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4-5 Graphing Other Trigonometric Functions
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
or . Find the location of
two consecutive vertical asymptotes.
and Create a table listing the coordinates of key points for y = 2 tan x for one period on
Function
Vertical
Asymptote
Intermediate
Point
x-intercept
Intermediate
Point
.
Vertical
Asymptote
y = tan x
x=
(0, 0)
x=
y = 2 tan x
x=
(0, 0)
x=
Sketch the curve through the indicated key points for the function. Then repeat the pattern.
3. SOLUTION: is the graph of y = cot x shifted
The graph of
units to the right. The period is or . Find the
location of two consecutive vertical asymptotes.
and Create a table listing the coordinates of key points for
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Function
Vertical
Asymptote
Intermediate
Point
for one period on
x-intercept
Intermediate
Point
.
Vertical
Asymptote
Page 1
4-5 Graphing Other Trigonometric Functions
3. SOLUTION: is the graph of y = cot x shifted
The graph of
units to the right. The period is or . Find the
location of two consecutive vertical asymptotes.
and Create a table listing the coordinates of key points for
Function
y = cot x
Vertical
Asymptote
Intermediate
Point
for one period on
x-intercept
.
Intermediate
Point
Vertical
Asymptote
x=π
x =0
x=
x=
Sketch the curve through the indicated key points for the function. Then repeat the pattern.
5. SOLUTION: The graph of
is the graph of y = cot x compressed vertically and reflected in the x-axis. The period is
or . Find the location of two consecutive vertical asymptotes.
and Create a table listing the coordinates of key points for
Vertical
Function
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Asymptote
y = cot x
x =0
Intermediate
Point
for one period on
x-intercept
.
Intermediate
Point
Vertical
Asymptote
x=π
Page 2
4-5 Graphing Other Trigonometric Functions
5. SOLUTION: is the graph of y = cot x compressed vertically and reflected in the x-axis. The period is
The graph of
or . Find the location of two consecutive vertical asymptotes.
and Create a table listing the coordinates of key points for
Vertical
Asymptote
Function
y = cot x
for one period on
Intermediate
Point
x-intercept
.
Intermediate
Point
Vertical
Asymptote
x =0
x=π
x =0
x=π
Sketch the curve through the indicated key points for the function. Then repeat the pattern.
7. y = –2 tan (6x − π)
SOLUTION: is the graph of y = tan x expanded vertically, compressed horizontally, shifted
The graph of
units to the right, and reflected in the x-axis. The period is
or . Find the location of two consecutive
vertical asymptotes.
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Create a table listing the coordinates of key points for y = tan 2x for one period on
Function
Vertical
Intermediate
x-intercept
Intermediate
Page 3
.
Vertical
4-5 Graphing Other Trigonometric Functions
7. y = –2 tan (6x − π)
SOLUTION: is the graph of y = tan x expanded vertically, compressed horizontally, shifted
The graph of
units to the right, and reflected in the x-axis. The period is
or . Find the location of two consecutive
vertical asymptotes.
and Create a table listing the coordinates of key points for y = tan 2x for one period on
Vertical
Asymptote
Function
y = tan x
x=
y = 2 tan x
x=
Intermediate
Point
x-intercept
Intermediate
Point
(0, 0)
.
Vertical
Asymptote
x=
x=
Sketch the curve through the indicated key points for the function. Then repeat the pattern.
9. y =
csc 2x
SOLUTION: The graph of y =
is
csc 2x is the graph of y = csc x compressed vertically and compressed horizontally. The period
or . Find the location of two vertical asymptotes.
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Create a table listing the coordinates of key points for y =
Page 4
csc 2x for one period on
.
4-5 Graphing Other Trigonometric Functions
9. y =
csc 2x
SOLUTION: The graph of y =
csc 2x is the graph of y = csc x compressed vertically and compressed horizontally. The period
or . Find the location of two vertical asymptotes.
is
and Create a table listing the coordinates of key points for y =
Function
y = csc x
y=
csc 2x
Vertical
Asymptote
Intermediate
Point
csc 2x for one period on
Vertical
Asymptote
Intermediate
Point
.
Vertical
Asymptote
x = −π
x =0
x=π
x=
x =0
x=
Sketch the curve through the indicated key points for the function. Then repeat the pattern.
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
or 2 . Find the
location of two vertical asymptotes.
and Create a table listing the coordinates of key points for y = sec (x +
) for one period on
.
