Download 4-5 Graphing Other Trigonometric Functions

Point
Vertical
Asymptote
4-5 Graphing Other Trigonometric Functions
Sketch the curve through the indicated key points for
the function. Then repeat the pattern.
Locate the vertical asymptotes, and sketch the
graph of each function.
1. y = 2 tan x
SOLUTION: The graph of y = 2 tan x is the graph of y = tan x
expanded vertically. The period is
= . y = a tan (bx + c), so a = 2, b = 1, and c = 0. Use
the tangent asymptote equations to find the location
of the asymptotes.
2. SOLUTION: is the graph of y = tan
The graph of
units to the left. The period is x shifted
or . y = a tan (bx + c), so a = 1, b = 1, and c =
. Use
the tangent asymptote equations to find the location
of the asymptotes.
Create a table listing the coordinates of key points
for y = 2 tan x for one period on
.
Function
Vertical
Asymptote
Intermediate
Point
x-int
y = tan x
y = 2 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.
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Create a table listing the coordinates of key points
for
for one period on
.
Function
Vertical
Asymptote
Intermediate
Point
x-int
Intermediate
Point
y = tan x
y = tan (x +
4)
(0, 0)
Page 1
(0, 1)
Function
y = tan x
4-5
Vertical
Asymptote
Graphing
Other
Intermediate
Point
x-int
y = tan (x +
4)
Create a table listing the coordinates of key points
Trigonometric Functions
for
for one period on
.
(0, 0)
Function
Intermediate
Point
Vertical Asymptote
(0, 1)
Sketch the curve through the indicated key points for
the function. Then repeat the pattern.
y = cot x
Vertical
Asymptote
Intermediate
Point
x-int
x=0
Intermediate
Point
Vertical
Asymptote
Sketch the curve through the indicated key points for
the function. Then repeat the pattern.
3. SOLUTION: is the graph of y = cot
The graph of
units to the right. The period is x shifted
or 4. y = –3 tan
. SOLUTION: y = a cot (bx + c), so a = 1, b = 1, and c =
. Use
the tangent asymptote equations to find the location
of two consecutive vertical asymptotes.
The graph of y = –3 tan
is the graph of y = tan x
expanded vertically, expanded horizontally, and
reflected in the x-axis. The period is
or 3 . y = a tan (bx + c), so a = –3, b = , and c = 0. Use
the tangent asymptote equations to find the location
of two consecutive vertical asymptotes.
Create a table listing the coordinates of key points
for
for one period on
.
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eSolutions
Function
y = cot x
Page 2
or . The period is
4-5 Graphing Other Trigonometric Functions
y = a cot (bx + c), so a =
, b = 1, and c = 0. Use
the tangent asymptote equations to find the location
of two consecutive vertical asymptotes.
Create a table listing the coordinates of key points
for y = tan 2x for one period on
.
Create a table listing the coordinates of key points
Function
Vertical
Asymptote
Intermediate
Point
x-int
for
y = tan x
for one period on
.
Function
(0, 0)
(0, 0)
Intermediate
Point
Vertical
Asymptote
y = cot x
Vertical
Asymptote
Intermediate
Point
x-int
x=0
x=0
Intermediate
Point
Vertical
Asymptote
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.
5. SOLUTION: 6. y = –tan 3x
is the graph of y = cot x
The graph of
compressed vertically and reflected in the x-axis.
The period is
or . y = a cot (bx + c), so a =
, b = 1, and c = 0. Use
to find the location
of two consecutive vertical asymptotes.
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the tangent
asymptote
equations
SOLUTION: The graph of y = – tan 3x is the graph of y = tan x
compressed horizontally and reflected in the x-axis.
The period is
or . y = a tan (bx + c), so a = –1, b = 3, and c = 0. Use
the tangent asymptote equations to find the location
Page 3
of two consecutive vertical asymptotes.
compressed horizontally and reflected in the x-axis.
or The period is
. 4-5 Graphing Other Trigonometric Functions
y = a tan (bx + c), so a = –1, b = 3, and c = 0. Use
the tangent asymptote equations to find the location
of two consecutive vertical asymptotes.
7. y = –2 tan (6x − π)
SOLUTION: The graph of
is the graph of y =
tan x expanded vertically, compressed horizontally,
shifted units to the right, and reflected in the x or axis. The period is
. y = a tan (bx + c), so a = –2, b = 6, and c = –π. Use
the tangent asymptote equations to find the location
of two consecutive vertical asymptotes.
Create a table listing the coordinates of key points
for y = tan 2x for one period on
.
Function
Vertical
Asymptote
Intermediate
Point
x-int
y = tan x
y = 2 tan
x
(0, 0)
(0, 0)
Intermediate
Point
Vertical
Asymptote
Create a table listing the coordinates of key points
for y = tan 2x for one period on
.
Function
Vertical
Asymptote
Intermediate
Point
x-int
Sketch the curve through the indicated key points for
the function. Then repeat the pattern.
y = tan x
y = 2 tan x
(0, 0)
Intermediate
Point
Vertical
Asymptote
7. y = –2 tan (6x − π)
Sketch the curve through the indicated key points for
the function. Then repeat the pattern.
SOLUTION: The graph of
is the graph of y =
tan x expanded vertically, compressed horizontally,
shifted
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Manual units to the right, and reflected in the x- Powered by Cognero
axis. The period is
or . Page 4
Point
Vertical
Asymptote
x=0
Asymptote
Intermediate
Point
x-int
4-5 Graphing Other Trigonometric Functions
Sketch the curve through the indicated key points for
the function. Then repeat the pattern.
x=0
Intermediate
Point
Vertical
Asymptote
Sketch the curve through the indicated key points for
the function. Then repeat the pattern.
8. SOLUTION: is the graph of y = cot x
The graph of
expanded horizontally. The period is
or 2 . 9. y =
csc 2x
y = a cot (bx + c), so a = 1, b = , and c = 0. Use
the tangent asymptote equations to find the location
of the asymptotes.
SOLUTION: compressed vertically and compressed horizontally.
Find the location of two consecutive vertical
asymptotes.
The period is
csc 2x is the graph of y = csc x
The graph of y =
or . y = a csc (bx + c), so a = , b = 2, and c = 0. Use
the asymptote equations to find the location of two
vertical asymptotes.
Create a table listing the coordinates of key points
for
for one period on
.
Function
y = cot x
Vertical
Asymptote
Intermediate
Point
x-int
x=0
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Point
Vertical
Create a table listing the coordinates of key points
x=0
for y =
csc 2x for one period on
.
Function
Vertical
Asymptote
Intermediate
Point
y = csc x
Page 5
for y =
csc 2x for one period on
.
4-5 Graphing
FunctionOther Trigonometric Functions
y = csc x
Vertical
Asymptote
Intermediate
Point
x-int
x=0
x=0
Create a table listing the coordinates of key points
Intermediate
Point
Vertical
Asymptote
for one period on for
.
Sketch the curve through the indicated key points for
the function. Then repeat the pattern.
Function
y = csc x
Vertical
Asymptote
Intermediate
Point
x = –π
x = x-int
x=0
x = x=π
x=
Intermediate
Point
Vertical
Asymptote
Sketch the curve through the indicated key points for
the function. Then repeat the pattern.
10. SOLUTION: The graph of
is the graph of y
= csc x compressed horizontally and shifted units to the left. The period is
or . y = a csc (bx + c), so a = 1, b = 4, and c = . Use
the asymptote equations to find the location of two
vertical asymptotes.
