Download 4-6 Inverse Trigonometric Functions

When t =
4-6 Inverse Trigonometric Functions
1. sin
= . Therefore, arcsin
.
Find the exact value of each expression, if it
exists.
–1
, sin t =
3. arcsin
0
SOLUTION: SOLUTION: Find a point on the unit circle on the interval
Find a point on the unit circle on the interval
with a y-coordinate of
with a y-coordinate of 0.
When t = 0, sin t = 0. Therefore, sin
–1
When t =
0 = 0.
, sin t =
.
. Therefore, arcsin =
.
2. arcsin
4. sin – 1
SOLUTION: Find a point on the unit circle on the interval
SOLUTION: with a y-coordinate of
.
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 = .
5. 3. arcsin
SOLUTION: SOLUTION: Find a point on the unit circle on the interval
Find a point on the unit circle on the interval
with a y-coordinate of
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.
with a y-coordinate of
.
Page 1
–1 4-6 When
Inverse
Functions
t = Trigonometric
, sin t = . Therefore,
sin
= When t =
.
, cos t = 0. Therefore, arccos 0 =
.
7. cos– 1
5. SOLUTION: SOLUTION: Find a point on the unit circle on the interval
Find a point on the unit circle on the interval
with a y-coordinate of
When t =
=
, sin t =
with an x-coordinate of
.
. Therefore, sin
–1
When t =
, cos t =
.
–1
.
.
8. arccos (–1)
6. arccos 0
SOLUTION: SOLUTION: Find a point on the unit circle on the interval
with an x-coordinate of 0.
Find a point on the unit circle on the interval
with an x-coordinate of –1.
When t =
=
. Therefore, cos
, cos t = 0. Therefore, arccos 0 =
When t =
.
, cos t = –1. Therefore, arccos (–1)=
.
9. –1
7. cos
SOLUTION: Find a point on the unit circle on the interval
SOLUTION: Find a point on the unit circle on the interval
with an x-coordinate of
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with an x-coordinate of
.
.
Page 2
4-6 Inverse Trigonometric Functions
When t =
, cos t = –1. Therefore, arccos (–1)=
When t =
.
–1
, cos t =
. Therefore, cos
=
.
11. arctan 1
9. SOLUTION: Find a point on the unit circle on the interval
SOLUTION: Find a point on the unit circle on the interval
with an x-coordinate of
such that .
When t =
When t =
, cos t =
=1.
. Therefore,
, tan t =
. Therefore, arctan 1=
=
.
.
12. arctan (–
10. cos– 1
)
SOLUTION: Find a point on the unit circle on the interval
SOLUTION: Find a point on the unit circle on the interval
with an x-coordinate of
When t =
, cos t =
such that =–
.
.
–1
. Therefore, cos
=
When t =
.
11. arctan 1
arctan (–
, tan t =
)=
. Therefore,
.
SOLUTION: Find a point on the unit circle on the interval
such that =1.
13. SOLUTION: Find a point on the unit circle on the interval
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such that =
.
Page 3
When t =
, tan t =
. Therefore,
4-6 arctan
Inverse
Functions
(– Trigonometric
)=
.
When t = 0, tan t =
–1
. Therefore, tan
0 = 0.
15. ARCHITECTURE The support for a roof is 13. shaped like two right triangles, as shown below. Find
θ.
SOLUTION: Find a point on the unit circle on the interval
such that =
.
SOLUTION: 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
When t =
SOLUTION: , sin t =
. Therefore, sin
–1
= .
Find a point on the unit circle on the interval
such that 16. RESCUE A cruise ship sailed due west 24 miles = 0.
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.
When t = 0, tan t =
. Therefore, tan
–1
0 = 0.
15. ARCHITECTURE The support for a roof is shaped like two right triangles, as shown below. Find
θ.
SOLUTION: 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
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.
Page 4
When t =
–1
4-6 When
Inverse
Functions
t = Trigonometric
, sin t = . Therefore,
sin
= .
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.
, cos t =
–1
. Therefore, cos
= .
Sketch the graph of each function.
17. y = arcsin x
SOLUTION: First, rewrite y = arcsin x in the form sin y = x.
to Next, assign values to y on the interval
make a table of values.
y
x = sin y
–1
SOLUTION: 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
0
0
1
.
Plot the points and connect them with a smooth
curve.
When t =
, cos t =
–1
. Therefore, cos
18. y = sin – 1 2x
= SOLUTION: First, rewrite y = sin
.