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Function
Vertical
Asymptote
Page 5
Intermediate
Point
Vertical
Asymptote
Intermediate
Point
Vertical
Asymptote
4-5 Graphing Other Trigonometric Functions
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
or 2 . Find the
location of two vertical asymptotes.
and Create a table listing the coordinates of key points for y = sec (x +
Function
Vertical
Asymptote
Intermediate
Point
) for one period on
Vertical
Asymptote
Intermediate
Point
.
Vertical
Asymptote
y = sec x
x=
x=
x=
y = sec (x + π)
x=
x=
x=
Sketch the curve through the indicated key points for the function. Then repeat the pattern.
13. SOLUTION: is the graph of y = sec x expanded vertically and shifted
The graph of
period is
units to the right. The or 2 . Find the location of two vertical asymptotes.
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Cognero
Create
a table
listing
coordinates
Function
Vertical
for one period on of key points for
Intermediate
Vertical
Intermediate
.
Vertical
Page 6
4-5 Graphing Other Trigonometric Functions
13. SOLUTION: is the graph of y = sec x expanded vertically and shifted
The graph of
period is
units to the right. The or 2 . Find the location of two vertical asymptotes.
and for one period on Create a table listing the coordinates of key points for
Function
y = sec x
Vertical
Asymptote
Intermediate
Point
Vertical
Asymptote
Intermediate
Point
.
Vertical
Asymptote
x=
x=
x=
x=
x=
x=
Sketch the curve through the indicated key points for the function. Then repeat the pattern.
15. SOLUTION: is the graph of y = csc x expanded vertically and shifted
The graph of
period is
units to the right. The
or 2 . Find the location of two vertical asymptotes.
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Page 7
4-5 Graphing Other Trigonometric Functions
15. SOLUTION: is the graph of y = csc x expanded vertically and shifted
The graph of
period is
units to the right. The
or 2 . Find the location of two vertical asymptotes.
and for one period on Create a table listing the coordinates of key points for
Function
y = csc x
Vertical
Asymptote
Intermediate
Point
x = −π
Vertical
Asymptote
Intermediate
Point
x =0
x=
.
x=
Vertical
Asymptote
x=π
x=
Sketch the curve through the indicated key points for the function. Then repeat the pattern.
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 =
x sin x is the product of the functions y =
calculator to graph f (x), –f (x), and y =
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x and y = sin x, so f (x) =
x. Use a graphing
x sin x in the same viewing window.
Page 8
4-5 Graphing Other Trigonometric Functions
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 =
x sin x is the product of the functions y =
calculator to graph f (x), –f (x), and y =
x and y = sin x, so f (x) =
x. Use a graphing
x 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.
19. y = 2x2 cos x
SOLUTION: 2
2
2
The function y = 2x cos x is the product of the functions y = 2x and y = cos x, so f (x) = 2x . Use a graphing
2
calculator to graph f (x), –f (x), and y = 2x cos 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 =
x sin 2x
SOLUTION: The function y =
x sin 2x is the product of the functions y =
calculator to graph f (x), –f (x), and y =
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x and y = sin 2x, so f (x) =
x. Use a graphing
x sin 2x in the same viewing window.
Page 9
4-5
The
amplitudeOther
of the function
is decreasing as
x approaches 0 from both directions, and increasing as x approaches
Graphing
Trigonometric
Functions
positive and negative infinity.
21. y =
x sin 2x
SOLUTION: The function y =
x sin 2x is the product of the functions y =
calculator to graph f (x), –f (x), and y =
x and y = sin 2x, so f (x) =
x. Use a graphing
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.
23. y = e 0.5x cos x
SOLUTION: 0.5x
0.5x
The function y = e
cos x is the product of the functions y = e
and y = cos x, so f (x) = e
0.5x
calculator to graph f (x), –f (x), and y = e
cos x in the same viewing window.
0.5x
. Use a graphing
The amplitude of the function is decreasing as x approaches negative infinity and increasing as x approaches positive
infinity.
25. y = |x| cos 3x
SOLUTION: The function y =
cos 3x is the product of the functions y =
calculator to graph f (x), –f (x), and y =
and y = cos 3x, so f (x) =
. Use a graphing
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
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27. MECHANICS When the car shown below hit a 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
4-5
The
amplitudeOther
of the function
is decreasing as
x approaches negative infinity and increasing as x approaches positive
Graphing
Trigonometric
Functions
infinity.