11. y = sec (x + π)
SOLUTION: The graph of y = sec (x + π) is the graph of y = sec
x shifted π units to the left. The period is
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or 2
. y = a sec (bx + c), so a = 1, b = 1, and c = π. Use
the asymptote equations to find the location of two
vertical asymptotes.
Page 6
x shifted π units to the left. The period is
4-5
or 2
. Graphing Other Trigonometric Functions 12. y = –2 csc 3x
y = a sec (bx + c), so a = 1, b = 1, and c = π. Use
SOLUTION: the asymptote equations to find the location of two
vertical asymptotes.
The graph of
is the graph of y = csc x
expanded vertically, compressed horizontally, and
reflected in the x-axis. The period is
or . y = a csc (bx + c), so a = –2, b = 3, and c = 0. Use
the asymptote equations to find the location of two
vertical asymptotes.
Create a table listing the coordinates of key points
for y = sec (x +
) for one period on
.
Function
Vertical
Asymptote
Intermediate
Point
x-int
Create a table listing the coordinates of key points
y = sec x
for one period on for
.
Function
(0, 1)
Vertical
Asymptote
Intermediate
Point
x-int
Intermediate
Point
Vertical
Asymptote
(0, –1)
y = csc x
y = –2 csc 3x
x=0
x=0
Intermediate
Point
Vertical
Asymptote
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.
12. y = –2 csc 3x
SOLUTION: TheManual
graph-of
is the graph of y
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= csc x
expanded vertically, compressed horizontally, and
reflected in the x-axis. The period is
or . Page 7
13. Intermediate
Point
Vertical
Asymptote
4-5 Graphing Other Trigonometric Functions
Sketch the curve through the indicated key points for
the function. Then repeat the pattern.
13. SOLUTION: is the graph of y =
The graph of
sec x expanded vertically and shifted
units to the
or 2 . right. The period is
y = a sec (bx + c), so a = 4, b = 1, and c =
. Use
the asymptote equations to find the location of two
vertical asymptotes.
14. SOLUTION: is the graph of y = sec
The graph of
x expanded horizontally and shifted π units to the
left. The period is
or 10 . y = a sec (bx + c), so a = 1, b = , and c = . Use
the asymptote equations to find the location of two
vertical asymptotes.
Create a table listing the coordinates of key points
for one period on for
.
Function
Vertical
Asymptote
Intermediate
Point
x-int
y=
sec
x
(0,
1)
Intermediate
Point
Vertical
Asymptote
Sketch
the- curve
through
the
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indicated key points for
the function. Then repeat the pattern.
Create a table listing the coordinates of key points
for
for one period on Page. 8
4-5 Graphing Other Trigonometric Functions
Create a table listing the coordinates of key points
for one period on for
y = a csc (bx + c), so a = , b = 1, and c =
. Use
the asymptote equations to find the location of two
vertical asymptotes.
.
Function
Vertical
Asymptote
Intermediate
Point
x-int
y = sec x
(0, 1)
Intermediate
Point
Vertical
Asymptote
Create a table listing the coordinates of key points fo
for one period on .
Sketch the curve through the indicated key points for
the function. Then repeat the pattern.
Function
Vertical
Asymptote
Intermediate
Point
x-int
y = csc
x
x=0
Intermediate
Point
Vertical
Asymptote
15. Sketch the curve through the indicated key points for
the function. Then repeat the pattern.
SOLUTION: is the graph of y =
The graph of
csc x expanded vertically and shifted
units to the
or 2 . right. The period is
y = a csc (bx + c), so a = , b = 1, and c =
. Use
the asymptote equations to find the location of two
vertical asymptotes.
16. y = –sec
SOLUTION: The graph of
is the graph of y = sec x
expanded horizontally and reflected in the x-axis.
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The period is
or 16 . Page 9
SOLUTION: 4-5 The
Graphing
graph of Other Trigonometric
is the graph of y = Functions
sec x
expanded horizontally and reflected in the x-axis.
or 16 . The period is
y = a sec (bx + c), so a = –1, b = , and c = 0. Use
the asymptote equations to find the location of two
vertical asymptotes.
Identify the damping factor f (x) of each
function. Then use a graphing calculator to
sketch the graphs of f (x), −f (x), and the given
function in the same viewing window. Describe
the behavior of the graph.
17. y =
x sin x
SOLUTION: The function y =
functions y =
x sin x is the product of the
x and y = sin x, so f (x) =
x. Use a
graphing calculator to graph f (x), –f (x), and y =
x
sin x in the same viewing window.
Create a table listing the coordinates of key points
for one period on [−4 , 12 ].
for
The amplitude of the function is decreasing as x
approaches 0 from both directions, and increasing as
x approaches positive and negative infinity.
Function
Vertical
Asymptote
Intermediate
Point
x-int
y = sec x
18. y = 4x cos x
(0, 1)
(0, –1)
Intermediate
Point
Vertical
Asymptote
SOLUTION: The function y = 4x cos x is the product of the
functions y = 4x and y = cos x, so f (x) = 4x. Use a
graphing calculator to graph f (x), –f (x), and y = 4x
cos x in the same viewing window.
Sketch the curve through the indicated key points for
the function. Then repeat the pattern.
The amplitude of the function is decreasing as x
approaches 0 from both directions, and increasing as
x approaches positive and negative infinity.
19. y = 2x2 cos x
SOLUTION: Identify
damping
factor f (x)
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of each
function. Then use a graphing calculator to
sketch the graphs of f (x), −f (x), and the given
2
The function y = 2x cos x is the product of thePage 10
2
2
functions y = 2x and y = cos x, so f (x) = 2x . Use a
2
4-5
The amplitude of the function is decreasing as x
approaches
0 from
bothTrigonometric
directions, and increasing
as
Graphing
Other
Functions
x approaches positive and negative infinity.
19. y = 2x2 cos x
The amplitude of the function is decreasing as x
approaches 0 from both directions, and increasing as
x approaches positive and negative infinity.
21. y =
x sin 2x
SOLUTION: 2
The function y = 2x cos x is the product of the
2
2
functions y = 2x and y = cos x, so f (x) = 2x . Use a
2
graphing calculator to graph f (x), –f (x), and y = 2x
cos x in the same viewing window.
SOLUTION: The function y =
functions y =
x sin 2x is the product of the
x and y = sin 2x, so f (x) =
x. Use
a graphing calculator to graph f (x), –f (x), and y =
x sin 2x in the same viewing window.
The amplitude of the function is decreasing as x
approaches 0 from both directions, and increasing as
x approaches positive and negative infinity.
The amplitude of the function is decreasing as x
approaches 0 from both directions, and increasing as
x approaches positive and negative infinity.
20. SOLUTION: The function y =
x sin x is the product of the
and y = sin x, so f (x) =
functions y =
. Use a
graphing calculator to graph f (x), –f (x), and y =
sin x in the same viewing window.
22. y = (x – 2)2 sin x
SOLUTION: 2
The function y = (x − 2) sin x is the product of the
2
functions y = (x − 2) and y = sin x, so f (x) = (x − 2)
2
. Use a graphing calculator to graph f (x), –f (x), and
2
y =(x − 2) sin x in the same viewing window.
The amplitude of the function is decreasing as x
approaches 0 from both directions, and increasing as
x approaches positive and negative infinity.
21. y =
23. y = e 0.5x cos x
x sin 2x
SOLUTION: SOLUTION: 0.5x
The function y =
functions y =
The amplitude of the function is decreasing as x
approaches 2 from both directions, and increasing as
x approaches positive and negative infinity.
x sin 2x is the product of the
x and y = sin 2x, so f (x) =
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x. Use
a graphing calculator to graph f (x), –f (x), and y =
x sin 2x in the same viewing window.
