Download Study Guide and Review - Chapter 4

Study Guide and Review - Chapter 4
State whether each sentence is true or false . If
false , replace the underlined term to make a
true sentence.
1. The sine of an acute angle in a right triangle is the
ratio of the lengths of its opposite leg to the
hypotenuse.
5. The rate at which an object moves along a circular
path is called its linear speed.
SOLUTION: The rate at which an object moves along a circular
path is called its linear speed. The rate at which the
object rotates about a fixed point is its angular speed.
SOLUTION: The sine of an acute angle in a right triangle is the
ratio of the lengths of its opposite leg to the
hypotenuse. The cosine is the ratio of the lengths of
its adjacent leg to the hypotenuse. The tangent is the
ratio of the lengths of its opposite leg to the adjacent
leg.
6. 0 ,
SOLUTION: 0 ,
7. The period of the graph of y = 4 sin 3x is 4.
The secant ratio is the reciprocal of the cosine ratio,
while the cosecant ratio is the reciprocal of the sine
ratio. The cotangent is the reciprocal of the tangent
ratio.
3. An angle of elevation is the angle formed by a
horizontal line and an observer’s line of sight to an
object below the line.
SOLUTION: The amplitude of the graph of y = 4 sin 3x is 4. The
period is
. The frequency is
.
8. For f (x) = cos bx, as b increases, the frequency
decreases.
SOLUTION: SOLUTION: An angle of elevation is the angle formed by a
horizontal line and an observer’s line of sight to an
object above the line, while an angle of depression is
the angle formed by a horizontal line and an
observer’s line of sight to an object below the line.
are examples of quadrantal angles. , and
Quadrantal angles lie on one of the coordinate axes.
These angles are not used as reference angles.
2. The secant ratio is the reciprocal of the sine ratio.
SOLUTION: are examples of reference angles.
, and
For f (x) = cos bx, as b increases, the frequency
increases while the period decreases. The frequency
and period are reciprocals.
9. The range of the arcsine function is [0,
].
SOLUTION: 4. The radian measure of an angle is equal to the ratio
The range of the arccosine function is [0,
], while
the range of the arcsine function is
.
of the length of its intercepted arc to the radius.
SOLUTION: The measure θ in radians of a central angle is equal
to the ratio of the length of the intercepted arc s to
the radius r of the circle. A central angle has a
measure of 1 radian if it intercepts an arc with the
same length as the radius of the circle.
10. The Law of Sines can be used to determine
unknown side lengths or angle measures of some
triangles.
SOLUTION: The Law of Sines can be used to determine
unknown side lengths or angle measures of some
triangles. Such occasions are when AAS, ASA, and
SSA are known. Sometimes, when SSS and SAS are
known, the Law of Cosines will be needed as well.
Find the exact values of the six trigonometric
functions of θ.
5. The rate at which an object moves along a circular
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Page 1
path is called its linear speed.
SOLUTION: 11. The Law of Sines can be used to determine
unknown side lengths or angle measures of some
triangles. Such occasions are when AAS, ASA, and
SSAGuide
are known.
SSS and4SAS are
Study
andSometimes,
Review when
- Chapter
known, the Law of Cosines will be needed as well.
Find the exact values of the six trigonometric
functions of θ.
11. Find the value of x. Round to the nearest tenth,
if necessary.
13. SOLUTION: SOLUTION: The length of the side opposite is 36, the length of the side adjacent to is 15, and the length of the hypotenuse is 39.
An acute angle measure and the length of the
hypotenuse are given, so the sine function can be
used to find the length of the side opposite .
14. SOLUTION: An acute angle measure and the length of a leg is
given, so the cosine function can be used to find the
length of the hypotenuse.
12. SOLUTION: The length of the side opposite θ is 9, the length of
the side adjacent to is 40, and the length of the hypotenuse is 41.
Find the measure of angle θ. Round to the
nearest degree, if necessary.
15. SOLUTION: Because the length of the side opposite θ and the
hypotenuse are given, use the sine function.
Find the value of x. Round to the nearest tenth,
if necessary.
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13. Page 2
Study Guide and Review - Chapter 4
Find the measure of angle θ. Round to the
nearest degree, if necessary.
18. 450
SOLUTION: To convert a degree measure to radians, multiply by
15. SOLUTION: Because the length of the side opposite θ and the
hypotenuse are given, use the sine function.
19. SOLUTION: To convert a radian measure to degrees, multiply by
16. SOLUTION: Because the lengths of the sides opposite and
adjacent to θ are given, use the tangent function.
20. SOLUTION: To convert a radian measure to degrees, multiply by
Write each degree measure in radians as a
multiple of π and each radian measure in
degrees.
17. 135
SOLUTION: To convert a degree measure to radians, multiply by
Identify all angles coterminal with the given
angle. Then find and draw one positive and one
negative angle coterminal with the given angle.
21. 342
SOLUTION: 18. 450
SOLUTION: To convert
a degreebymeasure
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to radians, multiply by
All angles measuring
are coterminal
with a 342 angle. Page 3
Sample answer: Let n = 1 and −1.
Study Guide and Review - Chapter 4
Identify all angles coterminal with the given
angle. Then find and draw one positive and one
negative angle coterminal with the given angle.