25. y = |x| cos 3x
SOLUTION: The function y =
cos 3x is the product of the functions y =
calculator to graph f (x), –f (x), and y =
and y = cos 3x, so f (x) =
. Use a graphing
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.
27. MECHANICS When the car shown below hit a 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.
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: −ct
a. The maximum displacement of the shock absorber 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 .
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.
−3t
b. Use a graphing calculator to determine the value of t when the graph of y = 8e cos 5πt is equal to 4. Graph y =
−3t
8e cos 5πt and y = 4 on the same coordinate plane. Find the intersection of the two graphs using the CALC
menu.
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Page 11
4-5
The
amplitudeOther
of the function
is decreasing as
x approaches 0 from both directions, and increasing as x approaches
Graphing
Trigonometric
Functions
positive and negative infinity.
27. MECHANICS When the car shown below hit a 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.
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: −ct
a. The maximum displacement of the shock absorber 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 .
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.
−3t
b. Use a graphing calculator to determine the value of t when the graph of y = 8e cos 5πt is equal to 4. Graph y =
−3t
8e cos 5πt and y = 4 on the same coordinate plane. Find the intersection of the two graphs using the CALC
menu.
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.
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 . Find the location of
two vertical asymptotes.
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and Page 12
intersection
of the two
graphs occurs when
x ≈ 0.06. So, the amount of time that it takes for the amplitude of 4-5 The
Graphing
Other
Trigonometric
Functions
the vibration to decrease to 4 inches is about 0.06 second.
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 . Find the location of
two vertical asymptotes.
and Create a table listing the coordinates of key points for y = sec x + 3 for one period on
Function
Vertical
Asymptote
Intermediate
Point
Vertical
Asymptote
Intermediate
Point
.
Vertical
Asymptote
y = sec x
x=
x=
x=
y = sec x + 3
x=
x=
x=
Sketch the curve through the indicated key points for the function. Then repeat the pattern.
31. y = csc
− 2
SOLUTION: is the graph of y = csc x expanded horizontally and shifted 2 units down. The period is
The graph of
or 6π. Find the location of two vertical asymptotes.
and Create a table listing the coordinates of key points for
eSolutions Manual - Powered by Cognero
Function
Vertical
Asymptote
Intermediate
Point
for one period on Vertical
Asymptote
Intermediate
Point
.
Vertical
Asymptote
Page 13
4-5 Graphing Other Trigonometric Functions
31. y = csc
− 2
SOLUTION: is the graph of y = csc x expanded horizontally and shifted 2 units down. The period is
The graph of
or 6π. Find the location of two vertical asymptotes.
and Create a table listing the coordinates of key points for
Function
y = csc x
Vertical
Asymptote
Intermediate
Point
for one period on Vertical
Asymptote
.
Intermediate
Point
Vertical
Asymptote
x = −π
x =0
x=π
x=
x=
x=
Sketch the curve through the indicated key points for the function. Then repeat the pattern.
33. y = cot (2x + π) − 3
SOLUTION: is the graph of y = cot x compressed horizontally, shifted
The graph of
shifted 3 units down. The period is
or units to the left, and . Find the location of two consecutive vertical asymptotes.
and Create a table listing the coordinates of key points for
Vertical
Asymptote
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Function
y = cot x
x =0
Intermediate
Point
for one period on
x-intercept
Intermediate
Point
.
Vertical
Asymptote
Page 14
x=π
4-5 Graphing Other Trigonometric Functions
33. y = cot (2x + π) − 3
SOLUTION: is the graph of y = cot x compressed horizontally, shifted
The graph of
shifted 3 units down. The period is
or units to the left, and . Find the location of two consecutive vertical asymptotes.
and Create a table listing the coordinates of key points for
Function
Vertical
Asymptote
y = cot x
Intermediate
Point
for one period on
x-intercept
Intermediate
Point
x =0
x=
.
Vertical
Asymptote
x=π
x=
Sketch the curve through the indicated key points for the function. Then repeat the pattern.
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.
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 Jeff when the angle of elevation is 45°?
SOLUTION: a.
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Page 15
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 Jeff when the angle of elevation is 45°?
4-5 SOLUTION: Graphing Other Trigonometric Functions
a.
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.
b. Evaluate d = 150 csc θ for several values of θ within the domain.
θ
d
212.1
162.4
150
Use these points to graph d = 150 csc θ.
