5-2 ing Trigonometric Identities.pdf 26625w
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SOLUTION:
5-2 ing Trigonometric Identities each identity.
1. (sec
2
– 1) cos
2
= sin
2
6. tan θ csc2
– tan
= cot
SOLUTION:
SOLUTION:
2. sec2
2
(1 – cos
2
) = tan
7.
–
= cot
SOLUTION: SOLUTION:
3. sin
– sin
2
cos
3
= sin
SOLUTION:
8. 4. csc
– cos
cot = sin
+
= 2 csc
SOLUTION:
SOLUTION:
4
5.
= cot
SOLUTION:
9.
+ tan
= sec
SOLUTION: 2
6. tan θ csc
– tan
= cot
SOLUTION:
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Page 1
5-2 ing Trigonometric Identities
9.
+ tan
12.
= sec
SOLUTION:
10.
+
= 2 sec
+
2
sin
SOLUTION:
= sin + cos
13. (csc
SOLUTION:
– cot
+ cot ) = 1
)(csc
SOLUTION:
14. cos4
4
– sin
2
= cos
2
– sin
SOLUTION:
11.
+
15.
= 1
+
= 2 sec
2
SOLUTION:
SOLUTION:
16.
+
= 2 sec
SOLUTION: 12.
+
= 2 sec
eSolutions Manual - Powered by Cognero SOLUTION:
2
sin Page 2
5-2 ing Trigonometric Identities 16.
19. FIREWORKS If a rocket is launched from ground
= 2 sec
+
level, the maximum height that it reaches is given by
SOLUTION:
h=
is the angle between the
, where
ground and the initial path of the rocket, v is the rocket’s initial speed, and g is the acceleration due to gravity, 9.8 meters per second squared.
17. csc4
4
– cot
2
= 2 cot
+ 1
SOLUTION:
a. that
=
.
b. Suppose a second rocket is fired at an angle of 80 from the ground with an initial speed of 110 meters per second. Find the maximum height of the rocket.
SOLUTION: a.
18.
=
SOLUTION: b. Evaluate the expression
for v = 110 m,
2
, and g = 9.8 m/s .
19. FIREWORKS If a rocket is launched from ground level, the maximum height that it reaches is given by h=
, where
is the angle between the
ground and the initial path of the rocket, v is the rocket’s initial speed, and g is the acceleration due to gravity, 9.8 meters per second squared.
The maximum height of the rocket is about 598.7 meters. each identity. 20. (csc + cot )(1 – cos
) = sin
SOLUTION:
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a. that
=
Page 3
.
maximumTrigonometric height of the rocket Identities is about 598.7 5-2 The ing meters.
each identity. + cot )(1 – cos
20. (csc
) = sin
22.
SOLUTION:
21. sin2
tan
2
SOLUTION:
2
= tan
2
– sin
SOLUTION:
23.
= cos
+ cot
SOLUTION:
22. SOLUTION:
24. (csc
– cot
2
) =
SOLUTION:
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Page 4
5-2 ing Trigonometric Identities 24. (csc
2
– cot
27. sec
) =
– cos
= tan sin
SOLUTION: SOLUTION:
28. 1 – tan4 θ = 2 sec2 θ – sec4 θ SOLUTION: 25.
=
SOLUTION: 29. (csc
– cot
2
) =
SOLUTION:
30. 26. tan2
2
cos
= 1 – cos
2
= sec
SOLUTION:
SOLUTION:
27. sec
– cos
= tan sin
SOLUTION:
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31.
= (sin
SOLUTION:
+ cos )
2
Page 5
5-2 ing Trigonometric Identities 31.
= (sin
+ cos )
2
GRAPHING CALCULATOR Test whether each equation is an identity by graphing. If it appears to be an identity, it. If not, find an x-value for which both sides are defined but not equal. 34. =
SOLUTION:
SOLUTION: Graph
and then graph .
32. OPTICS If two prisms of the same power are placed next to each other, their total power can be determined using z = 2p cos , where z is the combined power of the prisms, p is the power of the individual prisms, and θ is the angle between the two 2
prisms. that 2p cos θ = 2p (1 – sin
)sec
.
SOLUTION:
33. PHOTOGRAPHY The amount of light ing through a polarization filter can be modeled using I = 2 Im cos , where I is the amount of light ing through the filter, Im is the amount of light shined on the filter, and is the angle of rotation between the light source and the filter. that .
SOLUTION:
The graphs of the related functions do not coincide for all values of x for which both functions are defined. Using the intersect feature from the CALC menu on the graphing calculator to find that when x = , Y1 = –1 and Y2 is undefined. Therefore, the equation is not an identity.
35. sec x + tan x = SOLUTION: and then graph
Graph GRAPHING CALCULATOR Test whether each equation is an identity by graphing. If it appears to be an identity, it. If not, find an x-value for which both sides are defined but not equal. 34. =
SOLUTION: eSolutions Manual - Powered by Cognero Graph
and then graph .
