4-6 Inverse Trigonometric Functions

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1 Find the exact value of each expression, if it exists. 29. 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, =. 35. cos (tan 1 1) First, find tan 1 1. The inverse property applies, because 1 is on the interval. Therefore, tan 1 1 =. Next, find cos. On the unit circle, corresponds to. So, cos =. cos (tan 1 1) = 37. 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, corresponds to (0, 1). So, sin = 1, and = 1. esolutions Manual - Powered by Cognero Page 1

2 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-coordinate. When t =, cos t = sin t =. Therefore, tan 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, =. cos (tan 1 1 sin 1 1) = esolutions Manual - Powered by Cognero Page 2

3 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. From the Pythagorean Theorem, the length of the side opposite u is. Now, solve for tan u. So, tan (arccos x) =. esolutions Manual - Powered by Cognero Page 3

4 43. 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 u is. Now, solve for sin u. So, sin (cos 1 x) =. esolutions Manual - Powered by Cognero Page 4

5 45. 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 csc u. So, csc(sin 1 x) =. esolutions Manual - Powered by Cognero Page 5

6 47. 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 opposite u is. Now, solve for cot u. So, cot (arccos x) =. Describe how the graphs of g(x) and f (x) are related. 49. f (x) = sin 1 x and g(x) = sin 1 (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. 51. f (x) = cos 1 x and g(x) = 3(cos 1 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. 53. f (x) = arccos x and g(x) = 5 + arccos 2x g(x) is of the form f (2x) + 5. The 2 represents a horizontal compression while the 5 represents a translation up. 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. esolutions Manual - Powered by Cognero Page 6

7 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. 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. Use the tangent function to find θ. 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 pile would reach 4 feet if the diameter was about 2(5.5) or 11 feet. esolutions Manual - Powered by Cognero Page 7

8 Give the domain and range of each composite function. Then use your graphing calculator to sketch its graph. 57. y = sin (cos 1 x) The domain of sin x is {x x } and the range of cos 1 x falls within this domain, so there are no further restrictions on the domain. The domain of cos 1 x is [ 1, 1] so the domain of the composite function is restricted to {x 1 x 1}. The range of cos 1 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}. 59. y = sin 1 (cos x) The domain of sin 1 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, 1], so this becomes the domain of sin 1 x, or the limit of the input values for sin 1 x. The only corresponding output values for these input values is. Therefore, the range of the composite function is. 61. y = tan (arccos x) The domain of tan x is restricted to and the range of arccos x is [0, π], so the domain is. The domain of arccos x is {x 1 x 1}, so the domain of the composite function is further restricted to {x 1 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}. esolutions Manual - Powered by Cognero Page 8

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