4-6 Inverse Trigonometric Functions

Transcription

1 Find the exact value of each expression, if it exists. 1. sin 1 0 Find a point on the unit circle on the interval with a y-coordinate of arcsin Find a point on the unit circle on the interval with a y-coordinate of. When t = 0, sin t = 0. Therefore, sin 1 0 = arcsin Find a point on the unit circle on the interval with a y-coordinate of. When t =, sin t =. Therefore, arcsin =. 4. sin 1 Find a point on the unit circle on the interval with a y-coordinate of. When t =, sin t =. Therefore, arcsin =. When t =, sin t =. Therefore, sin 1 =. esolutions Manual - Powered by Cognero Page 1

2 5. Find a point on the unit circle on the interval with a y-coordinate of. 7. cos 1 Find a point on the unit circle on the interval with an x-coordinate of. When t =, sin t =. Therefore, sin 1 6. arccos 0 =. Find a point on the unit circle on the interval with an x-coordinate of 0. When t =, cos t =. Therefore, cos 1 =. 8. arccos ( 1) Find a point on the unit circle on the interval with an x-coordinate of 1. When t =, cos t = 0. Therefore, arccos 0 =. When t =, cos t = 1. Therefore, arccos ( 1)=. esolutions Manual - Powered by Cognero Page 2

3 9. Find a point on the unit circle on the interval with an x-coordinate of. 11. arctan 1 Find a point on the unit circle on the interval such that =1. When t =, cos t =. Therefore, =. 10. cos 1 Find a point on the unit circle on the interval with an x-coordinate of. When t =, tan t =. Therefore, arctan 1=. 12. arctan ( ) Find a point on the unit circle on the interval such that =. When t =, cos t =. Therefore, cos 1 =. When t =, tan t =. Therefore, arctan ( ) =. esolutions Manual - Powered by Cognero Page 3

4 13. Find a point on the unit circle on the interval 15. ARCHITECTURE The support for a roof is shaped like two right triangles, as shown below. Find θ. such that =. Use inverse trigonometric functions and the unit circle to solve. Find a point on the unit circle on the interval with a y-coordinate of. When t =, tan t =. Therefore, tan 1 =. 14. tan 1 0 Find a point on the unit circle on the interval When t =, sin t =. Therefore, sin 1 =. such that = 0. When t = 0, tan t =. Therefore, tan 1 0 = 0. esolutions Manual - Powered by Cognero Page 4

5 16. RESCUE A cruise ship sailed due west 24 miles before turning south. When the cruise ship became disabled and the crew radioed for help, the rescue boat found that the fastest route covered a distance of 48 miles. Find the angle θ at which the rescue boat should travel to aid the cruise ship. Sketch the graph of each function. 17. y = arcsin x First, rewrite y = arcsin x in the form sin y = x. Next, assign values to y on the interval to make a table of values. Use inverse trigonometric functions and the unit circle to solve. Find a point on the unit circle on the interval with an x-coordinate of. y x = sin y Plot the points and connect them with a smooth curve. When t =, cos t =. Therefore, cos 1 =. esolutions Manual - Powered by Cognero Page 5

6 18. y = sin 1 2x First, rewrite y = sin 1 2x in the form sin y = x. 19. y = sin 1 (x + 3) First, rewrite y = sin 1 (x + 3) in the form sin y = x. Next, assign values to y on the interval make a table of values. y to Next, assign values to y on the interval make a table of values. y x = sin y to Plot the points and connect them with a smooth curve. Plot the points and connect them with a smooth curve. esolutions Manual - Powered by Cognero Page 6

7 20. y = arcsin x 3 First, rewrite y = arcsin x 3 in the form sin y = x. 21. y = arccos x First, rewrite y = arccos x in the form cos y = x. Next, assign values to y on the interval make a table of values. y x = sin(y + 3) Plot the points and connect them with a smooth curve. to Next, assign values to y on the interval make a table of values. y x = cos y Plot the points and connect them with a smooth curve. to esolutions Manual - Powered by Cognero Page 7