.
Page 6
5-2
Using the intersect feature from the CALC menu on the graphing calculator to find that when x = , Y1 = –1 and Y2 is undefined. Therefore, the ing Identities equation is notTrigonometric an identity.
36. sec2 x – 2 sec x tan x + tan2 x =
35. sec x + tan x = SOLUTION:
SOLUTION: and then graph
Graph
2
.
graph Y2 =
The equation appears to be an identity because the graphs of the related functions coincide. this algebraically.
The graphs of the related functions do not coincide for all values of x for which both functions are defined. Using the intersect feature from the CALC menu on the graphing calculator to find that when x = 0, Y1 = 1 and Y2 = 0. Therefore, the equation is not an identity. 2
= 1 – 2 sin x
SOLUTION:
36. sec2 x – 2 sec x tan x + tan2 x =
Graph Y1 =
SOLUTION: 2
Graph Y1 = sec x – 2 sec x tan x + tan x and then graph Y2 =
.
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.
37.
2
2
Graph Y1 = sec x – 2 sec x tan x + tan x and then
and then graph Y2 = 1 – 2
2
sin x.
Page 7
5-2
defined. Using the intersect feature from the CALC menu on the graphing calculator to find that when x = 0, Y1 = 1 and Y2Trigonometric = 0. Therefore, the Identities equation is not an ing identity. 2
37.
= 1 – 2 sin x
38.
SOLUTION:
=
SOLUTION:
Graph Y1 =
and then graph Y2 = 1 – 2
Graph Y1 =
2
sin x.
and then graph Y2 = .
The equation appears to be an identity because the graphs of the related functions coincide. this algebraically.
The graphs of the related functions do not coincide for all values of x for which both functions are defined. Using the intersect feature from the CALC menu on the graphing calculator to find that when x = , Y1 –0.17 and Y2 = 0. Therefore, the equation is not an identity.
39. cos2 x – sin2 x = SOLUTION: 2
38.
.
SOLUTION: Graph Y1 =
2
Graph Y1 = cos x – sin x and then graph Y2 =
=
and then graph Y2 = .
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Page 8
Using the intersect feature from the CALC menu
5-2
on the graphing calculator to find that when x = , Y1 Y2 = 0. Therefore, –0.17 and the equation is ing Trigonometric Identities not an identity.
39. cos2 x – sin2 x =
each identity.
40.
SOLUTION: 2
2
SOLUTION:
Graph Y1 = cos x – sin x and then graph Y2 = .
The equation appears to be an identity because the graphs of the related functions coincide. this algebraically.
41.
=
SOLUTION:
42. ln |csc x + cot x| + ln |csc x – cot x| = 0 SOLUTION: each identity.
40. SOLUTION:
43. ln |cot x| + ln |tan x cos x| = ln |cos x| SOLUTION:
each identity. 44. sec2 + tan2 = sec4 eSolutions Manual - Powered by Cognero
4
– tan
SOLUTION: Start with the right side of the identity.
Page 9
5-2 ing Trigonometric Identities each identity. 44. sec2 + tan2 = sec4
48. sec4 x = 1 + 2 tan2 x + tan4 x
4
– tan
SOLUTION:
SOLUTION:
Start with the right side of the identity.
Start with the right side of the identity.
49. sec2 x csc2 x = sec2 x + csc2 x 45. –2 cos2
4
= sin
– cos
4
SOLUTION:
– 1
Start with the left side of the identity.
SOLUTION: Start with the right side of the identity.
46. sec2 θ sin2 θ = sec4 θ – (tan4 θ + sec2 θ) SOLUTION: Start with the right side of the identity.
47. 3 sec2
tan
2
6
+ 1 = sec
6
– tan
SOLUTION:
50. ENVIRONMENT A biologist studying pollution situates a net across a river and positions instruments at two different stations on the river bank to collect samples. In the diagram shown, d is the distance between the stations and w is width of the river.
a. Determine an equation in of tangent α that can be used to find the distance between the stations. b. that d = . c. Complete the table shown for d = 40 feet.
Start with the right side of the identity. d. If > 60 or < 20 , the instruments will not function properly. Use the table from part c to determine whether sites in which the width of the river is 5, 35, or 140 feet could be used for the experiment.
SOLUTION: a.
48. sec4 x = 1 + 2 tan2 x + tan4 x SOLUTION: Start with the right side of the identity. eSolutions Manual - Powered by Cognero
Page 10
b.
determine whether sites in which the width of the river is 5, 35, or 140 feet could be used for the experiment.