162.4
212.1
c. Substitute θ = 45° into the function you found in part a.
The hawk is about 212.1 feet away from Jeff when the angle of elevation is 45°.
GRAPHING CALCULATOR Find the values of θ on the interval –
true.
37. cot θ = 2 sec θ
<θ<
that make each equation
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 CALC menu.
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4-5 Graphing Other Trigonometric Functions
The hawk is about 212.1 feet away from Jeff when the angle of elevation is 45°.
GRAPHING CALCULATOR Find the values of θ on the interval –
true.
37. cot θ = 2 sec θ
<θ<
that make each equation
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 CALC menu.
The graphs appear to intersect when x
about 0.427 and 2.715.
0.427 and x
2.715. So, the values of θ that make the equation true are
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.
Find the points of intersection using the CALC menu.
The graphs appear to intersect when x −2.880, x ≈ −1.833, x 0.262, and x
make the equation true are about −2.880, −1.833, 0.262, and 1.309.
1.309. So, the values of θ that
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
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plane.
Page 17
graphs appear
to intersect
when x −2.880,
x ≈ −1.833, x
4-5 The
Graphing
Other
Trigonometric
Functions
0.262, and x
make the equation true are about −2.880, −1.833, 0.262, and 1.309.
1.309. So, the values of θ that
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 CALC menu.
The graphs appear to intersect when x
about −2.356 and 0.785.
−2.356 and x
0.785. So, the values of θ that make the equation true are
43. TENSION A helicopter is delivering a large mural 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
.
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 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 12.19 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.
Therefore, the downward force is 5331.2 N.
b. The tension T on each rope is equal to half the downward force times sec
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,T=
sec .
Page 18
a. The downward force F is equal to the mass m of the mural times gravity g, so F = mg.
4-5 Graphing Other Trigonometric Functions
Therefore, the downward force is 5331.2 N.
b. The tension T on each rope is equal to half the downward force times sec
sec ,T=
Therefore, an equation that represents the tension on each rope is T = 2665.6 sec
.
.
c.
d. Draw a diagram to model the situation.
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.
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.
Because the hypotenuse and the side opposite an acute angle are given, you can use the sine function to find θ.
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Page 19
4-5 Graphing Other Trigonometric Functions
Because the hypotenuse and the side opposite an acute angle are given, you can use the sine function to find θ.
Find the tension.
Therefore, θ is about 97º and the tension being applied to each rope is about 4023 N.
Match each function with its graph.
45. SOLUTION: The parent function y = sec x resembles either graph b or c. Find the location of two vertical asymptotes.
and The location of the asymptotes suggests that the correct answer is b.
47. SOLUTION: TheManual
tangent
function
has a graph
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comprised of branches that increase as x increases. The only graph that meetsPage
this20
condition is a.
GRAPHING CALCULATOR Graph each pair of functions on the same screen and make a conjecture as to
and 4-5 The
Graphing
Trigonometric
location ofOther
the asymptotes
suggests thatFunctions
the correct answer is b.
47. 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.
49. f (x) = sec2 x; g(x) = tan2 x + 1
SOLUTION: Graph f (x) and g(x) on the same coordinate plane.
Since sec x is undefined for x =
+ n , where n is an integer, f (x) is also undefined for x =
an integer. Since tan x is undefined for x =
+ n , where n is
+ 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. So, the expressions are equivalent.
51. SOLUTION: Graph f (x) and g(x) on the same coordinate plane.
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).
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53. function:
period:by Cognero
; ps:
SOLUTION: ; vs: –1
Page 21
Since sec x is undefined for x =
+ n , where n is an integer, f (x) is also undefined for these values of x.
of theOther
phase, f Trigonometric
(x) is undefined for xFunctions
= n , where n is an integer. g(x) is defined at these values. So, the
4-5 Because
Graphing
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).
53. function: tan; period:
; ps:
; vs: –1
SOLUTION: Start with the general form of the tangent function
, there is a phase shift of y = a tan (bx + c) + d. The period is
units to the right, and there is a vertical shift of 1 unit down, so d = –1.
For a tangent function, period =
.
The phase shift = –
. Use b to find the phase shift.
Therefore, a tangent function with a period
down is given by
55. function: cot; period: 3π; ps:
, a phase shift of
or
units to the right, and a vertical shift of 1 unit .
; 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.
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For a cotangent function, period =
.