8 22. y = cos 1 3x First, rewrite y = cos 1 3x in the form cos y = x. 23. y = arctan x First, rewrite y = arctan x in the form tan y = x. Next, assign values to y on the interval to Next, assign values to y on the interval make a table of values. y 0 0 to make a table of values. y x = tan y Plot the points and connect them with a smooth curve. Plot the points and connect them with a smooth curve. esolutions Manual - Powered by Cognero Page 8

9 24. y = tan 1 3x First, rewrite y = tan 1 3x in the form tan y = x. 25. y = tan 1 (x + 1) First, rewrite y = tan 1 (x + 1) in the form tan y = x. Next, assign values to y on the interval make a table of values. Note that x = tan y has to Next, assign values to y on the interval make a table of values. Note that x = tan y 1 has no x-values for y-values of and. to no x-values for y-values of and. y x = tan y 1 y x = tan y Plot the points and connect them with a smooth curve. Further investigation reveals that as x approaches negative infinity, y approaches, and as x approaches positive infinity, y approaches. Plot the points and connect them with a smooth curve. Further investigation reveals that as x approaches negative infinity, y approaches, and as x approaches positive infinity, y approaches. esolutions Manual - Powered by Cognero Page 9

10 26. y = arctan x 1 First, rewrite y = arctan x 1 in the form tan y = x. 27. DRAG RACE A television camera is filming a drag race. The camera rotates as the vehicles move past it. The camera is 30 meters away from the track. Consider θ and x as shown in the figure. The range of inverse tangent of x is. Since we are subtracting 1, the new range should be. Next, assign values to y within this range to make a table of values. x = tan (y y + 1) a. Write θ as a function of x. b. Find θ when x = 6 meters and x = 14 meters. a. The relationship between θ and the sides is opposite and adjacent, so tan θ = the inverse, θ = arctan. b.. After taking Plot the points and connect them with a smooth curve. Further investigation reveals that as x approaches negative infinity, y approaches , and as x approaches positive infinity, y approaches SPORTS Steve and Ravi want to project a pro soccer game on the side of their apartment building. They have placed a projector on a table that stands 5 feet above the ground and have hung a 12-foot-tall screen 10 feet above the ground. esolutions Manual - Powered by Cognero Page 10

11 screen 10 feet above the ground. b. The goal is to maximize θ, so if we graph, we can identify the maximum value of θ. The d-value that corresponds with this maximum is the distance. a. Write a function expressing θ in terms of distance d. b. Use a graphing calculator to determine the distance for the maximum projecting angle. a. Make a diagram of the situation. The d-value that corresponds with this maximum is about 9.2 feet. Find the exact value of each expression, if it exists. 29. We are asked to find θ in terms of d. However, we do not know the value of the angle α either. Using right triangle trigonometry, we can determine that tan α = and tan (θ + α) =. Now we have two equations, but still three variables. We need to find a way to eliminate α. If we can get α and θ + α isolated in each equation, we can eliminate α. The inverse property applies, because lies on the interval [ 1, 1]. Therefore, =. 30. The inverse property applies, because lies on the interval [ 1, 1]. Therefore, =. 31. The inverse property applies, because lies on the interval [ 1, 1]. Therefore, =. esolutions Manual - Powered by Cognero Page 11

12 32. cos 1 (cos π) The inverse property applies, because π lies on the interval. Therefore, cos 1 (cos π)= π First, find cos. On the unit circle, corresponds to (0, 1). So, cos = 0. The inverse property applies, because lies on the Next, find sin 1 0. The inverse property applies, because 0 is on the interval [ 1, 1]. Therefore, sin 1 interval. Therefore, =. 0 = 0, and = The inverse property applies, because lies on the interval. Therefore, =. 35. cos (tan 1 1) First, find tan 1 1. The inverse property applies, because 1 is on the interval. Therefore, First, find cos 1. To do this, find a point on the unit circle on the interval [0, 2π] with an x-coordinate of. When t =, cos t =. Therefore, cos 1 =. Next, find or sin. On the unit circle, tan 1 1 =. Next, find cos. On the unit circle, corresponds to. So, cos =. corresponds to (0, 1). So, sin = 1. = 1, and cos (tan 1 1) = esolutions Manual - Powered by Cognero Page 12