–1
2x in the form sin y = x.
Sketch the graph of each function.
17. y = arcsin x
SOLUTION: 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.
Next, assign values to y on the interval
to y
make a table of values.
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y
x = sin y
–1
0
0
Page 5
4-6 Inverse Trigonometric Functions
18. y = sin – 1 2x
19. y = sin – 1 (x + 3)
SOLUTION: SOLUTION: First, rewrite y = sin
–1
2x in the form sin y = x.
First, rewrite y = sin
–1
(x + 3) in the form sin y = x.
Next, assign values to y on the interval
Next, assign values to y on the interval
to to make a table of values.
make a table of values.
y
y
x = sin y
–3
–3.89
0
0
0
–3.85
–3
–2.15
–2.11
Plot the points and connect them with a smooth
curve.
Plot the points and connect them with a smooth
curve.
19. y = sin – 1 (x + 3)
SOLUTION: First, rewrite y = sin
–1
(x + 3) in the form sin y = x.
20. y = arcsin x – 3
SOLUTION: First, rewrite y = arcsin x – 3 in the form sin y = x.
Next, assign values to y on the interval
make a table of values.
x = sin y
– by
3 Cognero
y - Powered
eSolutions Manual
–3.89
to Next, assign values to y on the interval
make a table of values.
x = sin(y +
3)
y
to Page 6
4-6 Inverse Trigonometric Functions
20. y = arcsin x – 3
21. y = arccos x
SOLUTION: SOLUTION: First, rewrite y = arcsin x – 3 in the form sin y = x.
First, rewrite y = arccos x in the form cos y = x.
to Next, assign values to y on the interval
make a table of values.
x = sin(y +
3)
y
Next, assign values to y on the interval
make a table of values.
x = cos y
y
0
0.99
0
to 1
0
0.80
0.14
-1
−0.60
−0.99
Plot the points and connect them with a smooth
curve.
Plot the points and connect them with a smooth
curve.
21. y = arccos x
22. y = cos– 1 3x
SOLUTION: First, rewrite y = arccos x in the form cos y = x.
SOLUTION: –1
First, rewrite y = cos
3x in the form cos y = x.
Next, assign values to y on the interval
make a table of values.
x = cos y
y
0
1
to Next, assign values to y on the interval
make a table of values.
to y
0
0
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-1
Page 7
0
4-6 Inverse Trigonometric Functions
22. y = cos– 1 3x
23. y = arctan x
SOLUTION: SOLUTION: –1
First, rewrite y = cos
First, rewrite y = arctan x in the form tan y = x.
3x in the form cos y = x.
to Next, assign values to y on the interval
make a table of values.
x = tan y
y
Next, assign values to y on the interval
make a table of values.
to –3.08
y
–1
0
0
0
1
3.08
0
Plot the points and connect them with a smooth
curve.
Plot the points and connect them with a smooth
curve.
24. y = tan– 1 3x
SOLUTION: –1
First, rewrite y = tan
3x in the form tan y = x.
23. y = arctan x
SOLUTION: First, rewrite y = arctan x in the form tan y = x.
Next, assign values to y on the interval
make a table of values. Note that x =
Next, assign values to y on the interval
to tan y has
to no x-values for y-values of and − .
make a table of values.
x = tan y
y
–3.08
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0
–1
0
y
x=
tan y
Page 8
4-6 Inverse Trigonometric Functions
24. y = tan– 1 3x
25. y = tan– 1 (x + 1)
SOLUTION: SOLUTION: –1
First, rewrite y = tan
–1
3x in the form tan y = x.
First, rewrite y = tan
(x + 1) 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. Note that x =
tan y has
make a table of values. Note that x = tan y − 1 has
no x-values for y-values of and − .
no x-values for y-values of and − .
y
x=
0
to y
x = tan y –
1
0
–2
–1
tan y
0
0
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 .
.
25. y = tan– 1 (x + 1)
26. y = arctan x – 1
SOLUTION: –1
First, rewrite y = tan
(x + 1) in the form tan y = x.
SOLUTION: First, rewrite y = arctan x – 1 in the form tan y = x.
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Next, assign values to y on the interval
Page 9
to The range of inverse tangent of x is
. Since
4-6 Inverse Trigonometric Functions
26. y = arctan x – 1
27. DRAG RACE A television camera is filming a SOLUTION: First, rewrite y = arctan x – 1 in the form tan y = x.
The range of inverse tangent of x is
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.