21. 342
22. SOLUTION: SOLUTION: All angles measuring
All angles measuring
are coterminal
with a 342 angle. Sample answer: Let n = 1 and −1.
are coterminal with a
radian angle. Sample answer: Let n = 1 and −1.
22. Find the area of each sector.
SOLUTION: All angles measuring
are coterminal with a
radian angle. Sample answer: Let n = 1 and −1.
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23. SOLUTION: The measure of the sector’s central angle is 31°
and the radius is 14 inches. Convert the central angle
measure to radians.
Use the central angle and the radius to find the area
Page 4
of the sector.
Therefore, the area of the sector is about 246.1
square meters.
Study Guide and Review - Chapter 4
Find the area of each sector.
Sketch each angle. Then find its reference
angle.
25. 240
SOLUTION: 23. The terminal side of 240º lies in Quadrant III.
Therefore, its reference angle is ' = 240 – 180
or 60º.
SOLUTION: The measure of the sector’s central angle is 31°
and the radius is 14 inches. Convert the central angle
measure to radians.
Use the central angle and the radius to find the area
of the sector.
26. 75
SOLUTION: The terminal side of 75 lies in Quadrant I.
Therefore, its reference angle is ' = 75 – 0 or
75 .
Therefore, the area of the sector is about 53.0
square inches.
24. SOLUTION: The measure of the sector’s central angle
is and the radius is 10 meters.
27. SOLUTION: Use the central angle and the radius to find the area
of the sector.
The terminal side of
lies in Quadrant III. Therefore, its reference angle is
' =
.
Therefore, the area of the sector is about 246.1
square meters.
Sketch each angle. Then find its reference
angle.
25. 240
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SOLUTION: The terminal side of 240º lies in Quadrant III.
28. Page 5
Study Guide and Review - Chapter 4
Find the exact values of the five remaining
trigonometric functions of θ.
27. 29. cos
SOLUTION: The terminal side of
lies in Quadrant III. Therefore, its reference angle is
' =
= , where sin
> 0 and tan > 0
SOLUTION: To find the other function values, find the
coordinates of a point on the terminal side of .
Because sin θ and tan are positive, must lie in
Quadrant I. This means that both x and y are
positive.
.
= Because cos
or , x =2 and r = 5. Find y.
28. Use x = 2, y =
, and r = 5 to write the five
remaining trigonometric ratios.
SOLUTION: The terminal side of
lies in Quadrant II. Therefore, its reference angle is
' =
.
30. where sin
> 0 and cos < 0
SOLUTION: Find the exact values of the five remaining
trigonometric functions of θ.
29. cos
= , where sin
> 0 and tan > 0
To find the other function values, find the
coordinates of a point on the terminal side of .
Because sin θ is positive and cos is negative, must lie in Quadrant II. This means that x is
negative and y is positive.
Because tan
= or –
, use the point (–4, 3) to
find r.
SOLUTION: To find the other function values, find the
coordinates of a point on the terminal side of .
Because sin θ and tan are positive, must lie in
Quadrant I. This means that both x and y are
positive.
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Because
= by or ,x
=2 and r = 5. Find y.
Use x = –4, y = 3, and r = 5 to write the five
remaining trigonometric ratios.
Page 6
Study Guide and Review - Chapter 4
30. where sin
> 0 and cos < 0
SOLUTION: = or –
, use the point (–4, 3) to
find r.
To find the other function values, find the
coordinates of a point on the terminal side of .
Because cos θ is positive and cot θ is negative,
must lie in Quadrant IV. This means that x is
positive and y is negative.
= or –
, y = –5 and r = 13.
Find x.
Use x = –4, y = 3, and r = 5 to write the five
remaining trigonometric ratios.
where cos
Use x = 12, y = –5, and r = 13 to write the five
remaining trigonometric ratios.
> 0 and cot < 0
32. cot θ =
SOLUTION: where cos
> 0 and cot < 0
To find the other function values, find the
coordinates of a point on the terminal side of .
Because cos θ is positive and cot θ is negative,
must lie in Quadrant IV. This means that x is
positive and y is negative.
Because sin
> 0 and cot < 0
where cos
Because sin
31. > 0 and cot < 0
where cos
SOLUTION: To find the other function values, find the
coordinates of a point on the terminal side of .
Because sin θ is positive and cos is negative, must lie in Quadrant II. This means that x is
negative and y is positive.
Because tan
31. = or –
, y = –5 and r = 13.
Find x.
, where sin
< 0 and tan > 0
SOLUTION: To find the other function values, find the
coordinates of a point on the terminal side of .
Because sin θ is negative and tan is positive, must lie in Quadrant III. This means that both x
and y are negative.
Because cot
= or , use the point (–2, –3) to
find r.
Use x = 12, y = –5, and r = 13 to write the five
remaining trigonometric ratios.
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Use x = –2, y = –3, and r =
remaining trigonometric ratios.
to write the five Page 7
Study Guide and Review - Chapter 4
32. cot θ =
, where sin
< 0 and tan > 0
34. cot
SOLUTION: SOLUTION: Because the terminal side of
lies in Quadrant IV, the reference angle θ' of is To find the other function values, find the
coordinates of a point on the terminal side of .