13 38. sin (tan 1 1 sin 1 1) First, find tan 1 1. To do this, find a point on the unit circle on the interval [0, 2π] with an x-coordinate 39. cos (tan 1 1 sin 1 1) Find tan 1 1. Find a point on the unit circle on the interval [0, 2π] with an x-coordinate equal to the y- equal to the y-coordinate. When t = =. Therefore, tan 1 1 =., cos t = sin t coordinate. When t =, cos t = sin t =. Therefore, tan 1 1 =. Next, find sin 1 1. To do this, find a point on the unit circle on the interval [0, 2π] with a y-coordinate of 1. When t =, sin t = 1. Therefore, sin 1 1 =. Find sin 1 1. Find a point on the unit circle on the interval [0, 2π] with an y-coordinate of 1. When t =, sin t = 1. Therefore, sin 1 1 =. Find. On the unit circle, corresponds to. So, =. Find. On the unit circle, corresponds to. So, =. sin (tan 1 1 sin 1 1) = cos (tan 1 1 sin 1 1) = esolutions Manual - Powered by Cognero Page 13

14 40. Find cos 1 0. Find a point on the unit circle on the interval [0, 2π] with an x-coordinate equal 0. When t =, cos t = 0. Therefore,. Find sin 1. Find a point on the unit circle on the interval [0, 2π] with an y-coordinate of. When t =, sin t =. Write each trigonometric expression as an algebraic expression of x. 41. tan (arccos x) Let u = arccos x, so cos u = x. Because the domain of the inverse cosine function is restricted to Quadrants I and II, u must lie in one of these quadrants. The solution is similar for each quadrant, so we will solve for Quadrant I. Draw a diagram of a right triangle with an acute angle u, an adjacent side length x and a hypotenuse length 1. Therefore,. Find. On the unit circle, corresponds From the Pythagorean Theorem, the length of the side opposite u is. Now, solve for tan u. to. So,. =. So, tan (arccos x) =. esolutions Manual - Powered by Cognero Page 14

15 42. csc (cos 1 x) Let u = cos 1 x, so cos u = x. Because the domain of the inverse cosine function is restricted to Quadrants I and IV, u must lie in one of these quadrants. The solution is similar for each quadrant, so we will solve for Quadrant I. Draw a diagram of a right triangle with an acute angle u, an adjacent side length x and a hypotenuse length sin (cos 1 x) Let u = cos 1 x, so cos u = x. Because the domain of the inverse cosine function is restricted to Quadrants I and II, u must lie in one of these quadrants. The solution is similar for each quadrant, so we will solve for Quadrant I. Draw a diagram of a right triangle with an acute angle u, an adjacent side length x and a hypotenuse length 1. From the Pythagorean Theorem, the length of the side opposite to u is. Now, find csc u. From the Pythagorean Theorem, the length of the side opposite u is. Now, solve for sin u. So, csc (cos 1 x) =. So, sin (cos 1 x) =. esolutions Manual - Powered by Cognero Page 15

16 44. cos (arcsin x) Let u = arcsin x, so sin u = x. Because the domain of the inverse cosine function is restricted to Quadrants I and II, u must lie in one of these quadrants. The solution is similar for each quadrant, so we will solve for Quadrant I. Draw a diagram of a right triangle with an acute angle u, an opposite side length x and a hypotenuse length csc (sin 1 x) Let u = sin 1 x, so sin u = x. Because the domain of the inverse cosine function is restricted to Quadrants I and II, u must lie in one of these quadrants. The solution is similar for each quadrant, so we will solve for Quadrant I. Draw a diagram of a right triangle with an acute angle u, an opposite side length x and a hypotenuse length 1. From the Pythagorean Theorem, the length of the side adjacent to u is. Now, solve for cos u. From the Pythagorean Theorem, the length of the side adjacent to u is. Now, solve for csc u. So, cos (arcsin x) =. So, csc(sin 1 x) =. esolutions Manual - Powered by Cognero Page 16

