, 5 only Possible solution: Statement 1: True The value of a identifies the amplitude of the function. Statement 3: False The period is 2 2 r = r . Statement 5: True If the vertical displacement is greater than the amplitude, then the graph will not touch or cross the x- axis. Statement 2: False The value of d identifies the vertical displacement. The maximum value is ( d + 3). Statement 4: True The value of c identifies the phase shift. Alberta Education, Provincial Assessment Sector 73 Mathematics 30–1

SE SO 4 SO 5 Use the following information to answer question 20. The height of a point on a Ferris wheel, time, t , in seconds can be represented by a sinusoidal function. The maximum height of the Ferris wheel above the ground is 17 Ferris wheel 60 seconds to complete two full rotations. 20. Assuming that the particular point starts at the minimum height above the ground, write an equation for the height of this point on the Ferris wheel, h , as a function of time, t , in the form h a = - cos 7 b t ( ) c d A + . Possible solution: Amplitude is 8 m. Period is 30 s. 17 1 = 8 - = a 2 30 b 2 15 r r = = c = 15 s to the right or to the left = ! 15 17 1 9 + = = d 2 h t 8 cos ( ) 15 ` 15 + 9 r = < - F or h t 8 cos ( ) 15 + + 15 9 r = < F Note : This item is SE since all 4 parameters of the function must be determined. 21. The values of i , where 180° ° # i < 360 , in the equation 2 0 cos cos 2 i i + = , are A. i = 90°,120° *B. i = 240°, 270° C. i = 60°, 90°, 270°, 300° D. i = 90°,120°, 240°, 270° Alberta Education, Provincial Assessment Sector 74 Mathematics 30–1

SE SO 5 SE SO 5 Use the following information to answer question 22. cos An incorrect solution to the equation cos Step 1 cos cos csc 1 1 i = Step 2 csc Step 3 csc i = 1 Step 4 cos 1 i = Step 5 i = 0°,180° 22. The step that records the first mistake in the solution is *A. Step 2 B. Step 3 C. Step 4 D. Step 5 23. Determine a general solution of cot 1 0 2 i - = , expressed in radians. Possible solution: cot 1 2 i = cot i = ! 1 tan 1 i = ! r r r r 7 i = , , , , etc. 4 4 3 4 5 4 Therefore, the general solution is n

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4 2 + r r i = , n I ! . Note : This item is SE since a general solution for a second-degree equation is required. Alberta Education, Provincial Assessment Sector 75 Mathematics 30–1 SE SO 5 SE SO 5 24. Determine the solution for the equation 2 cos s x x in 1 0 2 + - = , where - r r # # x . Possible solution: 2 cos s x x in 1 0 2 + - = 2 1 1 sin s x x in 0 2 _ i - - + = 2 2 1 sin s x x in 0 2 - - + = 2 sin s x x in 1 0 2 - - = ( ) 2 1 sin s x x + ( ) in - = 1 0 sin x 2 1 = - or sin x = 1 x , , 6 5 6 2 r r r = - - Note: This item is SE since it involves a Pythagorean identity substitution and since the restricted domain is outside [0 2, ] r . 25. Graphically solve for i , where –180° # # i 0°, given ` 2 3 - sec s i i j _ ec + 3 0 i = . State answers to the nearest degree. Possible solution: i = –109° and –30° Enter the following function into the calculator. 2 3 1 1 = e - o d + 3 n cos cos y x x A window that could be used is x : [–180, 0, 30], y : [– 5, 5, 1]. The x- intercepts are the solutions to the original equation.

Alberta Education, Provincial Assessment Sector 76 Mathematics 30–1 SO 5 Use the following information to answer question 26. A Mathematics 30–1 class was asked to determine a general solution to the equation 2 sin c i i os - cos i = 0, in degrees. The answers provided by four different students are shown below. Student 1 i = 60° ° + n (120 ), n I d Student 2 i = 90° ( + n 360°), n I d , and Student 3 i = n ( ° 180 ), i = 60° ( + n 4 i = 90° ( + n 180°), i = 30° ( + n 360°), and 26. The two students who provided a correct general solution are numbered A. 1 and 3 B. 1 and 4 C. 2 and 3 *D. 2 and 4 Note: This item is SE since it involves determining a general solution for a second-degree equation.