Because sin θ is negative and tan is positive, must lie in Quadrant III. This means that both x
and y are negative.
.
Because cot
= or , use the point (–2, –3) to
find r.
In quadrant IV cot θ is negative.
Use x = –2, y = –3, and r =
remaining trigonometric ratios.
to write the five Find the exact value of each expression. If undefined, write undefined.
33. sin 180
35. sec 450
SOLUTION: Because the terminal side of lies on the positive yaxis, the reference angle ' is .
SOLUTION: Because the terminal side of lies on the negative x-axis, the reference angle ' is 180°.
34. cot
SOLUTION: Because the terminal side of
lies in Quadrant IV, the reference angle θ' of is eSolutions Manual - Powered by Cognero
.
36. SOLUTION: Page 8
Because the terminal side of θ lies in Quadrant III,
Study Guide and Review - Chapter 4
36. SOLUTION: Because the terminal side of θ lies in Quadrant III,
the reference angle θ' of is .
Describe how the graphs of f (x)and g(x) are
related. Then find the amplitude and period of
g(x) , and sketch at least one period of both
functions on the same coordinate axes.
37. f (x) = sin x; g(x) = 5 sin x
SOLUTION: The graph of g(x) is the graph of f (x) expanded
vertically. The amplitude of g(x) is |5| or 5, and the
period is
or 2 .
Create a table listing the coordinates of the xintercepts and extrema for f (x) = sin x for one
period, 2π, on the interval [0, 2 ]. Then use the
amplitude of g(x) to find corresponding points on its
graph.
Describe how the graphs of f (x)and g(x) are
related. Then find the amplitude and period of
g(x) , and sketch at least one period of both
functions on the same coordinate axes.
37. f (x) = sin x; g(x) = 5 sin x
Functions
f (x) = sin x
g(x) = 5 sin x
x-int
(0, 0)
(0, 0)
(π, 0)
(π, 0)
(2π, 0)
(2π, 0)
Max
x-int
SOLUTION: Min
The graph of g(x) is the graph of f (x) expanded
vertically. The amplitude of g(x) is |5| or 5, and the
x-int
period is
or 2 .
Create a table listing the coordinates of the xintercepts and extrema for f (x) = sin x for one
period, 2π, on the interval [0, 2 ]. Then use the
amplitude of g(x) to find corresponding points on its
graph.
Sketch the curve through the indicated points for
each function. Then repeat the pattern to complete a
second period.
Functions
f (x) = sin x
g(x) = 5 sin x
x-int
(0, 0)
(0, 0)
Max
38. f (x) = cos x; g(x) = cos 2x
x-int
(π, 0)
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Min
(π, 0)
SOLUTION: Page 9
The graph of g(x) is the graph of f (x) compressed
horizontally. The amplitude of g(x) is |1| or 1, and the
Study Guide and Review - Chapter 4
38. f (x) = cos x; g(x) = cos 2x
39. f (x) = sin x; g(x) =
sin x
SOLUTION: The graph of g(x) is the graph of f (x) compressed
horizontally. The amplitude of g(x) is |1| or 1, and the
SOLUTION: period is
vertically. The amplitude of g(x) is
or
.
Create a table listing the coordinates of the xintercepts and extrema for f (x) = cos x for one
period, 2π, on the interval [0, 2 ]. Then use g(x) to
find corresponding points on its graph.
Functions
f (x) = cos x
g(x) = cos 2x
Max
(0, 1)
(0, 1)
x-int
Min
(π, –1)
x-int
Max
(2π, 1)
(π, 1)
The graph of g(x) is the graph of f (x) compressed
the period is
or , and
or 2 .
Create a table listing the coordinates of the xintercepts and extrema for f (x) = sin x for one
period, 2π, on the interval [0, 2π]. Then use the
amplitude of g(x) to find corresponding points on its
graph.
g(x) =
Functions
f (x) = sin x
x-int
(0, 0)
(0, 0)
(π, 0)
(π, 0)
(2π, 0)
(2π, 0)
sin x
Max
x-int
Min
Sketch the curve through the indicated points for
each function. Then repeat the pattern to complete a
second period.
x-int
Sketch the curve through the indicated points for
each function. Then repeat the pattern to complete a
second period.
39. f (x) = sin x; g(x) =
sin x
SOLUTION: The graph of g(x) is the graph of f (x) compressed
vertically. The amplitude of g(x) is
or , and
the period
or by
2 Cognero
.
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Create a table listing the coordinates of the x-
40. f (x) = cos x; g(x) = −cos
SOLUTION: x
Page 10
The graph of g(x) is the graph of f (x) expanded
horizontally and reflected in the x-axis. The
Study Guide and Review - Chapter 4
40. f (x) = cos x; g(x) = −cos
State the amplitude, period, frequency, phase
shift, and vertical shift of each function. Then
graph two periods of the function.
41. y = 2 cos (x – )
x
SOLUTION: The graph of g(x) is the graph of f (x) expanded
horizontally and reflected in the x-axis. The
SOLUTION: amplitude of g(x) is |–1| or 1, and the period is
or
In this function, a = 2, b = 1, c = , and d = 0.
Because d = 0, there is no vertical shift.
6 .