The function y = e
cos x is the product of the
0.5x
0.5x
functions y = e
and y = cos x, so f (x) = e . Use
a graphing calculator to graph f (x), –f (x), and y =
e
0.5x
cos x in the same viewing window.
Page 11
4-5
The amplitude of the function is decreasing as x
approaches
2 from
bothTrigonometric
directions, and increasing
as
Graphing
Other
Functions
x approaches positive and negative infinity.
23. y = e 0.5x cos x
25. y = |x| cos 3x
SOLUTION: SOLUTION: 0.5x
The function y = e
cos x is the product of the
0.5x
0.5x
functions y = e
and y = cos x, so f (x) = e . Use
a graphing calculator to graph f (x), –f (x), and y =
e
0.5x
cos x in the same viewing window.
The function y =
cos 3x is the product of the
functions y = and y = cos 3x, so f (x) = . Use a
graphing calculator to graph f (x), –f (x), and y =
cos 3x in the same viewing window.
The amplitude of the function is decreasing as x
approaches 0 from both directions, and increasing as
x approaches positive and negative infinity.
The amplitude of the function is decreasing as x
approaches negative infinity and increasing as x
approaches positive infinity.
24. y = 3x sin x
26. y = ln x cos x
SOLUTION: SOLUTION: x
The function y = 3 sin x is the product of the
x
x
functions y = 3 and y = sin x, so f (x) = 3 . Use a
x
graphing calculator to graph f (x), –f (x), and y = 3
sin x in the same viewing window.
The amplitude of the function is decreasing as x
approaches negative infinity and increasing as x
approaches infinity.
The function y = ln x cos x is the product of the
functions y = ln x and y = cos x, so f (x) = ln x. Use a
graphing calculator to graph f (x), –f (x), and y = ln x
cos x in the same viewing window.
The
and the amplitude of the function is increasing as x approaches infinity.
27. MECHANICS When the car shown below hit a
25. y = |x| cos 3x
SOLUTION: The function y =
The amplitude of the function is decreasing as x
approaches negative infinity and increasing as x
approaches infinity.
cos 3x is the product of the
functions y = and y = cos 3x, so f (x) = . Use a
graphing calculator to graph f (x), –f (x), and y =
cos 3x in the same viewing window.
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bump in the road, the shock absorber was
compressed 8 inches, released, and then began to
vibrate in damped harmonic motion with a frequency
of 2.5 cycles per second. The damping constant for
the shock absorber is 3.
Page 12
a. Write a trigonometric function that models the
displacement of the shock absorber y as a function
of time t. Let t = 0 be the instant the shock absorber
27. MECHANICS When the car shown below hit a
4-5
bump in the road, the shock absorber was
compressed 8 inches, released, and then began to
vibrate
in damped
harmonic
motion with a frequency
Graphing
Other
Trigonometric
Functions
of 2.5 cycles per second. The damping constant for
the shock absorber is 3.
The intersection of the two graphs occurs when x ≈ 0.06. So, the amount of time that it takes for the
amplitude of the vibration to decrease to 4 inches is
about 0.06 second.
28. DIVING The end of a diving board is 20.3 centimeters above its resting position at the moment
a diver leaves the board. Two seconds later, the
board has moved down and up 12 times. The
damping constant for the board is 0.901.
a. Write a trigonometric function that models the
displacement of the shock absorber y as a function
of time t. Let t = 0 be the instant the shock absorber
is released.
b. Determine the amount of time t that it takes for
the amplitude of the vibration to decrease to 4
inches.
SOLUTION: a. The maximum displacement of the shock absorber
−ct
occurs when t = 0, so y = k e cos ωt can be used
to model the motion of the shock absorber because
the graph of y = cos t has a y-intercept other than 0.
The maximum displacement occurs when the shock
absorber is compressed 8 inches. The total
displacement k is the maximum displacement minus
the minimum displacement. So, the total
displacement is 8 − 0 or 8 inches. The damping
constant c for the shock absorber is 3.
Use the value of the frequency to find .
a. Write a trigonometric function that models the
motion of the diving board y as a function of time t.
b. Determine the amount of time t that it takes the
diving board to be damped so that –0.5 ≤ y ≤ 0.5.
SOLUTION: a. The maximum displacement of the diving board
−ct
occurs when t = 0, so y = k e cos ωt can be used
to model the motion of the diving board because the
graph of y = cos t has a y-intercept other than 0. The
maximum displacement occurs when the diving
board is 20.3 centimeters above its resting position.
The total displacement k is the maximum
displacement minus the minimum displacement. So,
the total displacement is 20.3 centimeters. The
damping constant c for the diving board is 0.901.
Write a function using the values of k, ω , and c.
−3t
y = 8e cos 5πt is one model that describes the
motion of the shock absorber.
Since after 2 seconds the board has moved down
and up 12 times, it will have moved down and up 6
times after 1 second. Use the value of the frequency
to find .
b. Use a graphing calculator to determine the value
of t when the graph of y = 8e
−3t
cos 5πt is equal to
−3t
4. Graph y = 8e cos 5πt and y = 4 on the same
coordinate plane. Find the intersection of the two
graphs using the intersect function from the CALC
menu.
Write a function using the values of k,
, and c.
−0.901t
y = 20.3e
cos 12πt is one model that describes
the motion of the diving board.
b. Use a graphing calculator to determine the value
−0.901t
of t when the graph of y = 20.3e
cos 12πt is
oscillating between y = −0.5 and y = 0.5.
−0.901t
Graph y = 20.3e
cos 12πt, y = 0.5, and y =
−0.5 on the same coordinate plane.
The intersection of the two graphs occurs when x ≈ 0.06. So, the amount of time that it takes for the
amplitude of the vibration to decrease to 4 inches is
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about
0.06- second.
28. DIVING The end of a diving board is 20.3 Page 13
b. Use a graphing calculator to determine the value
−0.901t
of t when the graph of y = 20.3e
cos 12πt is
oscillating between y = −0.5 and y = 0.5.
4-5 Graphing Other
−0.901t Trigonometric Functions
Graph y = 20.3e
cos 12πt, y = 0.5, and y =
−0.5 on the same coordinate plane.
Create a table listing the coordinates of key points
From this window, it is unclear as to when the graph
−0.901t
of y = 20.3e
cos 12πt oscillates within the
interval −0.5 ≤ y ≤ 0.5. Adjust the window.
for y = sec x + 3 for one period on
.
Function
y = sec x
y = sec x
+3
(0, 1)
(0, 4)
Vertical
Asymptote
Intermediate
Point
x-int
Using the intersect function from the CALC menu,
−0.901t
Intermediate
Point
Vertical
Asymptote
it can be found that the graph of y = 20.3e
cos 12πt intersects y = −0.5 for the last time when x
−0.901t
≈ 4.09. When x > 4.09, the graph of y = 20.3e
cos 12πt will stay between y = −0.5 and y = 0.5. So,
it takes approximately 4.09 seconds for the graph of
Sketch the curve through the indicated key points for
the function. Then repeat the pattern.
−0.901t
y = 20.3e
cos 12πt to oscillate within the
interval −0.5 ≤ y ≤ 0.5.
Locate the vertical asymptotes, and sketch the
graph of each function.
29. y = sec x + 3
SOLUTION: The graph of y = sec x + 3 is the graph of y = sec x
shifted 3 units up. The period is
or 2 . y = a sec (bx + c), so a = 1, b = 1, and c = 3. Use
the asymptote equations to find the location of two
vertical asymptotes.