5-2 SOLUTION: ing Trigonometric Identities
of 5 and 140 feet could not be used because > 60 and < 20 , respectively. The site with a width of 35 feet could be used because 20 < < 60 .
a. HYPERBOLIC FUNCTIONS The hyperbolic trigonometric functions are defined in the following ways.
b.
x
–x
sinh x =
(e – e )
cosh x =
(e + e )
x
–x
tanh x = csch x =
, x
0
sech x =
c.
coth x = , x 0 each identity using the functions shown above. 51. cosh2 x – sinh2 x = 1
SOLUTION:
d. If w = 5 then will be greater than 63.4 since 5 < 20. If w = 140, then will be less than 18.4 since 140 > 120. If w = 35, then 45 < < 63.4 since 35 is between 20 and 40. The sites with widths of 5 and 140 feet could not be used because > 60 and < 20 , respectively. The site with a width of 35 feet could be used because 20 < < 60 .
52. sinh (–x) = –sinh x SOLUTION:
HYPERBOLIC FUNCTIONS The hyperbolic trigonometric functions are defined in the following ways. x
(e – e )
cosh x =
(e + e )
x
53. sech2 x = 1 – tanh2 x
–x
sinh x =
SOLUTION:
–x
tanh x = csch x =
, x
0
sech x = eSolutions Manual - Powered by Cognero
coth x = , x 0 each identity using the functions shown above.
Page 11
54. cosh (–x) = cosh x
5-2 ing Trigonometric Identities 53. sech2 x = 1 – tanh2 x
GRAPHING CALCULATOR Graph each side of each equation. If the equation appears to be an identity, it algebraically.
SOLUTION: 55.
SOLUTION: Graph Y1 =
–
and Y2 = 1.
54. cosh (–x) = cosh x SOLUTION:
GRAPHING CALCULATOR Graph each side of each equation. If the equation appears to be an identity, it algebraically.
55.
SOLUTION: Graph Y1 =
The graphs appear to be the same, so the equation appears to be an identity. this algebraically.
–
and Y2 = 1.
56. sec x – cos2 x csc x = tan x sec x SOLUTION:
The graphs appear to be the same, so the equation appears to be an identity. this algebraically.
2
Graphs are not the same so sec x – cos x csc x tan x secx.
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Page 12
57. (tan x + sec x)(1 – sin x) = cos x
5-2 ing Trigonometric Identities 56. sec x – cos2 x csc x = tan x sec x 58.
SOLUTION:
–
= –1
SOLUTION:
2
Graphs are not the same so sec x – cos x csc x tan x secx.
57. (tan x + sec x)(1 – sin x) = cos x SOLUTION:
The graphs are not the same, so
59. MULTIPLE REPRESENTATIONS In this 58.
–
SOLUTION:
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= –1
problem, you will investigate methods used to solve trigonometric equations. Consider 1 = 2 sin x. a. NUMERICAL Isolate the trigonometric function in the equation so that sin x is the only expression on one side of the equation. b. GRAPHICAL Graph the left and right sides of the equation you found in part a on the same graph over [0, 2 ). Locate any points of intersection and express the values in of radians. c. GEOMETRIC Use the unit circle to the answers you found in part b. d. GRAPHICAL Graph the left and right sides of the equation you found in part a on the same graph over –2 < x < 2 . Locate any points of intersection and express the values in of radians. e. VERBAL Make a conjecture as to the solutions of 1 = 2 sin x. Explain your reasoning.
SOLUTION:
Page 13
5-2
over –2 < x < 2 . Locate any points of intersection and express the values in of radians. e. VERBAL Trigonometric ing Make a conjecture asIdentities to the solutions of 1 = 2 sin x. Explain your reasoning.
The graphs of y = sin x and y =
SOLUTION:
,
,
and
intersect at
over (–2π, 2π).
e. Since sine is a periodic function, with period 2 , the solutions repeat every 2n from and
. Thus, the solutions of sin x =
are x
b.
=
+ 2nπ and x =
+ 2n , where n is an
integer.
60. REASONING Can substitution be used to determine whether an equation is an identity? Explain your reasoning.
SOLUTION: The graphs of y = sin x and y = and
intersect at
Substitution can be used to determine whether an equation is not an identity. However, this method cannot be used to determine whether an equation is an identity, because there is no way to prove that the identity is true for the entire domain.
over [0, 2 ).
61. CHALLENGE that the area A of a triangle
c. On the unit circle, sin x refers to the y-coordinate.
is given by A=
The y coordinate is
at
and
,
.
where a, b, and c represent the sides of the triangle and , , and are the respective opposite angles.
SOLUTION:
Use the Law of Sines to write an equation for b in of side a and angles Β and α.
d.
=
, so b =
.
The graphs of y = sin x and y = eSolutions Manual - Powered by Cognero
,
,
and
intersect at
over (–2π, 2π).
Page 14
5-2
Substitution can be used to determine whether an equation is not an identity. However, this method cannot be used to determine whether an equation is an identity, because there is no wayIdentities to prove that the ing Trigonometric identity is true for the entire domain.
61. CHALLENGE that the area A of a triangle is given by A=
62. Writing in Math Use the properties of logarithms to explain why the sum of the natural logarithm of the six basic trigonometric functions for any angle θ is 0.