Create a table listing the coordinates of the xintercepts and extrema for f (x) = cos x for one
period, 2π, on the interval [0, 2 ]. Then use g(x) to
find corresponding points on its graph.
g(x) = –cos
Functions
f (x) = cos x
x
Max
(0, 1)
(0, −1)
(π, –1)
(3π, 1)
(2π, 1)
(6π, −1)
x-int
Min
x-int
Max
Graph y = 2 cos x shifted
units to the right.
Sketch the curve through the indicated points for
each function. Then repeat the pattern to complete a
second period.
42. y = −sin 2x + 1
SOLUTION: In this function, a =
, b = 2, c = 0, and d = 1.
State the amplitude, period, frequency, phase
shift, and vertical shift of each function. Then
graph two periods of the function.
41. y = 2 cos (x – )
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In this function, a = 2, b = 1, c = , and d = 0.
Because d = 0, there is no vertical shift.
Graph y = – sin 2x shifted 1 unit up.
Page 11
Study Guide and Review - Chapter 4
42. y = −sin 2x + 1
43. SOLUTION: In this function, a =
, b = 2, c = 0, and d = 1.
SOLUTION: In this function, a =
, b = 1, c = , and d = 0.
Because d = 0, there is no vertical shift.
Graph y = – sin 2x shifted 1 unit up.
Graph y =
cos x shifted
units to the left.
43. SOLUTION: In this function, a =
, b = 1, c = , and d = 0.
Because d = 0, there is no vertical shift.
44. SOLUTION: In this function, a = 3, b = 1, c = , and d = 0.
Because d = 0, there is no vertical shift.
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Graph y =
cos x shifted
Page 12
units to the left.
Study Guide and Review - Chapter 4
44. Locate the vertical asymptotes, and sketch the
graph of each function.
45. y = 3 tan x
SOLUTION: The graph of y = 3 tan x is the graph of y = tan x
SOLUTION: In this function, a = 3, b = 1, c = expanded vertically. The period is
, and d = 0.
or . Find
the location of two consecutive vertical asymptotes.
Because d = 0, there is no vertical shift.
and Create a table listing the coordinates of key points
for y = 3 tan x for one period on
.
Function
Graph y = 3 sin x shifted
Vertical
Asymptote
Intermediate
Point
x-int
units to the left.
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 to sketch at
least one more cycle to the left and right of the first
curve.
Locate the vertical asymptotes, and sketch the
graph of each function.
45. y = 3 tan x
SOLUTION: The graph of y = 3 tan x is the graph of y = tan x
expanded vertically. The period is
or . Find
the location of two consecutive vertical asymptotes.
and Create a table listing the coordinates of key points
for y = 3 tan x for one period on
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Function
.
46. Page 13
SOLUTION: Asymptote
Sketch the curve through the indicated key points for
the function. Then repeat the pattern to sketch at
least one more cycle to the left and right of the first
curve.
Study Guide and Review - Chapter 4
46. SOLUTION: is the graph of y =
The graph of
tan x compressed vertically and translated
units or . Find the location of
right. The period is
two consecutive vertical asymptotes.
47. and SOLUTION: Create a table listing the coordinates of key points
for
for one period on [0,
].
x shifted
Function
Vertical
Asymptote
Intermediate
Point
x-int
units to the left. The period is
or .
Find the location of two consecutive vertical
asymptotes.
y = tan x
x =0
(0, 0)
Intermediate
Point
Vertical
Asymptote
is the graph of y = cot
The graph of
and Create a table listing the coordinates of key points
for
x=π
for one period on
.
Function
Sketch the curve through the indicated key points for
the function. Then repeat the pattern to sketch at
least one more cycle to the left and right of the first
curve.
Vertical
Asymptote
Intermediate
Point
x-int
y = cot x
x =0
Intermediate
Point
Vertical
Asymptote
x=π
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47. Sketch the curve through the indicated key points for
the function. Then repeat the pattern to sketch at
least one more cycle to the left and right of the first
curve.
Page 14
Study Guide and Review - Chapter 4
48. y = −cot (x –
47. )
SOLUTION: The graph of
is the graph of y = cot
x reflected in the x-axis and shifted π units to the
SOLUTION: is the graph of y = cot
The graph of
or . Find the location of two
left. The period is
x shifted
units to the left. The period is
or .
consecutive vertical asymptotes.
Find the location of two consecutive vertical
asymptotes.
and Create a table listing the coordinates of key points
for
for one period on [ , 2 ].
and Function
Create a table listing the coordinates of key points
for
for one period on
Vertical
Asymptote
Intermediate
Point
x-int
.
Function
y = cot x
Vertical
Asymptote
Intermediate
Point
x-int
x =0
Intermediate
Point
Vertical
Asymptote
x =0
x=π
x = 2π
Sketch the curve through the indicated key points for
the function. Then repeat the pattern to sketch at
least one more cycle to the left and right of the first
curve.
Intermediate
Point
Vertical
Asymptote
y = cot x
x=π
Sketch the curve through the indicated key points for
the function. Then repeat the pattern to sketch at
least one more cycle to the left and right of the first
curve.