. Since
we are subtracting 1, the new range should be
. Next, assign values to y within this
a. Write θ as a function of x.
b. Find θ when x = 6 meters and x = 14 meters.
range to make a table of values.
y
–2.5
SOLUTION: x = tan (y
+ 1)
a. The relationship between θ and the sides is
–14.1
–0.64
opposite and adjacent, so tan θ =
0
0.22
1.56
the inverse, θ = arctan
0.5
–4.59
14.1
. After taking
.
b.
Plot the points and connect them with a smooth
curve. Further investigation reveals that as x
approaches negative infinity, y approaches
− 1
≈ −2.57, and as x approaches positive infinity, y
approaches − 1 ≈ 0.57.
28. 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.
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.
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Page 10
28. SPORTS Steve and Ravi want to project a pro 4-6
soccer game on the side of their apartment building.
They
have Trigonometric
placed a projector onFunctions
a table that stands 5
Inverse
feet above the ground and have hung a 12-foot-tall
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.
The d-value that corresponds with this maximum is
about 9.2 feet.
SOLUTION: a. Make a diagram of the situation.
Find the exact value of each expression, if it
exists.
29. SOLUTION: The inverse property applies, because
lies on the interval [–1, 1]. Therefore,
=
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.
30. SOLUTION: The inverse property applies, because
lies on the interval [–1, 1]. Therefore,
=
.
31. SOLUTION: The inverse property applies, because
lies on the interval [–1, 1]. Therefore,
=
.
32. cos– 1 (cos π)
b. The goal is to maximize θ, so if we graph
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, we can identify the maximum
value of θ. The d-value that corresponds with this
SOLUTION: The inverse property applies, because π lies onPage
the 11
interval
–1
. Therefore, cos
(cos π)= π.
corresponds to The inverse property applies, because
lies on the . 4-6 interval
Inverse
Functions
[–1,Trigonometric
1]. Therefore,
= .
–1
cos (tan
32. cos– 1 (cos π)
=
. So, cos
1) =
36. SOLUTION: SOLUTION: The inverse property applies, because π lies on the
–1
interval
. Therefore, cos
(cos π)= π.
First, find cos
33. = 0.
to (0, 1). So, cos
–1
Next, find sin 0. The inverse property applies,
SOLUTION: The inverse property applies, because
interval
corresponds
. On the unit circle,
. Therefore,
–1
lies on the =
because 0 is on the interval [–1, 1]. Therefore, sin
= 0.
0 = 0, and
.
37. 34. SOLUTION: SOLUTION: The inverse property applies, because
lies on the –1
First, find cos
. To do this, find a point on the
unit circle on the interval [0, 2π] with an x-coordinate
interval
. Therefore,
=
.
of 35. cos (tan– 1 1)
–1
1=
, cos t =
. Therefore,
.
–1
First, find tan 1. The inverse property applies,
because 1 is on the interval
. Therefore,
–1
= cos
SOLUTION: tan
. When t = . Next, find cos
corresponds to Next, find
. On the unit circle,
. So, cos
=
or sin
. On the unit circle,
corresponds to (0, 1). So, sin . = 1, and = 1.
–1
cos (tan
38. sin (tan– 1 1 – sin– 1 1)
1) =
SOLUTION: –1
36. First, find tan 1. To do this, find a point on the unit
circle on the interval [0, 2π] with an x-coordinate
SOLUTION: First, find cos
equal to the y-coordinate. When t =
. On the unit circle,
corresponds
=
to (0, 1). So, cos
= 0.
–1
. Therefore, tan
1=
, cos t = sin t
. –1
Next, find sin 1. To do this, find a point on the unit
circle on the interval [0, 2π] with a y-coordinate of 1.
–1
Next,
find -sin
0. The
inverse property applies,
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because 0 is on the interval [–1, 1]. Therefore, sin
–1
When t =
, sin t = 1 . Therefore, sin
–1
1=
. to
corresponds to (0, 1). So, sin . So,
=
.
= 1, and = 1.