,
SOLUTION: where a, b, and c represent the sides of the triangle and , , and are the respective opposite angles. According to the Product Property of Logarithms, the sum of the logarithms of the basic trigonometric functions is equal to the logarithm of the product. Since the product of the absolute values of the functions is 1, the sum of the logarithms is ln 1 or 0.
SOLUTION:
Use the Law of Sines to write an equation for b in of side a and angles Β and α. =
, so b =
.
63. OPEN ENDED Create identities for sec x and csc x in of two or more of the other basic trigonometric functions.
SOLUTION: Start with the identities sec x = sec x and csc x = csc x. Manipulate one side of each equation until you come up with an identity with two or more trigonometric functions.
Two possible identities are: tan x sin x + cos x = sec x and sin x + cot x cos x = csc x.
62. Writing in Math Use the properties of logarithms to explain why the sum of the natural logarithm of the six basic trigonometric functions for any angle θ is 0.
SOLUTION:
According to the Product Property of Logarithms, the sum of the logarithms of the basic trigonometric functions is equal to the logarithm of the product. Since the product of the absolute values of the functions is 1, the sum of the logarithms is ln 1 or 0. eSolutions Manual - Powered by Cognero
Page 15
63. OPEN ENDED Create identities for sec x and csc x in of two or more of the other basic
64. REASONING If two angles
and
are
5-2
According to the Product Property of Logarithms, the sum of the logarithms of the basic trigonometric functions is equal to the logarithm of the product. Since the product of the absolute values of the ing Trigonometric Identities functions is 1, the sum of the logarithms is ln 1 or 0.
63. OPEN ENDED Create identities for sec x and csc
64. REASONING If two angles
and
are
x in of two or more of the other basic trigonometric functions.
complementary, is cos α + cos β = 1? Explain your reasoning. Justify your answers.
SOLUTION:
SOLUTION:
Start with the identities sec x = sec x and csc x = csc x. Manipulate one side of each equation until you come up with an identity with two or more trigonometric functions.
Two possible identities are: tan x sin x + cos x = sec x and sin x + cot x cos x = csc x.
2
2
and are complementary angles, then + = 90° and hence =90° – . Use this to make substitution in the equation. If
2
2
+ cos
cos
2
= cos
2
= cos
2
+ cos (90° – + sin
2
)
= 1.
65. Writing in Math Explain how you would a trigonometric identity in which both sides of the equation are equally complex.
SOLUTION: You can break the identity into two steps. For example consider the identity:
First manipulate one side of the equation until it is simplified reasonably.
64. REASONING If two angles 2
and
are
2
complementary, is cos α + cos β = 1? Explain your reasoning. Justify your answers.
SOLUTION: and are complementary angles, then + = 90° and hence =90° – . Use this to make substitution in the equation.
Since all steps performed were legitimate operations, we know that the following identity holds:
If
2
+ cos
cos
2
= cos
2
= cos
2
2
+ cos (90° – + sin
2
Now perform operations on the left side to .
)
= 1.
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65. Writing in Math Explain how you would a trigonometric identity in which both sides of the
Page 16
68. sin
cot
SOLUTION:
5-2 ing Trigonometric Identities Now perform operations on the left side to . 69. SOLUTION:
70.
SOLUTION:
We have shown
and by the transitive property
of equality, we have verified that . Simplify each expression. 66. cos csc
71. SOLUTION:
SOLUTION:
72. BALLOONING As a hot-air balloon crosses over
67. tan
cot
SOLUTION:
68. sin
a straight portion of interstate highway, its pilot eyes two consecutive mileposts on the same side of the balloon. When viewing the mileposts, the angles of depression are 64 and 7 . How high is the balloon to the nearest foot?
cot
SOLUTION:
SOLUTION:
69. SOLUTION:
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First, find the measures of and DAE.
CAD,
BCA,
ACD
Page 17
SOLUTION:
5-2 ing Trigonometric Identities 72. BALLOONING As a hot-air balloon crosses over a straight portion of interstate highway, its pilot eyes two consecutive mileposts on the same side of the balloon. When viewing the mileposts, the angles of depression are 64 and 7 . How high is the balloon to the nearest foot?
Locate the vertical asymptotes, and sketch the graph of each function.
73. y =
tan x
SOLUTION: The graph of y =
tan x is the graph of y = tan x or . Find
compressed vertically. The period is
the location of two consecutive vertical asymptotes.
SOLUTION:
and
First, find the measures of and DAE.
CAD,
BCA,
ACD
Create a table listing the coordinates of key points for y =
tan x for one period on
.
Function
Vertical Asymptote Intermediate Point x-int
In ACD, use the law of sines to find the length of .
y= tan x
(0, 0)
tan x
y=
(0, 0)
Intermediate Point Vertical Asymptote
Next, use right triangle ADE and the cosine function to find the length of .
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.
73. y =
74. y = csc 2x
tan x
SOLUTION:
SOLUTION: eSolutions Manual - Powered by Cognero
The graph of y =
tan x is the graph of y = tan x
The graph of y = csc 2x is the graph of y = csc x
Page 18
compressed horizontally. The period is
or .