Page 22
Therefore, a tangent function with a period
, a phase shift of
is given by
or
4-5 down
Graphing
Other Trigonometric
Functions
55. function: cot; period: 3π; ps:
units to the right, and a vertical shift of 1 unit .
; 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 =
The phase shift = –
.
. Use b to find the phase shift.
Therefore, a cotangent function with period 3π, a phase shift of
is given by
or
units to the right, and a vertial shift of 4 units up .
57. PROOF Verify that the y-intercept for the graph of any function of the form y = k e–ct cos ωt is k.
SOLUTION: To find the y-intercept, let t = 0.
REASONING Determine whether each statement is true or false . Explain your reasoning.
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59. If x = θ is an asymptote of y = csc x, then x = θ is also an asymptote of y = cot x.
SOLUTION: Page 23
4-5 Graphing Other Trigonometric Functions
REASONING Determine whether each statement is true or false . Explain your reasoning.
59. If x = θ is an asymptote of y = csc x, then x = θ is also an asymptote of y = cot x.
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. Also, 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.
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
adding or subtracting
units to the right or left. So, from x in the cosecant function should translate the graph of the function so that it has the
and y = sec x on the same coordinate plane using a graphing
same graph as y = sec x. Graph
calculator.
The cotangent function resembles the tangent function except that it is reflected in the x-axis and translated
to the right or left. So, adding or subtracting
units 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
and y
= tan x on the same coordinate plane using a graphing calculator.
State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two
periods of the function.
63. eSolutions
Manual - Powered by Cognero
SOLUTION: In this function, a = 3, b = 2, c = Page 24
, and d = 10.
4-5 Graphing Other Trigonometric Functions
State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two
periods of the function.
63. SOLUTION: In this function, a = 3, b = 2, c = , and d = 10.
Graph y = 3 sin 2x shifted
65. y =
units to the right and 10 units up.
cos (4x − π) + 1
SOLUTION: In this function, a =
, b = 4, c = , and d = 1.
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Graph y =
cos 4x shifted
Page 25
units to the left and 1 units up.
4-5 Graphing Other Trigonometric Functions
65. y =
cos (4x − π) + 1
SOLUTION: In this function, a =
, b = 4, c = , and d = 1.
Graph y =
cos 4x shifted
units to the left and 1 units up.
Find the exact values of the five remaining trigonometric functions of θ.
67. cos θ =
, 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 cos 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 r=
to find y.
Use x = 6, y = 1, and r =
to write the five remaining trigonometric ratios.
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Page 26
4-5 Graphing Other Trigonometric Functions
Find the exact values of the five remaining trigonometric functions of θ.
67. cos θ =
, 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 cos 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 r=
Use x = 6, y = 1, and r =
to find y.
to write the five remaining trigonometric ratios.
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: t
The formula for exponential growth is N = N 0(1 + r) . 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.
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Page 27
4-5 Graphing Other Trigonometric Functions
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: t
The formula for exponential growth is N = N 0(1 + r) . 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%.
Factor each polynomial completely using the given factor and long division.
71. x3 + 2x2 − x − 2; x − 1
SOLUTION: 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).
73. x3– x2 – 10x – 8; x + 1
SOLUTION: eSolutions Manual - Powered by Cognero
Page 28
3
4-5
2
2
So, x + 2x − x − 2= (x – 1)(x + 3x + 2).
3
2
Graphing
Other Trigonometric
Factoring
the quadratic
expression yields x +Functions
2x − x − 2= (x + 2)(x + 1)(x – 1).
73. x3– x2 – 10x – 8; x + 1
SOLUTION: 3
2
2
So, x – x – 10x – 8 = (x + 1)(x − 2x − 8).
3
2
Factoring the quadratic expression yields x – x – 10x – 8 = (x – 4)(x + 2)(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?
A5
B5
C5
D 10
E 10
SOLUTION: Draw ∆ACD as shown.
Since C is on circle A,
Theorem to solve for
must be 10. If is the diameter of circle D, then
must be 5. Use the Pythagorean
.
Since it cannot be negative,
is 5
. The correct answer is C.
77. Which equation is represented by the graph?
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Page 29
4-5 Graphing Other Trigonometric Functions
Since it cannot be negative,
is 5
. The correct answer is C.
77. Which equation is represented by the graph?
A
B
C
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=
function shifted
. Thus, it is the graph of the cotangent
units to the right. The equation represented by the graph is .
The correct answer is B.
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