17 46. sec (arcsin x) Let u = arcsin x, so sin u = x. Because the domain of the inverse cosine function is restricted to Quadrants I and IV, u must lie in one of these quadrants. The solution is similar for each quadrant, so we will solve for Quadrant I. Draw a diagram of a right triangle with an acute angle u, an opposite side length x and a hypotenuse length cot (arccos x) Let u = arccos x, so cos u = x. Because the domain of the inverse cosine function is restricted to Quadrants I and II, u must lie in one of these quadrants. The solution is similar for each quadrant, so we will solve for Quadrant I. Draw a diagram of a right triangle with an acute angle u, an adjacent side length x and a hypotenuse length 1. From the Pythagorean Theorem, the length of the side adjacent to u is. Now, solve for sec u. From the Pythagorean Theorem, the length of the side opposite u is. Now, solve for cot u. So, sec(arcsin x) =. So, cot (arccos x) =. esolutions Manual - Powered by Cognero Page 17

18 48. cot (arcsin x) Let u = arcsin x, so sin u = x. Because the domain of the inverse cosine function is restricted to Quadrants I and II, u must lie in one of these quadrants. The solution is similar for each quadrant, so we will solve for Quadrant I. Draw a diagram of a right triangle with an acute angle u, an opposite side length x and a hypotenuse length f (x) = cos x and g(x) = 3(cos x 2) g(x) is of the form 3f [(x) 2]. The 2 represents a translation down while the 3 represents a vertical expansion after the translation. Therefore, the translation down will be 6 units. 52. f (x) = arcsin x and g(x) = arcsin (x + 2) g(x) is of the form f (x + 2). The 2 represents a translation to the left while the represents a vertical compression. 53. f (x) = arccos x and g(x) = 5 + arccos 2x From the Pythagorean Theorem, the length of the side adjacent to u is. Now, solve for cot u. g(x) is of the form f (2x) + 5. The 2 represents a horizontal compression while the 5 represents a translation up. 54. f (x) = tan x and g(x) = tan 3x 4 So, cot(arcsin x) =. Describe how the graphs of g(x) and f (x) are related. 49. f (x) = sin x and g(x) = sin (x 1) 2 g(x) is of the form f (x 1) 2. The 1 represents a translation to the right while the 2 represents a translation down. g(x) is of the form f (3x) 4. The 3 represents a horizontal compression while the 4 represents a translation down. 55. SAND When piling sand, the angle formed between the pile and the ground remains fairly consistent and is called the angle of repose. Suppose Jade creates a pile of sand at the beach that is 3 feet in diameter and 1.1 feet high. 50. f (x) = arctan x and g(x) = arctan 0.5x 3 g(x) is of the form 0.5f (x) 3. The 0.5 represents a horizontal expansion while the 3 represents a translation down. a. What is the angle of repose? b. If the angle of repose remains constant, how many feet in diameter would a pile need to be to reach a height of 4 feet? a. Draw a diagram to model this situation. esolutions Manual - Powered by Cognero Page 18

19 Draw a diagram to model this situation. Use the tangent function to find θ. Give the domain and range of each composite function. Then use your graphing calculator to sketch its graph. 56. y = cos (tan x) The domain of cos x is {x x } and the range of tan x falls within this domain, so there are no further restrictions on the domain. The domain of tan x is also {x x }, so the domain of the composite function is {x x }. The range of tan x is, so this becomes Therefore, the angle of repose is about 36º. b. Draw a diagram to model this situation, where the height of the triangle is 4 ft and angle of repose is 36º. Use the tangent function to find x. the domain of cos x, or the limit of the input values for cos x. The only corresponding output values for these input values is {y 0 < y 1}. Therefore, the range of the composite function is {y 0 < y 1}. 57. y = sin (cos x) The domain of sin x is {x x } and the range of cos x falls within this domain, so there are no further restrictions on the domain. The domain of cos x is [, 1] so the domain of the composite function is restricted to {x x 1}. The range of cos x is [0, π], so this becomes the domain of sin x, or the limit of the input values for sin x. The only corresponding output values for these input values is {y 0 y 1}. Therefore, the range of the composite function is {y 0 y 1}. The pile would reach 4 feet if the diameter was about 2(5.5) or 11 feet. esolutions Manual - Powered by Cognero Page 19

