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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=
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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) =
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43. sin (cos– 1 x)
SOLUTION: . Now, solve for tan u.
.
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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
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–1
So, sin (cos
x) =
.
. Now, solve for sin u.
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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.
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So, csc(sin
x) =
.
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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.
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So, cot (arccos x) =
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.
–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.
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a. What is the angle of repose?
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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
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graph.
57. y = sin (cos−1 x)
of each composite function. Then use your graphing calculator to sketch its
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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}.
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