5-2 ing Trigonometric Identities 74. y = csc 2x
75. y =
sec 3x
SOLUTION: The graph of y = csc 2x is the graph of y = csc x or .
compressed horizontally. The period is
Find the location of two vertical asymptotes.
SOLUTION: is the graph of y = sec x
The graph of
compressed vertically and horizontally. The period is or
and
. Find the location of two vertical
asymptotes. Create a table listing the coordinates of key points for y = csc 2x for one period on
. and
Function
Vertical Asymptote Intermediate Point x-int
Intermediate Point Vertical Asymptote
y = csc x
y = csc 2x
x = −π
Create a table listing the coordinates of key points for one period on
for x =0
x =0
.
Function
Vertical Asymptote Intermediate Point x-int
x=π
Sketch the curve through the indicated key points for the function. Then repeat the pattern.
y = sec x
(0, 1)
Intermediate Point Vertical Asymptote
Sketch the curve through the indicated key points for the function. Then repeat the pattern.
75. y =
sec 3x
SOLUTION: The graph of
is the graph of y = sec x
compressed vertically and horizontally. The period is or
. Find the location of two vertical
asymptotes. eSolutions Manual - Powered by Cognero
Write each degree measure in radians as a multiple of π and each radian measure in degrees. Page 19 76. 660
SOLUTION:
5-2 ing Trigonometric Identities Write each degree measure in radians as a multiple of π and each radian measure in degrees. 76. 660
80. SOLUTION:
SOLUTION:
81. 9 SOLUTION: 77. 570 SOLUTION:
Solve each inequality. 82. x2 – 3x – 18 > 0
78. 158
SOLUTION:
SOLUTION: f(x) has real zeros at x = –3 and x = 6. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.
79. SOLUTION:
80.
SOLUTION:
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81. 9 SOLUTION:
Page 20
2
5-2 ing Trigonometric Identities Solve each inequality. 82. x2 – 3x – 18 > 0
The solutions of x – 3x – 18 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the solution set is .
83. x2 + 3x – 28 < 0 SOLUTION:
SOLUTION:
f(x) has real zeros at x = –3 and x = 6. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.
f(x) has real zeros at x = –7 and x = 4. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.
2
The solutions of x – 3x – 18 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the solution set is .
2
The solutions of x + 3x – 28 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution set is .
84. x2 – 4x ≤ 5 83. x2 + 3x – 28 < 0
SOLUTION:
SOLUTION:
First, write x – 4x ≤ 5 as x – 4x – 5 ≤ 0.
f(x) has real zeros at x = –7 and x = 4. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.
f(x) has real zeros at x = –1 and x = 5. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.
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2
2
Page 21
2
5-2
The solutions of x + 3x – 28 < 0 are x-values such that f (x) is negative. From the sign chart, you can ing Trigonometric. Identities see that the solution set is
84. x2 – 4x ≤ 5
2
The solutions of x – 4x –5 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see that the solution set is .
85. x2 + 2x ≥ 24
SOLUTION:
SOLUTION:
2
2
2
2
First, write x – 4x ≤ 5 as x – 4x – 5 ≤ 0.
First, write x + 2x ≥ 24 as x + 2x – 24 ≥ 0.
f(x) has real zeros at x = –1 and x = 5. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.
f(x) has real zeros at x = –6 and x = 4. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.
2
The solutions of x – 4x –5 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see that the solution set is .
85. x2 + 2x ≥ 24 SOLUTION: 2
2
First, write x + 2x ≥ 24 as x + 2x – 24 ≥ 0.
2
The solutions of x + 2x – 24 ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see that the solution set is .
86. −x2 − x + 12 ≥ 0 SOLUTION: 2
2
First, write – x – x + 12 ≥ 0 as x + x – 12 ≤ 0. f(x) has real zeros at x = –6 and x = 4. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or eSolutions Manualat- that Powered by Cognero negative point.
f(x) has real zeros at x = –4 and x = 3. Set up aPage sign22 chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.
2
5-2
The solutions of x + 2x – 24 ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see that the ing Trigonometric. Identities solution set is
86. −x2 − x + 12 ≥ 0
2
The solutions of x + x – 12 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see that the solution set is .
87. −x2 – 6x + 7 ≤ 0
SOLUTION:
SOLUTION:
2
2
2
2
First, write – x – x + 12 ≥ 0 as x + x – 12 ≤ 0.
First, write – x – 6x +7 ≤ 0 as x + 6x – 7 ≥ 0.
f(x) has real zeros at x = –4 and x = 3. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.
f(x) has real zeros at x = –7 and x = 1. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.
2
The solutions of x + x – 12 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see that the solution set is .
87. −x2 – 6x + 7 ≤ 0
88. FOOD The manager of a bakery is randomly
SOLUTION: 2
2
First, write – x – 6x +7 ≤ 0 as x + 6x – 7 ≥ 0.
f(x) has real zeros at x = –7 and x = 1. Set up a sign chart. Substitute anbyx-value eSolutions Manual - Powered Cogneroin each test interval into the polynomial to determine if f (x) is positive or negative at that point.