20 58. y = arctan (sin x) The domain of arctan x is {x x } and the range of sin x falls within this domain, so there are no further restrictions on the domain. The domain of sin x is also {x x }, so the domain of the composite function is {x x }. The range of sin x is [, 1], so this becomes the domain of arctan x, or the limit of the input values for arctan x. The only corresponding output values for these input values is. Therefore, the range of the composite function is. 59. y = sin (cos x) The domain of sin x is {x x } and the range of cos x falls within this domain, so there are no further restrictions on the domain. The domain of cos x is also {x x }, so the domain of the composite function is {x x }. The range of cos x is [, 1], so this becomes the domain of sin x, or the limit of the input values for sin x. The only corresponding output values for these input values is. Therefore, the range of the composite function is. 60. y = cos (arcsin x) The domain of cos x is {x x } and the range of arcsin x falls within this domain, so there are no further restrictions on the domain. The domain of arcsin x is {x x 1}, so the domain of the composite function is restricted to {x x 1}. The range of arcsin x is, so this becomes the domain of cos x, or the limit of the input values for cos x. The only corresponding output values for these input values is {y 0 y 1}. Therefore, the range of the composite function is {y 0 y 1}. 61. y = tan (arccos x) The domain of tan x is and the range of arccos x is [0, π], so the domain is restricted to. The domain of arccos x is {x x 1}, so the domain of the composite function is further restricted to {x x 1, x 0}. The range of arccos x is[0, π], so this becomes the domain of tan x, or the limit of the input values for tan x. The corresponding output values for these input values is {y y 0}. Therefore, the range of the composite function is{y y 0}. 62. INVERSES The arcsecant function is graphed by restricting the domain of the secant function to the intervals and, and the arccosecant function is graphed by restricting the domain of the cosecant function to the intervals and. a. State the domain and range of each function. b. Sketch the graph of each function. c. Explain why a restriction on the domain of the secant and cosecant functions is necessary in order to graph the inverse functions. esolutions Manual - Powered by Cognero Page 20

21 a. If the domain of secant is restricted to, then the range of arcsecant must also be because they are inverses. The domain of arcsecant that produces this range is D = (, ] [1, ). This is determined by finding the values for secant produced by the endpoints of the domain of secant. So, sec π = and sec 0 = 1. Also, as θ approaches from the left, sec θ approaches, and as θ approaches from the right, sec θ approaches. c. If restrictions are not put on the domain of the secant and cosecant functions when graphing the inverses, similar to the inverses of sine, cosine, and tangent, the inverses will not be functions. Without domain restrictions, y = sec x and y = csc x are no longer one-to-one and, therefore, no longer have inverses. If the domain of cosecant is restricted to must also be, then the range of arccosecant because they are inverses. The domain of arccosecant that produces this range is D = (, ] [1, ). This is determined by finding the values for cosecant produced by the endpoints of the domain of cosecant. So, csc = 1 and. Also, as θ approaches 0 from the left, csc θ approaches, and as θ approaches 0 from the right, csc θ approaches. b. arcsecant arccosecant esolutions Manual - Powered by Cognero Page 21