4-6 Inverse Trigonometric
Functions
–1
sin (tan
38. sin (tan– 1 1 – sin– 1 1)
–1
1 – sin
1) =
39. cos (tan– 1 1 – sin– 1 1)
SOLUTION: SOLUTION: –1
–1
First, find tan 1. To do this, find a point on the unit
circle on the interval [0, 2π] with an x-coordinate
Find tan 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 =
coordinate. When t =
–1
. Therefore, tan
=
, cos t = sin t
. 1=
Therefore, tan
–1
Next, find sin 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=
. –1
Find sin 1. Find a point on the unit circle on the
interval [0, 2π] with an y-coordinate of 1. When t =
. 1=
–1
. , cos t = sin t =
, sin t = 1 . Therefore, sin
–1
1=
.
Find
. So,
. On the unit circle,
to
. So,
=
.
–1
cos (tan
–1
–1
1 – sin
.
–1
1 – sin
1) =
1) =
40. 39. cos (tan– 1 1 – sin– 1 1)
SOLUTION: SOLUTION: –1
Find tan 1. Find a point on the unit circle on the
interval [0, 2π] with an x-coordinate equal to the ycoordinate. When t =
Therefore, tan
=
sin (tan
corresponds corresponds . On the unit circle,
to
Find
–1
, cos t = sin t =
. –1
Find cos 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, 1=
. Find sin
–1
. Find a point on the unit circle on the
–1
Find sin 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=
interval [0, 2π] with an y-coordinate of
, sin t =
.
. Therefore,
. So,
=
. corresponds FindManual - Powered
. Onbythe
unit circle,
eSolutions
Cognero
to
. When t =
.
Page 13
Find
. On the unit circle,
corresponds to
. So,
=
.
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)
40. SOLUTION: SOLUTION: –1
Find cos 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
, sin t =
. When t =
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,
From the Pythagorean Theorem, the length of the
Find
. On the unit circle,
to
. So,
corresponds side opposite u is
. Now, solve for tan u.
.
=
.
So, tan (arccos x) =
Write each trigonometric expression as an
algebraic expression of x.
41. tan (arccos x)
SOLUTION: .
42. csc (cos– 1 x)
SOLUTION: –1
Let u = arccos x, so cos u = x.
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.
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 1.
Draw a diagram of a right triangle with an acute
angle u, an adjacent side length x and a hypotenuse
length 1.
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Page 14
tan (arccos
x) =
. Functions
4-6 So,
Inverse
Trigonometric
42. csc (cos– 1 x)
–1
x) =
.
43. sin (cos– 1 x)
SOLUTION: Let u = cos
–1
So, csc (cos
SOLUTION: x, so cos u = x.
–1
Let u = cos
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.
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.
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
From the Pythagorean Theorem, the length of the
side opposite to u is
side opposite u is
. Now, find csc u.
. Now, solve for sin u.
–1
–1
So, csc (cos
So, sin (cos
x) =
x) =
.
.
44. cos (arcsin x)
43. sin (cos– 1 x)
SOLUTION: 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.
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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 1.
Page 15
From the Pythagorean Theorem, the length of the
side adjacent to u is
. Now, solve for cos u.
4-6 Inverse Trigonometric
Functions
–1
So, sin (cos
x) =
.
So, cos (arcsin x) =
.
45. csc (sin– 1 x)
44. cos (arcsin x)
SOLUTION: SOLUTION: Let u = arcsin x, so sin u = x.
–1
Let u = sin
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.
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, cos (arcsin x) =
.
–1
So, csc(sin
45. csc (sin– 1 x)
–1
x) =
.
46. sec (arcsin x)
SOLUTION: Let u = sin
SOLUTION: x, so sin u = 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 1.
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eSolutions
From the Pythagorean Theorem, the length of the
side adjacent to u is
. Now, solve for csc u.
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 1.
Page 16
From the Pythagorean Theorem, the length of the
side adjacent to u is
. Now, solve for sec u.
–1
4-6 So,
Inverse
Functions
csc(sin Trigonometric
x) = .
46. sec (arcsin x)
So, sec(arcsin x) =
.
47. cot (arccos x)
SOLUTION: SOLUTION: Let u = arcsin x, so sin u = x.
Let u = arccos 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.
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.
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
From the Pythagorean Theorem, the length of the
side adjacent to u is
side opposite u is
. Now, solve for sec u.
So, sec(arcsin x) =
. Now, solve for cot u.
.
47. cot (arccos x)
So, cot (arccos x) =
48. cot (arcsin x)
SOLUTION: SOLUTION: Let u = arccos x, so cos u = 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 adjacent side length x and a hypotenuse
length 1.
.
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.