49. SOLUTION: 48. y = −cot (x –
)
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The graph of
is the graph of y = cot
x reflected in the x-axis and shifted π units to the
The graph of
is the graph of y = sec x
expanded vertically and expanded horizontally. Page
The 15
period is
or 4 . Find the location of two
Study Guide and Review - Chapter 4
50. y = –csc (2x)
49. SOLUTION: The graph of
is the graph of y = csc x
compressed horizontally and reflected in the x-axis.
SOLUTION: is the graph of y = sec x
The graph of
or . Find the location of two
The period is
expanded vertically and expanded horizontally. The
consecutive vertical asymptotes.
or 4 . Find the location of two
period is
and consecutive vertical asymptotes.
Create a table listing the coordinates of key points
for
and for one period on
.
Create a table listing the coordinates of key points
for
Function
for one period on [−π, 3π].
Vertical
Asymptote
Intermediate
Point
x-int
Function
Vertical
Asymptote
Intermediate
Point
x-int
y = csc x
y = sec x
(0, 1)
Intermediate
Point
Vertical
Asymptote
Intermediate
Point
Vertical
Asymptote
(0, 1)
x=π
y = –csc
2x
x = −π
x =0
x =0
x=π
x = 3π
Sketch the curve through the indicated key points for
the function. Then repeat the pattern to sketch at
least one more cycle to the left and right of the first
curve.
Sketch the curve through the indicated key points for
the function. Then repeat the pattern to sketch at
least one more cycle to the left and right of the first
curve.
51. y = sec (x –
)
SOLUTION: The graph of y = sec (x −
50. y = –csc (2x)
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SOLUTION: The graph of
is the graph of y = csc x
compressed horizontally and reflected in the x-axis.
) is the graph of y = sec
x translated π units to the right. The period is
or Page 16
2 . Find the location of two vertical asymptotes.
51. y = sec (x –
)
Study
Guide and Review - Chapter 4
SOLUTION: The graph of y = sec (x −
52. ) is the graph of y = sec
x translated π units to the right. The period is
or 2 . Find the location of two vertical asymptotes.
SOLUTION: is the graph of y =
The graph of
csc x compressed vertically and translated
the left. The period is
and units to
or 2 . Find the location of
two consecutive vertical asymptotes.
Create a table listing the coordinates of key points
for y = sec (x − π) for one period on
and .
Function
Vertical
Asymptote
Intermediate
Point
x-int
y = sec x
y = sec (x −
π)
Create a table listing the coordinates of key points
for
for one period on
.
(0, 1)
Function
Vertical
Asymptote
Intermediate
Point
x-int
Intermediate
Point
Vertical
Asymptote
Intermediate
Point
Vertical
Asymptote
Sketch the curve through the indicated key points for
the function. Then repeat the pattern to sketch at
least one more cycle to the left and right of the first
curve.
y=
csc x
x =0
x=π
Sketch the curve through the indicated key points for
the function. Then repeat the pattern to sketch at
least one more cycle to the left and right of the first
curve.
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Find the exact value of each expression, if Page
it 17
exists.
SOLUTION: The graph of
is the graph of y =
53. sin−1 (−1)
When t =
Study Guide and Review - Chapter 4
–1
, cos t =
. Therefore, cos
=
.
Find the exact value of each expression, if it
exists.
55. 53. sin−1 (−1)
SOLUTION: SOLUTION: Find a point on the unit circle on the interval
Find a point on the unit circle on the interval
such that with a y-coordinate of –1.
=
When t =
, sin t = –1. Therefore, sin
–1
When t =
–1 =
.
, tan t =
= . Therefore, tan
–1
.
56. arcsin 0
54. SOLUTION: SOLUTION: Find a point on the unit circle on the interval
Find a point on the unit circle on the interval
with a x-coordinate of
with a y-coordinate of 0.
.
When t = 0, sin t = 0. Therefore, arcsin 0 = 0.
When t =
, cos t =
–1
. Therefore, cos
.
=
57. arctan −1
SOLUTION: Find a point on the unit circle on the interval
such that =
55. SOLUTION: eSolutions
Manual - Powered by Cognero
Find a point on the unit circle on the interval
Page 18
When t =
Study Guide and Review - Chapter 4
, cos t =
. Therefore, arccos
=
.
When t = 0, sin t = 0. Therefore, arcsin 0 = 0.
57. arctan −1
59. SOLUTION: Find a point on the unit circle on the interval
such that =
SOLUTION: The inverse property applies, because
lies on the interval [–1, 1]. Therefore,
=
.
60. SOLUTION: The inverse property applies, because cos –3π lies
on the interval [–1, 1]. Therefore,
When t =
= , tan t =
. Therefore, arctan
Find all solutions for the given triangle, if
possible. If no solution exists, write no solution.
Round side lengths to the nearest tenth and
angle measurements to the nearest degree.
61. a = 11, b = 6, A = 22
.
58. arccos
SOLUTION: Draw a diagram of a triangle with the given
dimensions.
SOLUTION: Find a point on the unit circle on the interval
with a x-coordinate of
.
Notice that A is acute and a > b because 11 > 6.
Therefore, one solution exists. Apply the Law of
Sines to find B.
When t =
, cos t =
. Therefore, arccos
=
.