30. SOLUTION: is the graph of y =
The graph of
sec x shifted
period is
to the right and 4 units up. The or 2 . y = a sec (bx + c), so a = 1, b = 1, and c =
eSolutions Manual - Powered by Cognero
. Use
the asymptote equations to find the location of two
Page 14
vertical asymptotes.
period is
or 2 . 4-5 Graphing Other Trigonometric Functions
y = a sec (bx + c), so a = 1, b = 1, and c =
31. y = csc
. Use
− 2
SOLUTION: the asymptote equations to find the location of two
vertical asymptotes.
The graph of
expanded horizontally and shifted 2 units down. The
or 6π. period is
is the graph of y = csc x
y = a tan (bx + c), so a = 1, b = , and c = 2. Use
the asymptote equations to find the location of two
vertical asymptotes.
Create a table listing the coordinates of key points
for one period on for
.
Function
y = sec x
Vertical
Asymptote
Intermediate
Point
x-int
Create a table listing the coordinates of key points
x =0
for one period on for
(0, 1)
.
Function
Intermediate
Point
Vertical
Asymptote
Vertical
Asymptote
Intermediate
Point
x-int
Sketch the curve through the indicated key points for
the function. Then repeat the pattern.
y = csc x
x=0
x=0
Intermediate
Point
Vertical
Asymptote
Sketch the curve through the indicated key points for
the function. Then repeat the pattern.
31. y = csc
− 2
SOLUTION: The graph of
is the graph of y = csc x
eSolutions Manual - Powered by Cognero
Page 15
expanded horizontally and shifted 2 units down. The
period is
or 6π. 32. Point
Vertical
Asymptote
Function
4-5 Graphing Other Trigonometric Functions
y = csc x
Vertical
Asympt
Interm
Point
Sketch the curve through the indicated key points for
the function. Then repeat the pattern.
x=0
x-int
Interm
Point
Vertical
Asympt
Sketch the curve through the indicated key points for
the function. Then repeat the pattern.
32. SOLUTION: is the graph of y
The graph of
= csc x compressed horizontally, shifted
the left, and shifted 3 units up. The period is
units to . 33. y = cot (2x + π) − 3
y = a csc (bx + c), so a = 1, b = 3, and c = . Use
the asymptote equations to find the location of two
vertical asymptotes.
SOLUTION: is the graph of y =
The graph of
cot x compressed horizontally, shifted
units to the left, and shifted 3 units down. The period is
. y = a cot (bx + c), so a = 1, b = 2, and c = . Use th
asymptote equations to find the location of two
consecutive vertical asymptotes.
Create a table listing the coordinates of key points
for one period on for
.
Create a table listing the coordinates of key points fo
for one period on
Function
y = csc x
eSolutions Manual - Powered by Cognero
x-int
Interm
Function
Vertical
Asympt
Interm
Point
x=0
.
Vertical
Asymptote
Intermediate
Point
x-int
y = cot x
x=0
Page 16
for one period on
.
4-5 Graphing
FunctionOther Trigonometric Functions
y = cot x
Vertical
Asymptote
Intermediate
Point
x-int
x=0
Intermediate
Point
Vertical
Asymptote
Create a table listing the coordinates of key points
x=0
for
for one period on
.
Function
Sketch the curve through the indicated key points for
the function. Then repeat the pattern.
y = cot x
Vertical
Asympt
Interm
Point
x-int
x=0
Interm
Point
Vertical
Asympt
Sketch the curve through the indicated key points for
the function. Then repeat the pattern.
34. SOLUTION: The graph of
is the graph of y =
cot x expanded horizontally, shifted π units to the left,
and shifted 1 unit down. The period is
or 2π. y = a cot (bx + c), so a = 1, b = , and c = . Use
the asymptote equations to find the location of two
consecutive vertical asymptotes.
35. PHOTOGRAPHY Jeff is taking pictures of a hawk that is flying 150 feet above him. The hawk
will eventually fly directly over Jeff. Let d be the
distance Jeff is from the hawk and θ be the angle of
elevation to the hawk from Jeff’s camera.
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a. Write d as a function of θ.
b. Graph the function on the interval 0 < θ < π.
c. Approximately how far away is the hawk from
Page 17
Jeff when the angle of elevation is 45°?
SOLUTION: 4-5 a.
Graphing
Other Trigonometric Functions
Write d as a function of θ.
b. Graph the function on the interval 0 < θ < π.
c. Approximately how far away is the hawk from
Jeff when the angle of elevation is 45°?
c. Substitute θ = 45° into the function you found in part a.
SOLUTION: a.
The hawk is about 212.1 feet away from Jeff when
the angle of elevation is 45°.
Since d is the length of the hypotenuse, and you are
given the length of the side opposite θ, you can use
the sine function to write d as a function of θ.
36. DISTANCE A spider is slowly climbing up a wall. Brianna is standing 6 feet away from the wall
watching the spider. Let d be the distance Brianna is
from the spider and θ be the angle of elevation to the
spider from Brianna.
a. Write d as a function of θ.
b. Graph the function on the interval 0 < θ < .
c. Approximately how far away is the spider from
Brianna when the angle of elevation is 32°?
SOLUTION: a. Make a diagram of the situation.
b. Evaluate d = 150 csc θ for several values of θ
within the domain.
θ
d
212.1
162.4
150
162.4
Since d is the length of the hypotenuse, and you are
given the length of the side adjacent to θ, you can
use the cosine function to write d as a function of θ.
212.1
Use these points to graph d = 150 csc θ.
b. Evaluate d = 6 sec θ for several values of θ within
the domain.
θ
d
c. Substitute θ = 45° into the function you found in part a.
0
6
6.1
eSolutions Manual - Powered by Cognero
6.5
7.2
Page 18
θ
d
0
6
4-5 Graphing Other Trigonometric Functions
6.1
6.5
7.2
8.5
10.8
The spider is about 7.1 feet away from Brianna
when the angle of elevation is 32°.
GRAPHING CALCULATOR Find the values of θ on the interval – < θ < that make each
equation true.
37. cot θ = 2 sec θ
SOLUTION: Write the equation as two separate equations, y =
cot θ and y = 2 sec θ. Graph both equations on the
same coordinate plane.
15.7
30.8
-
Use these points to graph d = 6 sec θ.
Find the points of intersection using the intersect
function from the CALC menu.
The graphs appear to intersect when x 0.427 and x 2.715. So, the values of θ that make the
equation true are about 0.427 and 2.715.
c. Substitute θ = 32° into the function you found in part a.
38. sin θ = cot θ
SOLUTION: The spider is about 7.1 feet away from Brianna
when the angle of elevation is 32°.
Write the equation as two separate equations, y = sin
θ and y = cot θ. Graph both equations on the same
coordinate plane.
GRAPHING CALCULATOR Find the values of θ on the interval – < θ < that make each
equation true.
37. cot θ = 2 sec θ
SOLUTION: Write the equation as two separate equations, y =
cot θ and y = 2 sec θ. Graph both equations on the
same coordinate plane.
Find the points of intersection using the intersect
function from the CALC menu.
eSolutions Manual - Powered by Cognero
Find the points of intersection using the intersect
function from the CALC menu.
Page 19
The graphs appear to intersect when x
−0.905
4-5
The graphs appear to intersect when x 0.427 and x 2.715. So, the values of θ that make the
Graphing
Other
Functions
equation
true are
about Trigonometric
0.427 and 2.715.
38. sin θ = cot θ
The graphs appear to intersect when x −0.905
and x 0.905. So, the values of θ that make the
equation true are about −0.905 and 0.905.