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Page 17
From the Pythagorean Theorem, the length of the
From the Pythagorean Theorem, the length of the
SOLUTION: 4-6 So,
Inverse
Trigonometric
cot (arccos
x) =
. Functions
g(x) is of the form 0.5f (x) − 3. The 0.5 represents a
horizontal expansion while the 3 represents a
translation down.
51. f (x) = cos−1 x and g(x) = 3(cos−1 x − 2)
48. cot (arcsin x)
SOLUTION: SOLUTION: Let u = arcsin x, so sin u = x.
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.
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.
52. f (x) = arcsin x and g(x) =
arcsin (x + 2)
SOLUTION: 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
SOLUTION: side adjacent to u is
g(x) is of the form f (2x) + 5 . The 2 represents a
horizontal compression while the 5 represents a
translation up.
. Now, solve for cot u.
54. f (x) = tan−1 x and g(x) = tan−1 3x − 4
SOLUTION: So, cot(arcsin x) =
g(x) is of the form f (3x) − 4 . The 3 represents a
horizontal compression while the 4 represents a
translation down.
.
Describe how the graphs of g(x) and f (x) are
related.
−1
49. f (x) = sin
−1
x and g(x) = sin
(x − 1) − 2
SOLUTION: 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.
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.
50. f (x) = arctan x and g(x) = arctan 0.5x − 3
SOLUTION: g(x) is of the form 0.5f (x) − 3. The 0.5 represents a
horizontal expansion while the 3 represents a
translation down.
51. f (x) = cos−1 x and g(x) = 3(cos−1 x − 2)
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.
SOLUTION: eSolutions Manual - Powered by Cognero
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
Page 18
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?
4-6 SOLUTION: Inverse Trigonometric Functions
The pile would reach 4 feet if the diameter was
about 2(5.5) or 11 feet.
a. Draw a diagram to model this situation.
Give the domain and range of each composite
function. Then use your graphing calculator to
sketch its graph.
56. y = cos (tan−1 x)
SOLUTION: The domain of cos x is {x | x
} and the range of
−1
Use the tangent function to find θ.
tan x falls within this domain, so there are no
further restrictions on the domain. The domain of
−1
tan x is also {x | x }, so the domain of the
composite function is {x | x }.
The range of tan
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.
−1
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}.
57. y = sin (cos−1 x)
SOLUTION: The domain of sin x is {x | x } and the range of
−1
cos x falls within this domain, so there are no
further restrictions on the domain. The domain of
−1
cos x is [−1, 1] so the domain of the composite
function is restricted to {x | −1 ≤ x ≤ 1}.
Use the tangent function to find 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}.
58. y = arctan (sin x)
SOLUTION: 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.
56. y = cos (tan−1 x)
SOLUTION: The domain of cos x is {x | x
−1
} and the range of
tan x falls within this domain, so there are no
further restrictions on the domain. The domain of
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tan Manual
x is also
{x | x }, so the domain of the
composite function is {x | 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, 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
range of the composite function is
−1
. Therefore, the
Page 19
.
−1
4-6
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
valuesTrigonometric
is {y | 0 ≤ y ≤ 1}. Therefore, the range Inverse
Functions
of the composite function is {y | 0 ≤ y ≤ 1}.
58. y = arctan (sin x)
these input values is
. Therefore, the
range of the composite function is
.
60. y = cos (arcsin x)
SOLUTION: SOLUTION: 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 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 | −1 ≤ x ≤ 1}, so the domain of the composite function is restricted to {x | −1 ≤ x ≤ 1}.
The range of sin x is [−1, 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
The range of arcsin x is
these input values is
. Therefore, the
range of the composite function 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)
SOLUTION: 59. y = sin−1 (cos x)
The domain of tan x is
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
domain of sin x, or the limit of the input values for
−1
sin x. The only corresponding output values for
. Therefore, the
range of the composite function 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 cos x is [−1, 1], so this becomes the
these input values is
and the range of
arccos x is [0, π], so the domain is restricted to
.
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
60. y = cos (arcsin x)
intervals
SOLUTION: , 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 eSolutions
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range
of the
composite
function is {y | 0 ≤ y ≤ 1}.
61. y = tan (arccos x)
, and the arccosecant
function is graphed by restricting the domain of the
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 | −1 ≤ x ≤ 1}, so the domain of the composite function is restricted to {x | −1 ≤ x ≤ 1}.
The range of arcsin x is
and 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.