2
The solutions of x + 6x – 7 ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see that the solution set is .
checking slices of cake prepared by employees to ensure that the correct amount of flavor is in each slice. Each 12-ounce slice should contain half chocolate and half vanilla flavored cream. The amount of chocolate by which each slice varies can be represented by g(x) =
| x – 12|. Describe the Page 23
transformations in the function. Then graph the function.
2
5-2
The solutions of x + 6x – 7 ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see that the ing Trigonometric. Identities solution set is
88. FOOD The manager of a bakery is randomly
89. SAT/ACT
checking slices of cake prepared by employees to ensure that the correct amount of flavor is in each slice. Each 12-ounce slice should contain half chocolate and half vanilla flavored cream. The amount of chocolate by which each slice varies can be represented by g(x) =
| x – 12|. Describe the
transformations in the function. Then graph the function.
SOLUTION: The parent function of g(x) is f (x) = |x|. The factor of
will cause the graph to be compressed since and the subtraction of 12 will translate the
graph 12 units to the right. Make a table of values for x and g(x). x 4 8 12 16 g(x) 4 2 0 2 Plot the points and draw the graph of g(x).
a, b, a, b, b, a, b, b, b, a, b, b, b, b, a, … If the sequence continues in this manner, how many bs are there between the 44th and 47th appearances of the letter a? A 91 B 135 C 138 D 182 E 230
SOLUTION: The number of bs after each a is the same as the number of a in the list (i.e., after the 44th a there are 44 bs). Between the 44th and 47th appearances of a the number of bs will be 44 + 45 + 46 or 135. Therefore, the correct answer choice is B.
90. Which expression can be used to form an identity 20 4
, when
with tan −1? F sin G cos H tan J csc
SOLUTION:
89. SAT/ACT a, b, a, b, b, a, b, b, b, a, b, b, b, b, a, … If the sequence continues in this manner, how many
Therefore, the correct answer choice is J.
91. REVIEW Which of the following is not equivalent to cos
, when 0 < θ <
?
A
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B C cot D tan
Page 24
sin csc
5-2 ing Trigonometric Identities Therefore, the correct answer choice is J.
91. REVIEW Which of the following is not equivalent to cos
, when 0 < θ <
?
A
B C cot D tan
sin csc
SOLUTION:
Therefore, the correct answer choice is D.
92. REVIEW Which of the following is equivalent to sin + cot cos F 2 sin G
?
H cos2 J
SOLUTION:
Therefore, the correct answer choice is G.
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Page 25
5-2 ing Trigonometric Identities each identity.
1. (sec
2
– 1) cos
2
= sin
2
6. tan θ csc2
– tan
= cot
SOLUTION:
SOLUTION:
2. sec2
2
(1 – cos
2
) = tan
7.
–
= cot
SOLUTION: SOLUTION:
3. sin
– sin
2
cos
3
= sin
SOLUTION:
8. 4. csc
– cos
cot = sin
+
= 2 csc
SOLUTION:
SOLUTION:
4
5.
= cot
SOLUTION:
9.
+ tan
= sec
SOLUTION: 2
6. tan θ csc
– tan
= cot
SOLUTION:
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Page 1
5-2 ing Trigonometric Identities
9.
+ tan
12.
= sec
SOLUTION:
10.
+
= 2 sec
+
2
sin
SOLUTION:
= sin + cos
13. (csc
SOLUTION:
– cot
+ cot ) = 1
)(csc
SOLUTION:
14. cos4
4
– sin
2
= cos
2
– sin
SOLUTION:
11.
+
15.
= 1
+
= 2 sec
2
SOLUTION:
SOLUTION:
16.
+
= 2 sec
SOLUTION: 12.
+
= 2 sec
eSolutions Manual - Powered by Cognero SOLUTION:
2
sin Page 2
5-2 ing Trigonometric Identities 16.
19. FIREWORKS If a rocket is launched from ground
= 2 sec
+
level, the maximum height that it reaches is given by
SOLUTION:
h=
is the angle between the
, where
ground and the initial path of the rocket, v is the rocket’s initial speed, and g is the acceleration due to gravity, 9.8 meters per second squared.
17. csc4
4
– cot
2
= 2 cot
+ 1
SOLUTION:
a. that
=
.
b. Suppose a second rocket is fired at an angle of 80 from the ground with an initial speed of 110 meters per second. Find the maximum height of the rocket.
SOLUTION: a.
18.
=
SOLUTION: b. Evaluate the expression
for v = 110 m,
2
, and g = 9.8 m/s .
19. FIREWORKS If a rocket is launched from ground level, the maximum height that it reaches is given by h=
, where
is the angle between the
ground and the initial path of the rocket, v is the rocket’s initial speed, and g is the acceleration due to gravity, 9.8 meters per second squared.