39. 4 cos θ = csc θ
SOLUTION: SOLUTION: Write the equation as two separate equations, y = sin
θ and y = cot θ. Graph both equations on the same
coordinate plane.
Write the equation as two separate equations, y = 4
cos θ and y = csc θ. Graph both equations on the
same coordinate plane.
Find the points of intersection using the intersect
function from the CALC menu.
Find the points of intersection using the intersect
function from the CALC menu.
The graphs appear to intersect when x −0.905
and x 0.905. So, the values of θ that make the
equation true are about −0.905 and 0.905.
The graphs appear to intersect when x −2.880, x
≈ −1.833, x 0.262, and x 1.309. So, the values of θ that make the equation true are about −2.880,
−1.833, 0.262, and 1.309.
39. 4 cos θ = csc θ
SOLUTION: Write the equation as two separate equations, y = 4
cos θ and y = csc θ. Graph both equations on the
same coordinate plane.
40. tan = sin θ
SOLUTION: Write the equation as two separate equations, y =
tan and y = sin θ. Graph both equations on the
same coordinate plane.
Find the points of intersection using the intersect
function from the CALC menu.
eSolutions Manual - Powered by Cognero
Page 20
The graphs appear to intersect when x
−1.571, x
4-5
The graphs appear to intersect when x −2.880, x
≈ −1.833, x 0.262, and x 1.309. So, the values of
θ that makeOther
the equation
true are about −2.880,
Graphing
Trigonometric
Functions
−1.833, 0.262, and 1.309.
40. tan = sin θ
SOLUTION: Write the equation as two separate equations, y =
tan and y = sin θ. Graph both equations on the
The graphs appear to intersect when x −1.571, x
= 0, and x 1.571. So, the values of θ that make the
equation true are about −1.571, 0, and 1.571.
41. csc θ = sec θ
SOLUTION: Write the equation as two separate equations, y =
csc θ and y = sec θ. Graph both equations on the
same coordinate plane.
same coordinate plane.
Find the points of intersection using the intersect
function from the CALC menu.
The graphs appear to intersect when x −1.571, x
= 0, and x 1.571. So, the values of θ that make the
equation true are about −1.571, 0, and 1.571.
41. csc θ = sec θ
SOLUTION: Write the equation as two separate equations, y =
csc θ and y = sec θ. Graph both equations on the
same coordinate plane.
Find the points of intersection using the intersect
function from the CALC menu.
eSolutions Manual - Powered by Cognero
The graphs appear to intersect when x −2.356
and x 0.785. So, the values of θ that make the
Find the points of intersection using the intersect
function from the CALC menu.
The graphs appear to intersect when x −2.356
and x 0.785. So, the values of θ that make the
equation true are about −2.356 and 0.785.
42. tan θ = sec
SOLUTION: Write the equation as two separate equations, y =
csc θ and y = sec . Graph both equations on the
same coordinate plane.
Find the points of intersection using the intersect
function from the CALC menu.
Page 21
The graphs appear to intersect when x −2.050
and x 0.830. So, the values of θ that make the
4-5
The graphs appear to intersect when x −2.356
and
x 0.785. So, the values of θ
that makeFunctions
the
Graphing
Other Trigonometric
equation true are about −2.356 and 0.785.
42. tan θ = sec
SOLUTION: Write the equation as two separate equations, y =
csc θ and y = sec . Graph both equations on the
same coordinate plane.
find the downward force.
b. Write an equation that represents the tension T on
each rope.
c. Graph the equation from part b on the interval [0,
180°].
d. Suppose the mural is 9.14 meters long and the
ideal angle θ for tension purposes is a right angle.
Determine how much rope is needed to transport the
mural and the tension that is being applied to each
rope.
e . Suppose you have 12.2 meters of rope to use to
transport the mural. Find θ and the tension that is
being applied to each rope.
SOLUTION: a. The downward force F is equal to the mass m of
the mural times gravity g, so F = mg.
Find the points of intersection using the intersect
function from the CALC menu.
Therefore, the downward force is 5331.2 N.
The graphs appear to intersect when x −2.050
and x 0.830. So, the values of θ that make the
equation true are about −2.050 and 0.830.
Therefore, an equation that represents the tension on
each rope is T = 2665.6 sec .
43. TENSION A helicopter is delivering a large mural b. The tension T on each rope is equal to half the
downward force times sec .
c.
that is to be displayed in the center of town. Two
ropes are used to attach the mural to the helicopter,
as shown. The tension T on each rope is equal to
half the downward force times sec .
d. Draw a diagram to model the situation.
a. The downward force in newtons equals the mass
of the mural times gravity, which is 9.8 newtons per
kilogram. If the mass of the mural is 544 kilograms,
find the downward force.
b. Write an equation that represents the tension T on
each rope.
c. Graph the equation from part b on the interval [0,
180°].
d. Suppose the mural is 9.14 meters long and the
idealManual
angle- θPowered
for tension
purposes is a right angle.
eSolutions
by Cognero
Determine how much rope is needed to transport the
mural and the tension that is being applied to each
rope.
Use the Pythagorean Theorem to find x.
Therefore, 2(6.46) or about 12.9 meters of rope is
needed to transport the mural. Find the tension.
Page 22
4-5 Graphing Other Trigonometric Functions
Therefore, θ is about 97º and the tension being
applied to each rope is about 4023 N.
Therefore, 2(6.46) or about 12.9 meters of rope is
needed to transport the mural. Find the tension.
Match each function with its graph.
Therefore, the tension that is being applied to each
rope is about 3769.7 N.
e . Draw a diagram to model the situation. Because
12.2 meters of rope is used to transport the mural,
each rope will be 12.2 ÷ 2 or 6.1 meters long. When
the altitude is drawn, the triangle is separated into
two right triangles, as shown.
44. SOLUTION: The parent function y = csc x resembles either graph
b or c. Find the location of two vertical asymptotes.
and Because the hypotenuse and the side opposite an
acute angle are given, you can use the sine function
to find θ.
The location of the asymptotes suggests that the
correct answer is c.
45. SOLUTION: The parent function y = sec x resembles either graph
b or c. Find the location of two vertical asymptotes.
Find the tension.
Therefore, θ is about 97º and the tension being
applied to each rope is about 4023 N.
and The location of the asymptotes suggests that the
correct answer is b.
46. Match each function with its graph.
SOLUTION: The cotangent function has a graph comprised of
branches that decrease as x increases. The only
graph that meets this condition is d.
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47. Page 23
an integer, f (x) is also undefined for x =
SOLUTION: where n is an integer. However, g(x) is defined at
these values. So, the expressions are not equivalent
for all real numbers.
The cotangent function has a graph comprised of
branches
that Other
decrease Trigonometric
as x increases. The only
Graphing
Functions
graph that meets this condition is d.
4-5
+ n ,
49. f (x) = sec2 x; g(x) = tan2 x + 1
47. SOLUTION: Graph f (x) and g(x) on the same coordinate plane.
SOLUTION: The tangent function has a graph comprised of
branches that increase as x increases. The only
graph that meets this condition is a.
GRAPHING CALCULATOR Graph each pair
of functions on the same screen and make a
conjecture as to whether they are equivalent for all
real numbers. Then use the properties of the
functions to verify each conjecture.
48. f (x) = sec x cos x; g(x) = 1
Since sec x is undefined for x =
SOLUTION: + n , where n is
an integer, f (x) is also undefined for x =
Graph f (x) and g(x) on the same coordinate plane.
+ n ,
where n is an integer. Since tan x is undefined for x
= + n , where n is an integer, g(x) is also
undefined for x =
+ nπ, where n is an integer. f(x) and g(x) have the same domains and are
represented by identical graphs. So, the expressions
are equivalent.