SOLUTION: a. If the domain of secant is restricted to
Page 20
, then the range of arcsecant must
4-6
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
Inverse
secant
and Trigonometric
cosecant functions isFunctions
necessary in order
to graph the inverse functions.
arccosecant
SOLUTION: 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] ∪ [1, ). This is determined by
finding the values for secant produced by the
endpoints of the domain of secant. So, sec π = −1
and sec 0 = 1. Also, as θ approaches from the 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.
left, sec θ approaches ∞, and as θ approaches
from the right, sec θ approaches −∞.
Write each algebraic expression as a
trigonometric function of an inverse
trigonometric function of x.
If the domain of cosecant is restricted
to
, then the range of arccosecant
must also be
because they are
inverses. The domain of arccosecant that produces
this range is D = (− , −1] [1, ). This is
determined by finding the values for cosecant
produced by the endpoints of the domain of
cosecant. So, csc
= 1 and
63. SOLUTION: 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.
. Also, as
θ approaches 0 from the left, csc θ approaches ∞, and as θ approaches 0 from the right, csc θ
approaches − .
b.
arcsecant
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,
.
arccosecant
According to our new labels in the graph,
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c. If restrictions are not put on the domain of the
represents Page 21
, or tan u. Also,
4-6
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,Functions
no longer have
Inverse
Trigonometric
inverses.
Note that if we had switched the labels of the sides,
−1
we would have had cot (cos x).
Write each algebraic expression as a
trigonometric function of an inverse
trigonometric function of x.
64. SOLUTION: 63. 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.
SOLUTION: 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
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
us
, we can arbitrarily let one side of the
, we can arbitrarily let one side of the
triangle represent the numerator, x, and the other
triangle represent the numerator, x, and the other
side represent the denominator,
side represent the denominator,
.
.
According to our new labels in the graph,
According to our new labels in the graph,
represents represents u=
sin u =
= , or cot u. Also, sin
, or tan u. Also,
= −1
, so x = sin u and u = sin
, so x = sin u and u =
x.
−1
sin
x.
−1
−1
Finally, we have tan u = tan (sin
Finally, we have cot u = cot (sin
x) =
.
Note that if we had switched the labels of the sides,
−1
we would have had cot (cos x).
x) =
.
Note that if we had switched the labels of the sides,
−1
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 −1(x)
64. SOLUTION: eSolutions Manual - Powered by Cognero
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.
−1
= arcsin x. Describe the domain and range of f o f
Page 22
−1
and f
of .
b. GRAPHICAL Create a table of several values
for each composite function on the interval [–2, 2].
65. MULTIPLE REPRESENTATIONS In this
problem, you will explore the graphs of compositions
4-6 of
Inverse
Trigonometric
Functions
trigonometric
functions.
a. ANALYTICAL Consider f (x) = sin x and f −1(x)
f f −1
−1
= arcsin x. Describe the domain and range of f o f
−1
and f
of .
b. GRAPHICAL Create a table of several values
for each composite function on the interval [–2, 2].
−1
Then use the table to sketch the graphs of f f
−1
and f
f . Use a graphing calculator to check your graphs.
−1
c. ANALYTICAL Consider g(x) = cos x and g
(x) = arccos x. Describe the domain and range of g
−1
−1
g and g
g and make a conjecture as to
−1
f −1 o f
−1
what the graphs of g g and g
g will look
like. Explain your reasoning.
d. GRAPHICAL Sketch the graphs of g g −1
−1
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.
c. f f −1 represents cos (arccos x). The domain of
arccos x is [−1, 1], so the domain of cos (arccos x) is
also [−1, 1]. The range of cos x is [−1, 1], so the
range of cos (arccos x) is also [−1, 1].
SOLUTION: −1
a. f f
represents sin (arcsin x). The domain of
arcsin x is [−1, 1], so the domain of sin (arcsin x) is
also [−1, 1]. The range of sin x is [−1, 1], so the
range of sin (arcsin x) is also [−1, 1].
f −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 , so the range of arcsin (sin x) is
also .
b.
f −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].
The graph of g g −1 should be the line y = x for −1
≤ x ≤ 1. The inverse property of trigonometric functions states that on the closed interval [−1, 1],
cos (cos−1 x) = x. The graph of g −1 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.
d.
g g −1
f f −1
g −1 g
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Page 23
4-6 Inverse Trigonometric Functions
g −1 g
66. ERROR ANALYSIS Alisa and Trey are
discussing inverse trigonometric functions. Because
−1
tan x =
, Alisa conjectures that tan x =
. Trey disagrees. Is either of them
correct? Explain.
e . Sample answer: For f (x) = tan (tan−1 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−1
(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
SOLUTION: 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
−1
and tan
−1
=
x=
−1
, cos
x=
,
, and using these false
relationships to prove her incorrect assumption.