22 63. Write each algebraic expression as a trigonometric function of an inverse trigonometric function of x. Recall from example 8 in this lesson that we can draw a diagram of a right triangle with an acute angle u and a hypotenuse length Recall from example 8 in this lesson that we can draw a diagram of a right triangle with an acute angle u and a hypotenuse length 1. The six main trigonometric ratios are all ratios of some combination of the three sides of a right triangle. From our diagram, all we know is the hypotenuse is equal to 1. Since we need to find a trigonometric function of something that gives us, we can arbitrarily let one side of the triangle represent the numerator, x, and the other side represent the denominator,. The six main trigonometric ratios are all ratios of some combination of the three sides of a right triangle. From our diagram, all we know is the hypotenuse is equal to 1. Since we need to find a trigonometric function of something that gives us, we can arbitrarily let one side of the triangle represent the numerator, x, and the other side represent the denominator,. According to our new labels in the graph, represents, or tan u. Also, sin u = =, so x = sin u and u = sin x. Finally, we have tan u = tan (sin x) =. Note that if we had switched the labels of the sides, we would have had cot (cos x). According to our new labels in the graph, represents, or cot u. Also, sin u = =, so x = sin u and u = sin x. Finally, we have cot u = cot (sin x) =. Note that if we had switched the labels of the sides, we would have had tan (cos x). 65. MULTIPLE REPRESENTATIONS In this problem, you will explore the graphs of compositions of trigonometric functions. a. ANALYTICAL Consider f (x) = sin x and f (x) esolutions Manual - Powered by Cognero Page 22

23 = arcsin x. Describe the domain and range of f o f and f o f. b. GRAPHICAL Create a table of several values for each composite function on the interval [ 2, 2]. Then use the table to sketch the graphs of f and f f. Use a graphing calculator to check your graphs. c. ANALYTICAL Consider g(x) = cos x and g (x) = arccos x. Describe the domain and range of g g and g g and make a conjecture as to f what the graphs of g g and g g will look like. Explain your reasoning. d. GRAPHICAL Sketch the graphs of g g and g g. Use a graphing calculator to check your graphs. e. VERBAL Make a conjecture as to what the graphs of the two possible compositions of the tangent and arctangent functions will look like. Explain your reasoning. Then check your conjecture using a graphing calculator. a. f f represents sin (arcsin x). The domain of arcsin x is [, 1], so the domain of sin (arcsin x) is also [, 1]. The range of sin x is [, 1], so the range of sin (arcsin x) is also [, 1]. f represents arcsin (sin x). The domain of sin x is (, ), so the domain of arcsin (sin x) is also (, ). The range of arcsin is limited to f also., so the range of arcsin (sin x) is f f f o f c. f f represents cos (arccos x). The domain of arccos x is [, 1], so the domain of cos (arccos x) is also [, 1]. The range of cos x is [, 1], so the range of cos (arccos x) is also [, 1]. f represents arccos (cos x). The domain of cos x is (, ), so the domain of arccos (cos x) is also (, ). The range of arccos is limited to [0, π], so the range of arccos (cos x) is also [0, π]. f The graph of g g should be the line y = x for x 1. The inverse property of trigonometric functions states that on the closed interval [, 1], cos (cos x) = x. The graph of g g should be the line y = x for 0 x π. Once the graph reaches π, it will turn and decrease until it reaches the x-axis at the same rate. When it reaches the x-axis, it will turn again and increase until it reaches π. It will continue to do this as x approaches infinity. b. d. g g f f g g esolutions Manual - Powered by Cognero Page 23

24 4-6 Inverse Trigonometric Functions g g 66. ERROR ANALYSIS Alisa and Trey are discussing inverse trigonometric functions. Because tan x =, Alisa conjectures that tan x =. Trey disagrees. Is either of them correct? Explain. e. Sample answer: For f (x) = tan (tan x), due to the inverse property of trigonometric functions, for all values of x, f (x) = x. This should result in the line y = x for all real numbers. The graph of g(x) = tan (tan x) will be different because the tan x is undefined for multiples of π. As a result, asymptotes for multiples of π can be expected. We can also due to the definition of expect a range of Trey is correct. Alisa should not assume that all relationships of trigonometric functions apply to their inverses. She should also not make the mistake of thinking that sin and tan = x=, cos x=,, and using these false relationships to prove her incorrect assumption. Trigonometric values of angles are related to each other. However, when we find sin x, cos x, and tan x, we are calculating angle measures. Angle measures by themselves do not have this unique relationship. arctan. f(x) = tan (tan x) 67. CHALLENGE Use the graphs of y = sin x and y = cos x to find the value of sin x + cos the interval [, 1]. Explain your reasoning. x on g(x) = tan (tan x) a. Graph y = sin x and y = cos x. 66. ERROR ANALYSIS Alisa and Trey are discussing inverse trigonometric functions. Because tan x =, Alisa conjectures that tan x =. Trey disagrees. Is either of them x = 0.5: x = 1: correct? Explain. Trey is correct. Alisa should not assume that all relationships of trigonometric functions apply to their inverses. She should also not make the mistake of thinking that sin x=, cos x= x = :, esolutions Manual - Powered by Cognero and tan = Page 24, and using these false If we make a table of values, we can see that sin