Alberta Education, Provincial Assessment Sector 77 Mathematics 30–1 SO 6 1 2 i i - = - , where cos 0 i ! . si n cos entity cos i a) Student A substituted 3 r i = into both sides of the equation and got LS = RS. Student B entered LS into y 1 and RS into y 2 and concluded that the graphs are exactly the same. Explain why these methods are not considered a proof of this identity. Possible solution: Student A is verifying the identity using a single value of i , whereas a proof is valid for all permissible values of i . Student B is verifying that the graphs of each side of the identity are the same for all values of i , but is only examining a limited portion of both graphs. b) Student C correctly completed an algebraic process to show LS = RS. Show a process Student C might have used. sin 1 2 i i = - RS =- cos i Possible solution: LS cos cos 2 i = - cos i = - cos i ` LS = RS SE c) Which non-permissible values of i should be stated for this identity? Possible solution: cos 0 i ! , , etc. 2 2 3 ! r r i n 2 ! + r i r , n I !

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Note: This item is SE since since non-permissible values are required. Alberta Education, Provincial Assessment Sector 78 Mathematics 30–1 cot cs c x x SE 28. The expression sec + , for all permissible values of x , is equivalent to SO 6 A. sin x B. tan x C. csc x *D. cot x x cos + 1 x 1 cos x + 1 sin x + = Possible solution: sin x sin x 1 1 1 + cos x cos x + cos x cos x 1 x cos = sin 1 # + + x cos x cos = sin x x = cot x Note: This item is SE since it entails simplifying an identity involving rational operations.

Alberta Education, Provincial Assessment Sector 79 Mathematics 30–1 Use the following information to answer numerical-response question 29. Each trigonometric expression below can be simplified to a single numerical value. tan x x x - 1 sin sec 2 cot c x x sc 2 2 - 3 cos s x x in 7 1 7 2 2 1 + Numerical Response SO 6 29. When the numerical values of the simplified expressions are arranged in ascending order, the expression numbers are , , and . Solution: 213 sin cos x sin sin x x x x x x x 1 - = - = # - = 0 tan sin cos 1 sin sec x 2 cot c x x sc 1 2 2 - = - 3 cos s x x in cos s x x in 7 1 7 1 7 1 7 2 2 2 2 1 + + = = _ i

Alberta Education, Provincial Assessment Sector 80 Mathematics 30–1 SE SO 6 30. What is the exact value of tan 75°? 3 1 + Possible solution: 3 + 2 or 3 1 - tan 75° = tan(45° + 30°) - % % % % tan tan 45 + 30 = tan tan 1 45 30 = = 1 3 1 + 1 1 3 1 - e o 3 1 + 3 - 3 1 3 3 1 + = 3 3 1 + = 3 1 3 # - 3 1 - or 3 + 2 (rationalize the denominator) Note: This item is SE since it involves the sum identity of a tangent. 31. Prove algebraically that ( ) sin 2 2 x 2 2 2 - = - , where x , n n I 4 2 ! + ! SE tan x r r . SO 6 1 Possible solution: tan x x x cos sin Left Side Right Side 2 tan x sin ( ) 2 x

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1 2 - tan x 2 2 - cos sin x x tan(2 x ) sin cos (2 ) x (2 ) x tan(2 x ) LS = RS Note: This item is SE since it involves a double-angle tangent identity. Alberta Education, Provincial Assessment Sector 81 Mathematics 30–1 SO 3 SO 6 Use the following information to answer question 32. 32. Given that sin 7 2 i = - and cot 0 i < , determine the exact value of cos 3 2 r b l i - . Possible solution: x 2 + (–2) 2 = 7 2 x 2 = 45 x = ! 45 x = ! 3 5

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3 5 i = . Since i is in Quadrant IV, x = 3 5 and cos 7 cos cos cos sin sin 3 2 3 2 3 r r 2 r b l i i - = + i 3 = e o b b - - l l + e o 3 5 7 1 2 2 7 2 2 3 = - - 3 5 14 14 3 5 2 3 = - - 14 Alberta Education, Provincial Assessment Sector 82 Mathematics 30–1

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