Trigonometric values of angles are related to each
−1
−1
other. However, when we find sin x, cos x, and
−1
tan x, we are calculating angle measures. Angle
measures by themselves do not have this unique
relationship.
arctan.
f(x) = tan (tan−1 x)
67. CHALLENGE Use the graphs of y = sin−1 x and
−1
−1
−1
y = cos x to find the value of sin x + cos
the interval [−1, 1]. Explain your reasoning.
x on
SOLUTION: g(x) = tan−1 (tan x)
−1
a. Graph y = sin
−1
x and y = cos
x.
66. ERROR ANALYSIS Alisa and Trey are
discussing inverse trigonometric functions. Because
−1
tan x =
, Alisa conjectures that tan x =
. Trey disagrees. Is either of them
x = 0.5:
x = 1:
correct? Explain.
SOLUTION: 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
−1
x=
−1
, cos
x=
eSolutions Manual - Powered by Cognero
−1
and tan
=
, and using these false
x = −1:
,
Page 24
If we make a table of values, we can see that sin
−1
−1
x = 1:
4-6 Inverse Trigonometric Functions
x = −1:
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
−1
x + cos
−1
−1
cos
= . Explain your reasoning.
in the interval [−1, 1].
x=
SOLUTION: False; sample answer:
x
−1
−0.5
0
sin −1 x
range of the inverse. Recall that arccos is restricted
to the upper half of the unit circle. The value of θ for
0
−1
cos
−1
cos
π
x
= θ is
.
sin−1 x
y = cos x is one-to-one only for [0, π]. Only within
this domain does the inverse of y = cos x exist.
−1
+ cos
x
x
does not fall within the 0.5
1
REASONING Determine whether each function is odd, even, or neither. Justify your
answer.
-
sin −1 x
69. y = sin−1 x
-
SOLUTION: −1
cos
0
x
-
Sample answer: Suppose y = sin 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
−1
can get sin (−x) = −u.
–1
sin−1 x
−1
−1
+ cos
x
–1
–1
The graph of y = sin x + cos x supports this
conjecture. Every y-value appears to be for the given domain.
68. REASONING Determine whether the following
statement is true or false : If cos
−1
cos
= , then
–1
Also, as shown below, the graph of y = sin
symmetric with respect to the origin.
x is
70. y = cos−1 x
SOLUTION: –1
= . Explain your reasoning.
SOLUTION: eSolutions Manual - Powered by Cognero
False; sample answer:
does not fall within the range of the inverse. Recall that arccos is restricted
Sample answer: The graph of y = cos x is not
symmetric with respect to the y-axis or origin.
–1
Therefore, y = cos x is neither even nor odd.
Page 25
4-6 Inverse Trigonometric Functions
70. y = cos−1 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.
SOLUTION: –1
Sample answer: The graph of y = cos x is not
symmetric with respect to the y-axis or origin.
–1
Therefore, y = cos x is neither even nor odd.
SOLUTION: 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.
−1
For example, when −1 ≤ x ≤ 1, sin (sin x) = x. In
this case, the range of sin x, [−1, 1], limits the domain
−1
of sin
71. y = tan−1 x
SOLUTION: –1
Sample answer: Suppose y = tan x is odd. The definition of an odd function states for every x in the
−1
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,
−1
we can get tan (−x) = −u. Graphically, it can be
seen that for every x in the domain of f , f (−x) = −f
(x).
–1
Also, as shown below, the graph of y = tan
symmetric with respect to the origin.
x.
Locate the vertical asymptotes, and sketch the
graph of each function.
73. y = 3 tan θ
SOLUTION: The graph of y = 3 tan x is the graph of y = tan x
expanded vertically. The period is
or π. Find the
location of two consecutive vertical asymptotes.
x is
and Create a table listing the coordinates of key points
for y = 3 tan x for one period on
.
Function
72. Writing in Math Explain how the restrictions on the
Vertical
Asymptote
Intermediate
Point
x-int
sine, cosine, and tangent functions dictate the domain
and range of their inverse functions.
SOLUTION: 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.