25 x = : 68. REASONING Determine whether the following statement is true or false: If cos =, then If we make a table of values, we can see that sin x + cos x = in the interval [, 1]. x sin x 0 cos x sin x + cos x The graph of y = sin 1 x + cos 1 x supports this conjecture. Every y-value appears to be for the given domain. π x sin x - cos x 0 - sin x + cos x - cos =. Explain your reasoning. False; sample answer: does not fall within the range of the inverse. Recall that arccos is restricted to the upper half of the unit circle. The value of θ for cos = θ is. y = cos x is one-to-one only for [0, π]. Only within this domain does the inverse of y = cos x exist. REASONING Determine whether each function is odd, even, or neither. Justify your answer. 69. y = sin x Sample answer: Suppose y = sin 1 x is odd. The definition of an odd function states for every x in the domain of f, f ( x) = f (x). If we let sin x = u, we have x = sin u. From Lesson 4-3, we know that the sine function is odd, so x = sin ( u). From here, we can get sin ( x) = u. Also, as shown below, the graph of y = sin 1 x is symmetric with respect to the origin. esolutions Manual - Powered by Cognero Page 25

26 70. y = cos x Sample answer: The graph of y = cos 1 x is not symmetric with respect to the y-axis or origin. Therefore, y = cos 1 x is neither even nor odd. 71. y = tan x 72. Writing in Math Explain how the restrictions on the sine, cosine, and tangent functions dictate the domain and range of their inverse functions. Sample answer: The restricted domains of the cosine, sine, and tangent functions become the ranges of the arccosine, arcsine, and arctangent functions respectively. Additionally, the ranges of the cosine, sine, and tangent functions under these restrictions become the domains of their inverses. For example, when x 1, sin (sin x) = x. In this case, the range of sin x, [, 1], limits the domain of sin x. Sample answer: Suppose y = tan 1 x is odd. The definition of an odd function states for every x in the domain of f, f ( x) = f (x). If we let tan x = u, we have x = tan u. From Lesson 4-3, we know that the tangent function is odd, so x = tan ( u). From here, we can get tan ( x) = u. Graphically, it can be seen that for every x in the domain of f, f ( x) = f (x). Also, as shown below, the graph of y = tan 1 x is symmetric with respect to the origin. esolutions Manual - Powered by Cognero Page 26

27 Locate the vertical asymptotes, and sketch the graph of each function. 73. y = 3 tan θ The graph of y = 3 tan x is the graph of y = tan x 74. y = cot 5θ The graph of is the graph of y = cot x expanded vertically. The period is or π. Find the expanded vertically. The period is or π. Find the location of two consecutive vertical asymptotes. location of two consecutive vertical asymptotes. and Create a table listing the coordinates of key points for y = 3 tan x for one period on. Function Vertical Asymptote Intermediate Point x-int Intermediate Point Vertical Asymptote y = tan x y = 3 tan x (0, 0) (0, 0) and Create a table listing the coordinates of key points for for one period on [0, π]. Function Vertical Asymptote Intermediate Point x-int Intermediate Point Vertical Asymptote y = cot x y = cot 5θ x = 0 x = 0 x = π x = π Sketch the curve through the indicated key points for the function. Then repeat the pattern. Sketch the curve through the indicated key points for the function. Then repeat the pattern. esolutions Manual - Powered by Cognero Page 27