Since sec x is undefined for x =
+ n , where n is
an integer, f (x) is also undefined for x =
50. f (x) = cos x csc x; g(x) = cot x
SOLUTION: + n ,
Graph f (x) and g(x) on the same coordinate plane.
where n is an integer. However, g(x) is defined at
these values. So, the expressions are not equivalent
for all real numbers.
49. f (x) = sec2 x; g(x) = tan2 x + 1
SOLUTION: Graph f (x) and g(x) on the same coordinate plane.
Since csc x is undefined for x = n π, where n is an
integer, f (x) is also undefined for x = n π, where n is
an integer. Since cot x is undefined for x = n π,
where n is an integer, g(x) is also undefined for x =
n π, where n is an integer. f (x) and g(x) have the
same domains and are represented by identical
graphs. So, the expressions are equivalent.
Since sec x is undefined for x =
+ n , where n is
an integer, f (x) is also undefined for x =
+ n ,
where n is an integer. Since tan x is undefined for x
= Manual
+ n -, Powered
where nbyisCognero
an integer, g(x) is also
eSolutions
undefined for x =
51. + nπ, where n is an integer. SOLUTION: Page 24
Graph f (x) and g(x) on the same coordinate plane.
4-5
an integer. Since cot x is undefined for x = n π,
where n is an integer, g(x) is also undefined for x =
n π, where n is an integer. f (x) and g(x) have the
same
domainsOther
and are represented
by identical
Graphing
Trigonometric
Functions
graphs. So, the expressions are equivalent.
an integer, f (x) is also undefined for these values of
x. Because of the phase, f (x) is undefined for x =
n , where n is an integer. g(x) is defined at these
values. So, the expressions are not equivalent for all
real numbers.
Write an equation for the given function given
the period, phase shift (ps), and vertical shift
(vs).
52. function: sec; period: 3π; ps: 0; vs: 2
51. SOLUTION: Graph f (x) and g(x) on the same coordinate plane.
SOLUTION: Start with the general form of the secant function
y = a sec (bx + c) + d. The period is 3π and there is
a vertical shift of 2 units up, so d = 2.
For a secant function, period =
.
Since sec x is undefined for x =
+ n , where n is
an integer, f (x) is also undefined for these values of
x. Because of the phase, f (x) is undefined for x =
n , where n is an integer. g(x) is defined at these
values. So, the expressions are not equivalent for all
real numbers.
Write an equation for the given function given
the period, phase shift (ps), and vertical shift
(vs).
52. function: sec; period: 3π; ps: 0; vs: 2
Therefore, a secant function with period 3
vertical shift of 2 units up is given by
or
and a
.
SOLUTION: Start with the general form of the secant function
y = a sec (bx + c) + d. The period is 3π and there is
a vertical shift of 2 units up, so d = 2.
For a secant function, period =
53. function: tan; period:
; ps:
; vs: –1
SOLUTION: Start with the general form of the tangent function
y = a tan (bx + c) + d. The period is , there is a
.
phase shift of units to the right, and there is a vertical shift of 1 unit down, so d = –1.
For a tangent function, period =
Therefore, a secant function with period 3
vertical shift of 2 units up is given by
or
and a
.
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53. function: tan; period:
.
; ps:
Page 25
; vs: –1
Therefore, a secant function with period 3
vertical shift of 2 units up is given by
and a
of 1 unit down is given by
or
.
4-5 Graphing Other
Trigonometric
Functions
53. function: tan; period:
; ps:
.
54. function: csc; period:
; vs: –1
or
; ps: –π; vs: 0
SOLUTION: SOLUTION: Start with the general form of the tangent function
y = a tan (bx + c) + d. The period is , there is a
Start with the general form of the cosecant function
y = a csc (bx + c) + d. The period is , and there is
phase shift of units to the right, and there is a a phase shift of π units to the left.
vertical shift of 1 unit down, so d = –1.
For a cosecant function, period =
For a tangent function, period =
.
.
The phase shift = –
. Use b to find c.
The phase shift = –
. Use b to find c.
Therefore, a cosecant function with period
and a phase shift of units to the left is given by
y = csc (8x + 8 ) or y = csc (–8x + 8 ).
Therefore, a tangent function with a period
phase shift of
,a
units to the right, and a vertical shift
of 1 unit down is given by
or
.
54. function: csc; period:
55. function: cot; period: 3π; ps:
; vs: 4
SOLUTION: Start with the general form of the cotangent function
y = a cot (bx + c) + d. The period is 3 , there is a
phase shift of units to the right, and a vertical shift
of 4 units up.
For a cotangent function, period =
; ps: –π; vs: 0
.
SOLUTION: Start with the general form of the cosecant function
y = a csc (bx + c) + d. The period is , and there is
a phase shift of π units to the left.
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For a cosecant function, period =
Page 26
.
Therefore, a cosecant function with period
and a of 4 units up is given by
shift of Other
units to the left is given by
4-5 phase
Graphing
Trigonometric Functions
.
y = csc (8x + 8 ) or y = csc (–8x + 8 ).
55. function: cot; period: 3π; ps:
; vs: 4
56. function: csc; period:
SOLUTION: Start with the general form of the cotangent function
y = a cot (bx + c) + d. The period is 3 , there is a
phase shift of units to the right, and a vertical shift
For a cotangent function, period =
; vs: –3
;
SOLUTION: Start with the general form of the cosecant function
y = a csc (bx + c) + d. The period is , there is a
phase shift of
of 4 units up.
or
units to the left, and a vertical shift of 3 units down.
.
For a cosecant function, period =
The phase shift = –
.
. Use b to find c.
The phase shift = –
. Use b to find c.
Therefore, a cotangent function with period 3π, a
phase shift of units to the right, and a vertial shift of 4 units up is given by
or
Therefore, a cosecant function with period
phase shift of
,a
units to the left, and a vertical shift of 3 units down is given by y = csc (6x + 3π) – 3 or
y = csc (–6x + 3π) – 3.
.
57. PROOF Verify that the y-intercept for the graph of
–ct
56. function: csc; period:
;
; vs: –3
any function of the form y = k e
cos ωt is k.
SOLUTION: SOLUTION: To find the y-intercept, let t = 0.
Start with the general form of the cosecant function
y = a csc (bx + c) + d. The period is , there is a
phase shift of
units to the left, and a vertical shift of 3 units down.
eSolutions Manual - Powered by Cognero
For a cosecant function, period =
Page 27
.
REASONING Determine whether each statement is true or false . Explain your
Therefore, a cosecant function with period
phase shift of
,a
units to the left, and a vertical shift 3 units downOther
is givenTrigonometric
by y = csc (6x + 3π)Functions
– 3 or
4-5 of
Graphing
y = csc (–6x + 3π) – 3.
57. PROOF Verify that the y-intercept for the graph of
–ct
any function of the form y = k e
cos ωt is k.
The cotangent and cosecant functions are both
undefined when x = n , where n is an integer.
Therefore, x = θ is an asymptote for both y = csc x
and y = cot x.
60. ERROR ANALYSIS Mira and Arturo are
studying the graph shown. Mira thinks that it is the
graph of
SOLUTION: To find the y-intercept, let t = 0.
, and Arturo thinks that it is
the graph of y =
cot 2x. Is either of them correct?
Explain your reasoning.
REASONING Determine whether each statement is true or false . Explain your
reasoning.
58. If b ≠ 0, then y = a + b sec x has extrema of ±(a +
b).