−1
For example, when −1 ≤ x ≤ 1, sin (sin x) = x. In
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case, the
rangeby
ofCognero
sin x, [−1, 1], limits the domain
−1
of sin
x.
y = tan x
y = 3 tan x
(0, 0)
(0, 0)
Intermediate
Point
Vertical
Asymptote
Sketch the curve through the indicated key points for
the function. Then repeat the pattern.
Page 26
−1
For example, when −1 ≤ x ≤ 1, sin (sin x) = x. In
this case, the range of sin x, [−1, 1], limits the domain
4-6 Inverse
Trigonometric Functions
−1
of sin
x.
Locate the vertical asymptotes, and sketch the
graph of each function.
73. y = 3 tan θ
74. y = cot 5θ
SOLUTION: is the graph of y = cot x
The graph of
SOLUTION: The graph of y = 3 tan x is the graph of y = tan 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
for one period on [0, π].
and Function
Create a table listing the coordinates of key points
for y = 3 tan x for one period on
Vertical
Asymptote
Intermediate
Point
x-int
y = tan x
.
y = 3 tan x
(0, 0)
Intermediate
Point
Vertical
Asymptote
x =0
x =0
x=π
x=π
Intermediate
Point
Vertical
Asymptote
(0, 0)
y = cot 5θ
Vertical
Asymptote
Intermediate
Point
x-int
Function
y = cot x
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.
75. y = 3 csc
θ
SOLUTION: 74. y = cot 5θ
The graph of y = 3 csc
SOLUTION: The graph of
θ is the graph of y = csc x
expanded vertically and expanded horizontally. The
is the graph of y = cot x
expanded vertically. The period is
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period is
or 4π. Find the location of two vertical
or π. Find the
location of two consecutive vertical asymptotes.
asymptotes.
Page 27
4-6 Inverse Trigonometric Functions
75. y = 3 csc
76. WAVES A leaf floats on the water bobbing up and
θ
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.
SOLUTION: θ is the graph of y = csc x
The graph of y = 3 csc
expanded vertically and expanded horizontally. The
SOLUTION: or 4π. Find the location of two vertical
period is
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.
asymptotes.
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
Find the value of x. Round to the nearest tenth,
if necessary.
77. SOLUTION: x=π
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.
x = 2π
Sketch the curve through the indicated key points for
the function. Then repeat the pattern.
78. 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
eSolutions Manual - Powered by Cognero
point every 10 seconds. Write a cosine function that
models the movement of the leaf in relationship to
the equilibrium point.
SOLUTION: An acute angle measure and the length of the
opposite leg are given, so the sine function can be
Page 28
used to find the length of the hypotenuse x.
4-6 Inverse Trigonometric Functions
For each pair of functions, find [ f
f ](x), and [ f g](4).
g](x), [g
80. f (x) = x2 + 3x − 6; g(x) = 4x + 1
SOLUTION: 78. SOLUTION: 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. SOLUTION: 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) = SOLUTION: For each pair of functions, find [ f
f ](x), and [ f g](4).
g](x), [g
80. f (x) = x2 + 3x − 6; g(x) = 4x + 1
SOLUTION: eSolutions Manual - Powered by Cognero
82. f (x) =
2
; g(x) = x + 1
SOLUTION: Page 29
4-6 Inverse Trigonometric Functions
82. f (x) =
2
; g(x) = x + 1
SOLUTION: 10 questions
84. 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
SOLUTION: The side adjacent to θ is 24 − 8 − 8 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%?
SOLUTION: 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 0.70.
The correct choice is B.
85. Which of the following represents the exact value of
?
F
G
H
J
10 questions
SOLUTION: 84. 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?
First, find tan
–1
. To do this, find a point on the
unit circle on the interval [0, 2π] to represent this
ratio.
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A 25
Page 30
−1
Therefore, sin θ =
4-6 Inverse Trigonometric Functions
The correct choice is B.
85. Which of the following represents the exact value of
.
The correct choice is H.
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.
?
F
G
SOLUTION: H
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.
J
SOLUTION: 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.
87. REVIEW Two trucks, A and B, start from the
−1
In this diagram, θ = tan
. We can use the
Pythagorean theorem to find the hypotenuse.
Now we can evaluate
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
SOLUTION: Therefore, sin θ =
.
The correct choice is H.
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.
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.
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C 62.5 in.
D 71.8 in.
Page 31
4-6 Inverse Trigonometric Functions
The correct choice is A.
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
SOLUTION: 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.
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