28 75. y = 3 csc θ The graph of y = 3 csc θ is the graph of y = csc x expanded vertically and expanded horizontally. The period is or 4π. Find the location of two vertical asymptotes. 76. WAVES A leaf floats on the water bobbing up and down. The distance between its highest and lowest points is 4 centimeters. It moves from its highest point down to its lowest point and back to its highest point every 10 seconds. Write a cosine function that models the movement of the leaf in relationship to the equilibrium point. The distance between the highest and lowest points is 4 centimeters, so the magnitude is half of that, or 2. Therefore, a = 2. The period is from highest point to highest point, or 10 seconds. and Create a table listing the coordinates of key points for y = 3 csc x for one period on [ 2π, 2π]. The function is y = a cos bt. Function Vertical Asymptote Intermediate Point x-int Intermediate Point Vertical Asymptote y = 3 csc y = csc x x x = π x = 2π x = 0 x = 0 x = π x = 2π 77. Find the value of x. Round to the nearest tenth, if necessary. An acute angle measure and the length of a leg are given, so the tangent function can be used to find the length of the side opposite x. Sketch the curve through the indicated key points for the function. Then repeat the pattern. esolutions Manual - Powered by Cognero Page 28

29 For each pair of functions, find [ f g](x), [g f ](x), and [ f g](4). 80. f (x) = x 2 + 3x 6; g(x) = 4x An acute angle measure and the length of the opposite leg are given, so the sine function can be used to find the length of the hypotenuse x. 79. An acute angle measure and the length of a leg are given, so the cosine function can be used to find the length of the hypotenuse x. 81. f (x) = 6 5x; g(x) = esolutions Manual - Powered by Cognero Page 29

30 82. f (x) = ; g(x) = x SAT/ACT To the nearest degree, what is the angle of depression θ between the shallow end and the deep end of the swimming pool? A 25 B 37 C 41 D 53 E 73 The side adjacent to θ is or 8 feet. The side opposite θ is 10 4 or 6 feet. We can use tangent to find θ. 83. EDUCATION Todd has answered 11 of his last 20 daily quiz questions correctly. His baseball coach told him that he must raise his average to at least 70% if he wants to play in the season opener. Todd vows to study diligently and answer all of the daily quiz questions correctly in the future. How many consecutive daily quiz questions must he answer correctly to raise his average to a 70%? The correct choice is B. He needs to correctly answer a total of d questions such that the total number he has correctly answered, 11 + d, divided by the total number of questions, 20 + d, is questions esolutions Manual - Powered by Cognero Page 30

31 85. Which of the following represents the exact value of? F G H J 86. REVIEW The hypotenuse of a right triangle is 67 inches. If one of the angles has a measure of 47, what is the length of the shortest leg of the triangle? A 45.7 in. B 49.0 in. C 62.5 in. D 71.8 in. The shortest leg of a right triangle is opposite the smallest angle of the triangle. One angle is 90, another is 47, and the last angle is 43. We can use sine and this angle to find the length of the shortest side. First, find tan 1. To do this, find a point on the unit circle on the interval [0, 2π] to represent this ratio. The correct choice is A. In this diagram, θ = tan. We can use the Pythagorean theorem to find the hypotenuse. Now we can evaluate. Therefore, sin θ =. The correct choice is H. esolutions Manual - Powered by Cognero Page 31

32 87. REVIEW Two trucks, A and B, start from the intersection C of two straight roads at the same time. Truck A is traveling twice as fast as truck B, and after 4 hours, the two trucks are 350 miles apart. Find the approximate speed of truck B in miles per hour. F 39 G 44 H 51 J 78 If truck A is traveling twice as fast, then it has traveled twice the distance as truck B. Therefore, in the diagram, the length of CB is x and the length of CA is 2x.We can use the Pythagorean theorem to determine the length of CB. This is the distance that truck B has traveled in 4 hours.the approximate speed is about 39.1 miles per hour. The correct choice is F. esolutions Manual - Powered by Cognero Page 32