SOLUTION: The parent function y = sec x has extrema of −1 and
1. Substitute these values for sec x into the original
equation.
y = a + b(1) y = a + b y = a + b(−1)
y=a−b
Half of the extrema will be equal to a + b, but the
other half will be equal to a − b. Thus, the statement
is false.
SOLUTION: The graph of y = cot x has no y-intercept, while the
graph of y = tan x has a y-intercept of 0. Because
neither of the equations in the problem indicate a
horizontal shift of the parent function and the curve
in the graph does not pass through the origin, Mira’s
answer must be incorrect.
To determine whether Arturo’s equation is correct,
find the period, asymptotes, x-intercepts, and a
couple of intermediate points for y =
The period is
cot 2x.
or . Find the location of two
consecutive vertical asymptotes.
and 59. If x = θ is an asymptote of y = csc x, then x = θ is
also an asymptote of y = cot x.
There is an x-intercepts at and, by subtracting the
SOLUTION: Since y = csc x =
, asymptotes will occur for
values of x when sin x = 0. Since y = cot x or
, asymptotes will also occur for values of x
when sin x = 0. The cotangent and cosecant functions are both
undefined when x = n , where n is an integer.
Therefore, x = θ is an asymptote for both y = csc x
and y = cot x.
60. ERROR ANALYSIS Mira and Arturo are
studying the graph shown. Mira thinks that it is the
graph of
, and Arturo thinks that it is
eSolutions Manual - Powered by Cognero
the graph of y =
cot 2x. Is either of them correct?
period from , another x-intercept is at
. By
substituting values for x, there are intermediate
points at
and . Since all of these
characteristics match the graph, Arturo is correct.
61. CHALLENGE Write a cosecant function and a
cotangent function that have the same graphs as y =
sec x and y = tan x respectively. Check your
answers by graphing.
SOLUTION: The cosecant function resembles the secant function
except that it is translated units to the right or left.
So, adding or subtracting
Page 28
from x in the cosecant
function should translate the graph of the function so
that it has the same graph as y = sec x. Graph
substituting values for x, there are intermediate
and points at
. Since all of these
4-5 Graphing Other Trigonometric Functions
characteristics match the graph, Arturo is correct.
61. CHALLENGE Write a cosecant function and a
cotangent function that have the same graphs as y =
sec x and y = tan x respectively. Check your
answers by graphing.
SOLUTION: The cosecant function resembles the secant function
except that it is translated units to the right or left.
So, adding or subtracting
62. Writing in Math A damped trigonometric function
oscillates between the positive and negative graphs
of the damping factor. Explain why a damped
trigonometric function oscillates between the
positive and negative graphs of the damping factor
and why the amplitude of the function depends on
the damping factor.
SOLUTION: from x in the cosecant
Sample answer: Whether graphing a damped
trigonometric function using the sine or cosine
functions, the oscillation is a natural characteristic of
those parent functions. The amplitude of both graphs,
however, is determined by the damping factor.
Consider the graph y = a sin x. The greatest value
that the sin x can ever reach is 1. Thus, the amplitud
of this graph is a. This reasoning can be used to
imply that the amplitude of the damped trigonometric
function y = f (x) sin x will be f (x). As f (x) increases
or decreases, so does the amplitude of the damped
function.
function should translate the graph of the function so
that it has the same graph as y = sec x. Graph
and y = sec x on the same
coordinate plane using a graphing calculator.
The cotangent function resembles the tangent
function except that it is reflected in the x-axis and
translated units to the right or left. So, adding or subtracting
State the amplitude, period, frequency, phase
shift, and vertical shift of each function. Then
graph two periods of the function.
63. from x in the cotangent function and
multiplying the function by −1 should transform the
graph of the function so that it has the same graph as
y = tan x. Graph
SOLUTION: In this function, a = 3, b = 2, c = , and d = 10.
and y = tan x on
the same coordinate plane using a graphing
calculator.
62. Writing in Math A damped trigonometric function
oscillates between the positive and negative graphs
of the damping factor. Explain why a damped
trigonometric function oscillates between the
positive and negative graphs of the damping factor
and why the amplitude of the function depends on
the damping factor.
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SOLUTION: Sample answer: Whether graphing a damped
Graph y = 3 sin 2x shifted
units to the right and 10
units up.
Page 29
4-5
of this graph is a. This reasoning can be used to
imply that the amplitude of the damped trigonometric
function y = f (x) sin x will be f (x). As f (x) increases
or
decreases, so
does the
amplitude of the damped
Graphing
Other
Trigonometric
Functions
function.
State the amplitude, period, frequency, phase
shift, and vertical shift of each function. Then
graph two periods of the function.
64. SOLUTION: 63. In this function, a = 2, b = 3, c = , and d = –6.
SOLUTION: In this function, a = 3, b = 2, c = , and d = 10.
Graph y = 2 cos 3x shifted
6 units down.
Graph y = 3 sin 2x shifted
units to the right and
units to the right and 10
units up.
65. y =
cos (4x − π) + 1
SOLUTION: 64. In this function, a =
, b = 4, c = , and d = 1.
SOLUTION: In this function, a = 2, b = 3, c = , and d = –6.
Graph y =
units up.
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Graph y = 2 cos 3x shifted
6 units down.
units to the right and
cos 4x shifted
units to the left and 1 Page 30
4-5 Graphing Other Trigonometric Functions
65. y =
Find the exact values of the five remaining
trigonometric functions of θ.
cos (4x − π) + 1
66. sin θ =
SOLUTION: In this function, a =
, b = 4, c = , and d = 1.
, cos θ > 0
SOLUTION: To find the other function values, you must find the
coordinates of a point on the terminal side of θ. You
know that sin θ and cos θ are positive, so θ must lie
in Quadrant I. This means that both x and y are
positive.
or
Because sinθ =
, use the point (x, 4) and r =
5 to find x.
Graph y =
cos 4x shifted
units to the left and 1 units up.
Use x = 1, y = 2, and r = 5 to write the five
remaining trigonometric ratios.
Find the exact values of the five remaining
trigonometric functions of θ.
66. sin θ =
67. cos θ =
, sin θ > 0
, cos θ > 0
SOLUTION: SOLUTION: To find the other function values, you must find the
coordinates of a point on the terminal side of θ. You
know that sin θ and cos θ are positive, so θ must lie
in Quadrant I. This means that both x and y are
positive.
Because sinθ =
or
, use the point (x, 4) and r =
5 to find x.
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To find the other function values, you must find the
coordinates of a point on the terminal side of θ. You
know that sin θ and cos θ are positive, so θ must lie
in Quadrant I. This means that both x and y are
positive.
Because cosθ =
r=
or
, use the point (6, y) and
to find y.
Page 31
4-5 Graphing Other Trigonometric Functions
67. cos θ =
68. tan θ =
, sin θ > 0
, sin θ > 0
SOLUTION: SOLUTION: To find the other function values, you must find the
coordinates of a point on the terminal side of θ. You
know that sin θ and cos θ are positive, so θ must lie
in Quadrant I. This means that both x and y are
positive.
To find the other function values, you must find the
coordinates of a point on the terminal side of θ. You
know that sin θ and tan θ are positive, so θ must lie
in Quadrant I. This means that both x and y are
positive.
Because cosθ =
or
, use the point (6, y) and
Because tan θ =
or
, use the point (7, 24) to
find the value of r.
r=
to find y.
Use x = 7, y = 24, and r = 25 to write the five
remaining trigonometric ratios.
Use x = 6, y = 1, and r =
to write the five
remaining trigonometric ratios.
69. POPULATION The population of a city 10 years
68. tan θ =
ago was 45,600. Since then, the population has
increased at a steady rate each year. If the
population is currently 64,800, find the annual rate of
growth for this city.
