1. No category

4-6 Inverse Trigonometric Functions
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.
a. Write θ as a function of x.
b. Find θ when x = 6 meters and x = 14 meters.
SOLUTION: a. The relationship between θ and the sides is opposite and adjacent, so tan θ =
arctan
. After taking the inverse, θ =
.
b.
Find the exact value of each expression, if it exists.
29. SOLUTION: The inverse property applies, because
lies on the interval [–1, 1]. Therefore,
=
.
31. SOLUTION: eSolutions
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The inverse property applies, because
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lies on the interval [–1, 1]. Therefore,
=
.
SOLUTION: The inverse property applies, because
lies on the interval [–1, 1]. Therefore,
=
.
lies on the interval [–1, 1]. Therefore,
=
.
4-6 Inverse Trigonometric Functions
31. SOLUTION: The inverse property applies, because
33. SOLUTION: The inverse property applies, because
lies on the interval . Therefore,
=
.
35. cos (tan– 1 1)
SOLUTION: First, find tan
–1
Next, find cos
1) =
1. The inverse property applies, because 1 is on the interval
. On the unit circle,
corresponds to . Therefore, tan
. So, The cos
=
–1
1=
.
. Therefore, 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 –1
, cos t =
= . 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: First, find tan
–1
1. To do this, 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=
. Next, find sin
on the unit circle on the interval [0, 2π] with an y-coordinate of 1. When t =
–1
–1
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Then
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1=
. Finally find
–1
–1
–1
1. To do this, find a point
, sin t = 1 . Therefore, sin
. On the unit circle,
corresponds to –1
1=
.
Page
. 2
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: First, find tan
–1
1. To do this, 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=
. Next, find sin
on the unit circle on the interval [0, 2π] with an y-coordinate of 1. When t =
Then tan
–1
–1
1 – sin
So,
=
1=
. Finally find
–1
, and cos (tan
–1
1 – sin
1) =
–1
1. To do this, find a point
, sin t = 1 . Therefore, sin
. On the unit circle,
–1
1=
corresponds to .
.
.
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 line 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.
The length of the side opposite u is
or . Now, solve for tan u.
So, tan (arccos x) =
.
43. sin (cos– 1 x)
SOLUTION: –1
Let u = cos x, so cos u = x.
Because
domainbyof
the inverse cosine function is restricted to Quadrants I and II, u must line in one of thesePage 3
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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.
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 line 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.
The length of the side opposite u is
or . Now, solve for sin u.
–1
So, sin (cos
x) =
.
45. csc (sin– 1 x)
SOLUTION: –1
Let u = sin x, so sin u = x.
Because the domain of the inverse cosine function is restricted to Quadrants I and II, u must line 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.
The length of the side adjacent to u is
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or . Now, solve for csc u.
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4-6 Inverse Trigonometric
Functions
–1
So, sin (cos
x) =
.
45. csc (sin– 1 x)
SOLUTION: –1
Let u = sin x, so sin u = x.
Because the domain of the inverse cosine function is restricted to Quadrants I and II, u must line 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.
The length of the side adjacent to u is
or . 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 line 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.
The length of the side opposite u is
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or . Now, solve for cot u.
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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 line 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.
The length of the side opposite u is
or . Now, solve for cot u.
So, cot (arccos x) =
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.
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