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4-6 Inverse Trigonometric Functions Find the exact value of each expression, if it exists. 29. SOLUTION: The inverse property applies, because lies on the interval [–1, 1]. Therefore, = . lies on the interval [–1, 1]. Therefore, = . 31. SOLUTION: The inverse property applies, because 35. cos (tan– 1 1) SOLUTION: First, find tan –1 Next, find cos 1. The inverse property applies, because 1 is on the interval . On the unit circle, corresponds to . So, cos . Therefore, tan = –1 1= . . –1 cos (tan 1) = 37. SOLUTION: First, find cos When t = –1 . To do this, find a point on the unit circle on the interval [0, 2π] with an x-coordinate of , cos t = –1 = . Therefore, cos . . Next, find or sin . On the unit circle, corresponds to (0, 1). So, sin = 1, and = 1. 39. cos (tan– 1 1 – sin– 1 1) SOLUTION: –1 Find tan When t = 1. Find a point on the unit circle on the interval [0, 2π] with an x-coordinate equal to the y-coordinate. , cos t = sin t = –1 . Therefore, tan 1= . eSolutions Manual - Powered by Cognero Find sin –1 1. Find a point on the unit circle on the interval [0, 2π] with an y-coordinate of 1. When t = Page 1 , sin t = 1 . Next, find or sin . On the unit circle, corresponds to (0, 1). So, sin = 1, and = 4-6 Inverse Trigonometric Functions 1. 39. cos (tan– 1 1 – sin– 1 1) SOLUTION: –1 Find tan 1. Find a point on the unit circle on the interval [0, 2π] with an x-coordinate equal to the y-coordinate. When t = –1 . Therefore, tan , cos t = sin t = 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 = –1 Therefore, sin 1= , sin t = 1 . . Find . On the unit circle, corresponds to . So, = . –1 cos (tan –1 1 – sin 1) = Write each trigonometric expression as an algebraic expression of x. 41. tan (arccos x) SOLUTION: 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 So, tan (arccos x) = eSolutions Manual - Powered by Cognero 43. sin (cos– 1 x) SOLUTION: . Now, solve for tan u. . Page 2 Find . On the unit circle, corresponds to . So, = . –1 –1 4-6 cos Inverse Functions (tan Trigonometric 1 – sin 1) = Write each trigonometric expression as an algebraic expression of x. 41. tan (arccos x) SOLUTION: 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 So, tan (arccos x) = . Now, solve for tan u. . 43. sin (cos– 1 x) SOLUTION: –1 Let u = cos 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 eSolutions Manual - Powered by Cognero –1 So, sin (cos x) = . . Now, solve for sin u. Page 3 tan (arccos x) = . Functions 4-6 So, Inverse Trigonometric 43. sin (cos– 1 x) SOLUTION: –1 Let u = cos 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. –1 So, sin (cos x) = . 45. csc (sin– 1 x) SOLUTION: 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. eSolutions Manual -–1 Powered by Cognero So, csc(sin x) = . Page 4 4-6 Inverse Trigonometric Functions –1 So, sin (cos x) = . 45. csc (sin– 1 x) SOLUTION: 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) = . 47. cot (arccos x) SOLUTION: 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. eSolutions Manual - Powered by Cognero So, cot (arccos x) = Page 5 . –1 4-6 So, Inverse Functions csc(sin Trigonometric x) = . 47. cot (arccos x) SOLUTION: 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 SOLUTION: 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) SOLUTION: 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 SOLUTION: 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 a. What is the angle of repose? Page 6 SOLUTION: g(x) is of the form f (2x) + 5 . The 2 represents a horizontal compression while the 5 represents a translation up. 4-6 Inverse Trigonometric Functions 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. 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? SOLUTION: 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. Give the domain range eSolutions Manual - Powered and by Cognero graph. 57. y = sin (cos−1 x) of each composite function. Then use your graphing calculator to sketch its Page 7 4-6 Inverse Trigonometric Functions The pile would reach 4 feet if the diameter was about 2(5.5) or 11 feet. Give the domain and range of each composite function. Then use your graphing calculator to sketch its graph. 57. y = sin (cos−1 x) SOLUTION: −1 The domain of sin x is {x | x } and the range of cos x falls within this domain, so there are no further −1 restrictions on the domain. The domain of cos x is [−1, 1] so the domain of the composite function is restricted to {x | −1 ≤ x ≤ 1}. −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}. 59. y = sin−1 (cos x) SOLUTION: −1 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 }. −1 The range of cos x is [−1, 1], so this becomes the domain of sin only corresponding output values for these input values is function is −1 x, or the limit of the input values for sin x. The . Therefore, the range of the composite . 61. y = tan (arccos x) SOLUTION: and the range of arccos x is [0, π], so the domain is The domain of tan x is restricted to . 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