59. SOLUTION: The inverse property applies, because
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the interval [–1, 1]. Therefore,
lies on =
Because two angles are now known, C 180 –
(22 + 12 ) or about 146 . Apply the Law of Sines
to find c.
Page 19
SOLUTION: The inverse property applies, because cos –3π lies
on the interval [–1, 1]. Therefore,
Study Guide and Review - Chapter 4
62. a = 9, b = 10, A = 42
SOLUTION: Draw a diagram of a triangle with the given
dimensions.
Find all solutions for the given triangle, if
possible. If no solution exists, write no solution.
Round side lengths to the nearest tenth and
angle measurements to the nearest degree.
61. a = 11, b = 6, A = 22
SOLUTION: Notice that A is acute and a < b because 9 < 10.
Therefore, two solutions may exist. Find h.
Draw a diagram of a triangle with the given
dimensions.
Notice that A is acute and a > b because 11 > 6.
Therefore, one solution exists. Apply the Law of
Sines to find B.
9 > 6.69, so two solutions exist.
Apply the Law of Sines to find B.
Because two angles are now known, C 180 –
(22 + 12 ) or about 146 . Apply the Law of Sines
to find c.
Because two angles are now known, C 180 –
(42 + 48 ) or about 90 . Apply the Law of Sines
to find c.
Therefore, the remaining measures of
B 12 , C 146 , and c 16.4.
are
62. a = 9, b = 10, A = 42
SOLUTION: Draw a diagram of a triangle with the given
dimensions.
When,
then C 180 – (42 + 132 ) or
about 6º. Apply the Law of Sines to find c.
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Page 20
Study
Guide and Review - Chapter 4
When,
then C 180 – (42 + 132 ) or
about 6º. Apply the Law of Sines to find c.
Therefore, the remaining measures of
B 29 , C 73 , and c 19.5.
are
64. a = 2, b = 9, A = 88
SOLUTION: Notice that A is acute and a < b because 2 < 9. Find
h.
Therefore, the remaining measures of
are
B 48 , C 90 , and c 13.5 or B = 132 , C =
6 , and c = 1.4.
Because a < b and a < h, no triangle can be formed
with sides a = 2, b = 9, and A = 88 . Therefore, this
problem has no solution.
Solve each triangle. Round side lengths to the
nearest tenth and angle measures to the
nearest degree.
65. a = 13, b = 12, c = 8
63. a = 20, b = 10, A = 78
SOLUTION: Draw a diagram of a triangle with the given
dimensions.
SOLUTION: Use the Law of Cosines to find an angle measure.
Notice that A is acute and a > b because 20 > 10.
Therefore, one solution exists. Apply the Law of
Sines to find B.
Use the Law of Sines to find a missing angle
measure.
Because two angles are now known, C 180 –
(78 + 29 ) or about 73 . Apply the Law of Sines
to find c.
Find the measure of the remaining angle.
Therefore, A
78 , B
65 , and C
37 .
66. a = 4, b = 5, C = 96
Therefore, the remaining measures of
B 29 , C 73 , and c 19.5.
are
64. a = Manual
2, b = -9,Powered
A = 88by Cognero
eSolutions
SOLUTION: Notice that A is acute and a < b because 2 < 9. Find
SOLUTION: Use the Law of Cosines to find the missing side
measure.
Page 21
Find the measure of the remaining angle.
Study
GuideA and
Therefore,
, B 65 - ,Chapter
and C 374 .
78 Review
66. a = 4, b = 5, C = 96
SOLUTION: Use the Law of Cosines to find the missing side
measure.
Find the measure of the remaining angle.
Therefore, c = 6.7, A = 36 , and B = 48 .
67. CONSTRUCTION A construction company is installing a three-foot-high wheelchair ramp onto a
landing outside of an office. The angle of the ramp
must be 4°.
a. What is the length of the ramp?
b. What is the slope of the ramp?
SOLUTION: a. We are given the angle and the height opposite the
angle. Draw a diagram.
Use the Law of Sines to find a missing angle
measure.
We need to find the length of the ramp, so we can
use the tangent function.
Find the measure of the remaining angle.
Therefore, c = 6.7, A = 36 , and B = 48 .
67. CONSTRUCTION A construction company is installing a three-foot-high wheelchair ramp onto a
landing outside of an office. The angle of the ramp
must be 4°.
a. What is the length of the ramp?
b. What is the slope of the ramp?
b. The slope is equal to the rise over the run. One
coordinate is (0, 0) and another coordinate is (43, 3).
SOLUTION: a. We are given the angle and the height opposite the
angle. Draw a diagram.
68. NATURE For a photography project, Maria is photographing deer from a tree stand. From her sight
30 feet above the ground, she spots two deer in a
straight line, as shown below. How much farther
away is the second deer than the first?
We need to find the length of the ramp, so we can
use the tangent function.
SOLUTION: eSolutions Manual - Powered by Cognero
Draw a diagram of the situation.
Page 22
Study Guide and Review - Chapter 4
68. NATURE For a photography project, Maria is photographing deer from a tree stand. From her sight
30 feet above the ground, she spots two deer in a
straight line, as shown below. How much farther
away is the second deer than the first?