, sin θ > 0
SOLUTION: To find the other function values, you must find the
coordinates of a point on the terminal side of θ. You
know that sin θ and tan θ are positive, so θ must lie
in Quadrant I. This means that both x and y are
positive.
Because tan θ =
or
, use the point (7, 24) to
SOLUTION: The formula for exponential growth is N = N 0(1 + r)
t
. The value of t is 10. The current population N is
64,800 and the initial population N 0 was 45,600.
Substitute these values into the formula and solve for
r.
find the value of r.
Use x = 7, y = 24, and r = 25 to write the five
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remaining trigonometric ratios.
Page 32
4-5 Graphing Other Trigonometric Functions
The annual rate of growth is approximately 3.6%.
69. POPULATION The population of a city 10 years
ago was 45,600. Since then, the population has
increased at a steady rate each year. If the
population is currently 64,800, find the annual rate of
growth for this city.
SOLUTION: The formula for exponential growth is N = N 0(1 + r)
t
. The value of t is 10. The current population N is
64,800 and the initial population N 0 was 45,600.
Substitute these values into the formula and solve for
r.
The annual rate of growth is approximately 3.6%.
70. MEDICINE The half-life of a radioactive
substance is the amount of time it takes for half of
the atoms of the substance to disintegrate. Nuclear
medicine technologists use the iodine isotope I-131,
with a half-life of 8 days, to check a patient's thyroid
function. After ingesting a tablet containing the
iodine, the isotopes collect in the patient’s thyroid,
and a special camera is used to view its function.
Suppose a patient ingests a tablet containing 9
microcuries of I-131. To the nearest hour, how long
will it be until there are only 2.8 microcuries in the
patient’s thyroid?
SOLUTION: Use the continuous exponential decay formula. If N 0
is the initial value, then the final value N should equal
0.5N 0. The value of t is 8 days. Solve for k.
We now have the formula
The initial amount N 0 is 9 and the final amount N is
2.8. We need to determine the time t. Substitute and
solve for t.
70. MEDICINE The half-life of a radioactive
substance is the amount of time it takes for half of
the atoms of the substance to disintegrate. Nuclear
medicine technologists use the iodine isotope I-131,
with a half-life of 8 days, to check a patient's thyroid
function. After ingesting a tablet containing the
iodine, the isotopes collect in the patient’s thyroid,
and a special camera is used to view its function.
Suppose a patient ingests a tablet containing 9
microcuries of I-131. To the nearest hour, how long
will it be until there are only 2.8 microcuries in the
patient’s thyroid?
SOLUTION: Use the continuous exponential decay formula. If N 0
is the initial value, then the final value N should equal
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0.5NManual
0. The value of t is 8 days. Solve for k.
The time is about 13.48 days. This is equivalent to
about 324 hours.
Factor each polynomial completely using the
given factor and long division.
71. x3 + 2x2 − x − 2; x − 1
SOLUTION: Page 33
3
time is about
13.48Trigonometric
days. This is equivalent
to
4-5 The
Graphing
Other
Functions
about 324 hours.
Factor each polynomial completely using the
given factor and long division.
2
2
So, x + x − 16x − 16= (x + 4)(x − 3x − 4).
3
2
Factoring the quadratic expression yields x + x −
16x − 16 = (x + 4)(x – 4)(x + 1).
73. x3– x2 – 10x – 8; x + 1
SOLUTION: 71. x3 + 2x2 − x − 2; x − 1
SOLUTION: 3
2
2
So, x – x – 10x – 8 = (x + 1)(x − 2x − 8).
3
3
2
2
So, x + 2x − x − 2= (x – 1)(x + 3x + 2).
3
2
Factoring the quadratic expression yields x + 2x −
x − 2= (x + 2)(x + 1)(x – 1).
72. x3 + x2 − 16x − 16; x + 4
SOLUTION: 2
Factoring the quadratic expression yields x – x –
10x – 8 = (x – 4)(x + 2)(x + 1).
74. EXERCISE The American College of Sports
Medicine recommends that healthy adults exercise at
a target level of 60% to 90% of their maximum heart
rates. You can estimate your maximum heart rate by
subtracting your age from 220. Write a compound
inequality that models age a and target heart rate r.
SOLUTION: A person of age a has a maximum heart rate of (220
− a). The person’s target heart rate r must be at
least 60% of this number. This can be represented
by r ≥ 0.6(220 − a). The person’s target heart rate
must also not be more than 90% of this number. This
can be represented by r ≤ 0.9(220 − a). This two
inequalities can be combined to form the compound
inequality 0.6(220 – a) ≤ r ≤ 0.9(220 – a).
3
2
2
So, x + x − 16x − 16= (x + 4)(x − 3x − 4).
3
2
Factoring the quadratic expression yields x + x −
16x − 16 = (x + 4)(x – 4)(x + 1).
75. SAT/ACT In the figure, A and D are the centers of
the two circles, which intersect at points C and E.
is a diameter of circle D. If AB = CE = 10, what
is AD?
73. x3– x2 – 10x – 8; x + 1
SOLUTION: A5
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B5
C5
Page 34
must also not be more than 90% of this number. This
can be represented by r ≤ 0.9(220 − a). This two
inequalities can be combined to form the compound
4-5 Graphing Other Trigonometric Functions
inequality 0.6(220 – a) ≤ r ≤ 0.9(220 – a).
75. SAT/ACT In the figure, A and D are the centers of
the two circles, which intersect at points C and E.
is a diameter of circle D. If AB = CE = 10, what
is AD?
Since it cannot be negative,
correct answer is C.
is 5
. The
76. REVIEW Refer to the figure below. If c = 14, find
the value of b.
F
G 14
H7
A5
B5
C5
J7
SOLUTION: D 10
E 10
To find b, use cos θ =
.
SOLUTION: Draw ∆ACD as shown.
The correct answer is J.
Since C is on circle A,
must be 10. If is the diameter of circle D, then
must be 5. Use the Pythagorean Theorem to solve for
.
77. Which equation is represented by the graph?
A
Since it cannot be negative,
correct answer is C.
is 5
. The
76. REVIEW Refer to the figure below. If c = 14, find
the value of b.
B
C
D
SOLUTION: H7
Since the branches of the graph are decreasing as x
increases, the function is either the graph of the
cotangent function or the tangent function reflected
in the x-axis. This eliminates choices C and D. An
asymptote for the parent function y = cot x is atPage
x =35
0. This graph has an asymptote at x= . Thus, it is
J7
the graph of the cotangent function shifted
F
G 14Manual - Powered by Cognero
eSolutions
units to
the right. The equation represented by the graph is
.
4-5 Graphing Other Trigonometric Functions
The correct answer is J.
The correct answer is B.
77. Which equation is represented by the graph?
78. REVIEW If sin θ =
and
<θ<
, then θ
=?
F
G
A
H
B
J
C
SOLUTION: Solve for θ.
D
SOLUTION: Since the branches of the graph are decreasing as x
increases, the function is either the graph of the
cotangent function or the tangent function reflected
in the x-axis. This eliminates choices C and D. An
asymptote for the parent function y = cot x is at x =
0. This graph has an asymptote at x= . Thus, it is
the graph of the cotangent function shifted
units to
the right. The equation represented by the graph is
Since π < θ <
, add π to θ. So, θ =
+ or
.
The correct answer is G.
.
The correct answer is B.
78. REVIEW If sin θ =
and
<θ<
, then θ
=?
F
G
H
J
SOLUTION: Solve for θ.
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Page 36