The maximum height of the rocket is about 598.7 meters. each identity. 20. (csc + cot )(1 – cos
) = sin
SOLUTION:
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a. that
=
Page 3
.
maximumTrigonometric height of the rocket Identities is about 598.7 5-2 The ing meters.
each identity. + cot )(1 – cos
20. (csc
) = sin
22.
SOLUTION:
21. sin2
tan
2
SOLUTION:
2
= tan
2
– sin
SOLUTION:
23.
= cos
+ cot
SOLUTION:
22. SOLUTION:
24. (csc
– cot
2
) =
SOLUTION:
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Page 4
5-2 ing Trigonometric Identities 24. (csc
2
– cot
27. sec
) =
– cos
= tan sin
SOLUTION: SOLUTION:
28. 1 – tan4 θ = 2 sec2 θ – sec4 θ SOLUTION: 25.
=
SOLUTION: 29. (csc
– cot
2
) =
SOLUTION:
30. 26. tan2
2
cos
= 1 – cos
2
= sec
SOLUTION:
SOLUTION:
27. sec
– cos
= tan sin
SOLUTION:
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31.
= (sin
SOLUTION:
+ cos )
2
Page 5
5-2 ing Trigonometric Identities 31.
= (sin
+ cos )
2
GRAPHING CALCULATOR Test whether each equation is an identity by graphing. If it appears to be an identity, it. If not, find an x-value for which both sides are defined but not equal. 34. =
SOLUTION:
SOLUTION: Graph
and then graph .
32. OPTICS If two prisms of the same power are placed next to each other, their total power can be determined using z = 2p cos , where z is the combined power of the prisms, p is the power of the individual prisms, and θ is the angle between the two 2
prisms. that 2p cos θ = 2p (1 – sin
)sec
.
SOLUTION:
33. PHOTOGRAPHY The amount of light ing through a polarization filter can be modeled using I = 2 Im cos , where I is the amount of light ing through the filter, Im is the amount of light shined on the filter, and is the angle of rotation between the light source and the filter. that .
SOLUTION:
The graphs of the related functions do not coincide for all values of x for which both functions are defined. Using the intersect feature from the CALC menu on the graphing calculator to find that when x = , Y1 = –1 and Y2 is undefined. Therefore, the equation is not an identity.
35. sec x + tan x = SOLUTION: and then graph
Graph GRAPHING CALCULATOR Test whether each equation is an identity by graphing. If it appears to be an identity, it. If not, find an x-value for which both sides are defined but not equal. 34. =
SOLUTION: eSolutions Manual - Powered by Cognero Graph
and then graph .
.
Page 6
5-2
Using the intersect feature from the CALC menu on the graphing calculator to find that when x = , Y1 = –1 and Y2 is undefined. Therefore, the ing Identities equation is notTrigonometric an identity.
36. sec2 x – 2 sec x tan x + tan2 x =
35. sec x + tan x = SOLUTION:
SOLUTION: and then graph
Graph
2
.
graph Y2 =
The equation appears to be an identity because the graphs of the related functions coincide. this algebraically.
The graphs of the related functions do not coincide for all values of x for which both functions are defined. Using the intersect feature from the CALC menu on the graphing calculator to find that when x = 0, Y1 = 1 and Y2 = 0. Therefore, the equation is not an identity. 2
= 1 – 2 sin x
SOLUTION:
36. sec2 x – 2 sec x tan x + tan2 x =
Graph Y1 =
SOLUTION: 2
Graph Y1 = sec x – 2 sec x tan x + tan x and then graph Y2 =
.
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.
37.
2
2
Graph Y1 = sec x – 2 sec x tan x + tan x and then
and then graph Y2 = 1 – 2
2
sin x.
Page 7
5-2
defined. Using the intersect feature from the CALC menu on the graphing calculator to find that when x = 0, Y1 = 1 and Y2Trigonometric = 0. Therefore, the Identities equation is not an ing identity. 2
37.
= 1 – 2 sin x
38.
SOLUTION:
=
SOLUTION:
Graph Y1 =
and then graph Y2 = 1 – 2
Graph Y1 =
2
sin x.
and then graph Y2 = .
The equation appears to be an identity because the graphs of the related functions coincide. this algebraically.
The graphs of the related functions do not coincide for all values of x for which both functions are defined. Using the intersect feature from the CALC menu on the graphing calculator to find that when x = , Y1 –0.17 and Y2 = 0. Therefore, the equation is not an identity.
39. cos2 x – sin2 x = SOLUTION: 2
38.
.
SOLUTION: Graph Y1 =
2
Graph Y1 = cos x – sin x and then graph Y2 =
=
and then graph Y2 = .
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Page 8
Using the intersect feature from the CALC menu
5-2
on the graphing calculator to find that when x = , Y1 Y2 = 0. Therefore, –0.17 and the equation is ing Trigonometric Identities not an identity.
39. cos2 x – sin2 x =
each identity.
40.