69. FIGURE SKATING An Olympic ice skater performs a routine in which she jumps in the air for
2.4 seconds while spinning 3 full revolutions.
a. Find the angular speed of the figure skater.
b. Express the angular speed of the figure skater in
degrees per minute.
SOLUTION: a. The rate at which an object rotates about a fixed
point is its angular speed. The angular speed is given
SOLUTION: Draw a diagram of the situation.
by
. Each rotation requires 2
6 .
The angular speed is 2.5
The 40° and 60° angles shown are equal to the angles in the original picture because they are
alternate interior angles.
radians, so θ =
radians per second.
b. Convert the seconds to minutes then convert the
radians to degrees.
We can use trigonometry to find x and y and then
find y − x, which is the distance between the deer.
The angular speed is about 27,000 degrees per
minute.
70. TIMEPIECES The length of the minute hand of a
pocket watch is 1.5 inches. What is the area swept
by the minute hand in 40 minutes? SOLUTION: After 40 minutes, the minute hand has traveled
of a revolution. Multiply by 360 to determine
the central angle θ.
69. FIGURE SKATING An Olympic ice skater performs a routine in which she jumps in the air for
2.4 seconds while spinning 3 full revolutions.
a. Find the angular speed of the figure skater.
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by Cognero
b. Express
angular
speed of the figure skater in
degrees per minute.
SOLUTION: The area of a sector is determined by A =
r
2
,
where r is the radius and is the measure of the central angle in radians. We need to convert thePage 23
degrees to radians.
TheGuide
angular speed
about 27,000
degrees per
Study
and is
Review
- Chapter
4
minute.
The area is about 4.71 square inches.
70. TIMEPIECES The length of the minute hand of a
71. WORLD’S FAIR The first Ferris wheel had a
pocket watch is 1.5 inches. What is the area swept
by the minute hand in 40 minutes? diameter of 250 feet and took 10 minutes to
complete one full revolution.
a. How many degrees would the Ferris wheel rotate
in 100 seconds?
b. How far has a person traveled if he or she has
been on the Ferris wheel for 7 minutes?
c. How long would it take for a person to travel 200
feet?
SOLUTION: SOLUTION: After 40 minutes, the minute hand has traveled
a. One complete revolution is done in 10 minutes, so
divide 100 seconds by 10 minutes to determine the
fraction of a revolution that has been completed in
100 seconds.
of a revolution. Multiply by 360 to determine
the central angle θ.
The area of a sector is determined by A =
r
2
,
where r is the radius and is the measure of the central angle in radians. We need to convert the
degrees to radians.
One sixth of a revolution is equal to
degrees.
b. If someone has been on it for 7 minutes, then they
have made
Substitute the known values of r and
A.
to determine
or 60 of a revolution. The wheel has a diameter of 250 feet, so the circumference is 250
feet. After 7 minutes,
of the circumference has been traveled.
They have traveled about 550 feet.
c. It takes 10 minutes to travel the circumference, or
250 feet. Use a proportion to solve for t.
The area is about 4.71 square inches.
71. WORLD’S FAIR The first Ferris wheel had a
diameter of 250 feet and took 10 minutes to
complete one full revolution.
a. How many degrees would the Ferris wheel rotate
in 100 seconds?
b. How far has a person traveled if he or she has
been on the Ferris wheel for 7 minutes?
c. How long would it take for a person to travel 200
feet?
SOLUTION: a. One complete revolution is done in 10 minutes, so
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divide 100 seconds by 10 minutes to determine the
fraction of a revolution that has been completed in
100 seconds.
It would take about 2.5 minutes to travel 200 feet.
72. AIR CONDITIONING An air-conditioning unit
turns on and off to maintain the desired temperature.
On one summer day, the air conditioner turns on at
Page 24
8:30 A.M. when the temperature is 80 Fahrenheit
and turns off at 8:55 A.M. when the temperature is
74°.
Study Guide and Review - Chapter 4
It would take about 2.5 minutes to travel 200 feet.
72. AIR CONDITIONING An air-conditioning unit
that, or 12 hours and 30 minutes.
b. The last high tide occurred at 10:45 in the
morning, so the next one will occur 12 hours and 30
minutes from then, or at 11:15 in the evening.
74. MUSIC When plucked, a bass string is displaced
turns on and off to maintain the desired temperature.
On one summer day, the air conditioner turns on at
8:30 A.M. when the temperature is 80 Fahrenheit
and turns off at 8:55 A.M. when the temperature is
74°.
a. Find the amplitude and period if you were going to
use a trigonometric function to model this change in
temperature, assuming that the temperature cycle
will continue.
b. Is it appropriate to model this situation with a
trigonometric function? Explain your reasoning.
1.5 inches, and its damping factor is 1.9. It produces
a note with a frequency of 90 cycles per second.
Determine the amount of time it takes the string’s
motion to be dampened so that −0.1 < y < 0.1.
SOLUTION: SOLUTION: First, write a trigonometric function that models that
motion of the string. The maximum displacement of
–ct
the string occurs when t = 0, so y = k e cos t
can be used to model the motion of the string
because the graph of y = cos t has a y-intercept
other than 0.
a. The range of temperatures is 6 degrees. Half of
this range is 3, so the amplitude is 3. The
temperature takes 25 minutes to go down and we
are assuming that it will take 25 minutes to go back
up, so one full period is 50 minutes.
The maximum displacement occurs when the string
is plucked 1.5 inches. The total displacement is the
maximum displacement M minus the minimum
displacement m, so k = M – m = 1.5 in.
You can use the value of the frequency to find
b. Sample answer: The function appears to not be
periodic because it might take more or less time for
the temperature to warm back up to 80 . The
temperature will increase at a different rate
depending on the time of day. For example, it will get
hotter much quicker between 10 in the morning and
2 in the afternoon.
73. TIDES In Lewis Bay, the low tide is recorded as 2 feet at 4:30 A.M., and the high tide is recorded as
5.5 feet at 10:45 A.M.
a. Find the period for the trigonometric model.
b. At what time will the next high tide occur?
SOLUTION: a. The period is from one low tide to the next low
tide. From low tide to high tide, the time elapsed is 6
hours and 15 minutes. Therefore, the period is twice
that, or 12 hours and 30 minutes.
b. The last high tide occurred at 10:45 in the
morning, so the next one will occur 12 hours and 30
minutes from then, or at 11:15 in the evening.
74. MUSIC When plucked, a bass string is displaced
1.5 inches, and its damping factor is 1.9. It produces
a note with a frequency of 90 cycles per second.
Determine the amount of time it takes the string’s
motion to be dampened so that −0.1 < y < 0.1.
SOLUTION: First, write a trigonometric function that models that
eSolutions
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Cognero
motion
of the
string.byThe
maximum displacement of
–ct
the string occurs when t = 0, so y = k e cos t
can be used to model the motion of the string
.
Write a function using the values of k,
, and c.
–1.9t
y = 1.5e
t is one model that describes
the motion of the string.
–1.
Use a graphing calculator to graph y = 1.5e
t, y = –0.1, and y = 0.1. Select the intersect
feature under the TRACE menu to find the point
–1.9t
where the graphs of y = 1.5e
t and
y = –0.1 or y = 0.1 first intersect within the interval
–0.1 ≤ y ≤ 0.1.
From the graph, you can see that it takes
approximately 1.41 seconds for the graph of y =
–1.9t
1.5e
t to oscillate within the interval –
0.1 ≤ y ≤ 0.1.
75. PAINTING A painter is using a 15-foot ladder to
paint the side of a house. If the angle the ladder
makes with the ground is less than 65 , it will slide
Page 25
out from under him. What is the greatest distance
that the bottom of the ladder can be from the side of
From the graph, you can see that it takes
approximately 1.41 seconds for the graph of y =
–1.9t
1.5eGuide andt Review
to oscillate -within
the interval
Study
Chapter
4 –
0.1 ≤ y ≤ 0.1.
75. PAINTING A painter is using a 15-foot ladder to
paint the side of a house. If the angle the ladder
makes with the ground is less than 65 , it will slide
out from under him. What is the greatest distance
that the bottom of the ladder can be from the side of
the house and still be safe for the painter?
At the minimum angle, the ladder is about 6.3 feet
away from the side of the house.
76. NAVIGATION A boat is 20 nautical miles from a port at a bearing 30° north of east. The captain sees a second boat and reports to the port that his boat is
15 nautical miles from the second boat, which is
located due east of the port. Can port personnel be
sure of the second boat’s position? Justify your
answer.
SOLUTION: Draw a diagram of the situation.
SOLUTION: The smaller the angle, the larger the distance from
the ladder to the house. Therefore, to find the
greatest safe distance from the house, we can use
the smallest safe angle. The minimum angle that we
can use is 65 degrees, so find the distance from the
ladder to the side of the house at this minimum angle.
Use the cosine function.
From the diagram, there are two possible positions
where the second boat can be located. This follows
the ambiguous case of the Law of Sines. 10 < 15 <
20, so there are two possible triangles with the given
dimensions.
77. GEOMETRY Consider quadrilateral ABCD.
At the minimum angle, the ladder is about 6.3 feet
away from the side of the house.
76. NAVIGATION A boat is 20 nautical miles from a port at a bearing 30° north of east. The captain sees a second boat and reports to the port that his boat is
15 nautical miles from the second boat, which is
located due east of the port. Can port personnel be
sure of the second boat’s position? Justify your
answer.
a. Find C.
b. Find the area of ABCD.
SOLUTION: a. Use SAS to find the length of BD.
SOLUTION: Draw a diagram of the situation.
Use SSS to find C.
From the diagram, there are two possible positions
where the second boat can be located. This follows
the ambiguous
casebyofCognero
the Law of Sines. 10 < 15 <
eSolutions
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20, so there are two possible triangles with the given
dimensions.
b. Use SAS to find the area of each triangle formed
Page 26
by BD.
From the diagram, there are two possible positions
where the second boat can be located. This follows
the ambiguous case of the Law of Sines. 10 < 15 <
20, so
there are
twoReview
possible triangles
with the
Study
Guide
and
- Chapter
4 given
dimensions.
77. GEOMETRY Consider quadrilateral ABCD.
a. Find C.
b. Find the area of ABCD.
SOLUTION: a. Use SAS to find the length of BD.
Use SSS to find C.
b. Use SAS to find the area of each triangle formed
by BD.
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