SOLUTION: 2
2
SOLUTION:
Graph Y1 = cos x – sin x and then graph Y2 = .
The equation appears to be an identity because the graphs of the related functions coincide. this algebraically.
41.
=
SOLUTION:
42. ln |csc x + cot x| + ln |csc x – cot x| = 0 SOLUTION: each identity.
40. SOLUTION:
43. ln |cot x| + ln |tan x cos x| = ln |cos x| SOLUTION:
each identity. 44. sec2 + tan2 = sec4 eSolutions Manual - Powered by Cognero
4
– tan
SOLUTION: Start with the right side of the identity.
Page 9
5-2 ing Trigonometric Identities each identity. 44. sec2 + tan2 = sec4
48. sec4 x = 1 + 2 tan2 x + tan4 x
4
– tan
SOLUTION:
SOLUTION:
Start with the right side of the identity.
Start with the right side of the identity.
49. sec2 x csc2 x = sec2 x + csc2 x 45. –2 cos2
4
= sin
– cos
4
SOLUTION:
– 1
Start with the left side of the identity.
SOLUTION: Start with the right side of the identity.
46. sec2 θ sin2 θ = sec4 θ – (tan4 θ + sec2 θ) SOLUTION: Start with the right side of the identity.
47. 3 sec2
tan
2
6
+ 1 = sec
6
– tan
SOLUTION:
50. ENVIRONMENT A biologist studying pollution situates a net across a river and positions instruments at two different stations on the river bank to collect samples. In the diagram shown, d is the distance between the stations and w is width of the river.
a. Determine an equation in of tangent α that can be used to find the distance between the stations. b. that d = . c. Complete the table shown for d = 40 feet.
Start with the right side of the identity. d. If > 60 or < 20 , the instruments will not function properly. Use the table from part c to determine whether sites in which the width of the river is 5, 35, or 140 feet could be used for the experiment.
SOLUTION: a.
48. sec4 x = 1 + 2 tan2 x + tan4 x SOLUTION: Start with the right side of the identity. eSolutions Manual - Powered by Cognero
Page 10
b.
determine whether sites in which the width of the river is 5, 35, or 140 feet could be used for the experiment.
5-2 SOLUTION: ing Trigonometric Identities
of 5 and 140 feet could not be used because > 60 and < 20 , respectively. The site with a width of 35 feet could be used because 20 < < 60 .
a. HYPERBOLIC FUNCTIONS The hyperbolic trigonometric functions are defined in the following ways.
b.
x
–x
sinh x =
(e – e )
cosh x =
(e + e )
x
–x
tanh x = csch x =
, x
0
sech x =
c.
coth x = , x 0 each identity using the functions shown above. 51. cosh2 x – sinh2 x = 1
SOLUTION:
d. If w = 5 then will be greater than 63.4 since 5 < 20. If w = 140, then will be less than 18.4 since 140 > 120. If w = 35, then 45 < < 63.4 since 35 is between 20 and 40. The sites with widths of 5 and 140 feet could not be used because > 60 and < 20 , respectively. The site with a width of 35 feet could be used because 20 < < 60 .
52. sinh (–x) = –sinh x SOLUTION:
HYPERBOLIC FUNCTIONS The hyperbolic trigonometric functions are defined in the following ways. x
(e – e )
cosh x =
(e + e )
x
53. sech2 x = 1 – tanh2 x
–x
sinh x =
SOLUTION:
–x
tanh x = csch x =
, x
0
sech x = eSolutions Manual - Powered by Cognero
coth x = , x 0 each identity using the functions shown above.
Page 11
54. cosh (–x) = cosh x
5-2 ing Trigonometric Identities 53. sech2 x = 1 – tanh2 x
GRAPHING CALCULATOR Graph each side of each equation. If the equation appears to be an identity, it algebraically.
SOLUTION: 55.
SOLUTION: Graph Y1 =
–
and Y2 = 1.
54. cosh (–x) = cosh x SOLUTION:
GRAPHING CALCULATOR Graph each side of each equation. If the equation appears to be an identity, it algebraically.
55.
SOLUTION: Graph Y1 =
The graphs appear to be the same, so the equation appears to be an identity. this algebraically.
–
and Y2 = 1.
56. sec x – cos2 x csc x = tan x sec x SOLUTION:
The graphs appear to be the same, so the equation appears to be an identity. this algebraically.
2
Graphs are not the same so sec x – cos x csc x tan x secx.
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Page 12
57. (tan x + sec x)(1 – sin x) = cos x
5-2 ing Trigonometric Identities 56. sec x – cos2 x csc x = tan x sec x 58.
SOLUTION:
–
= –1
SOLUTION:
2
Graphs are not the same so sec x – cos x csc x tan x secx.
57. (tan x + sec x)(1 – sin x) = cos x SOLUTION:
The graphs are not the same, so
59. MULTIPLE REPRESENTATIONS In this 58.
–
SOLUTION:
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= –1
problem, you will investigate methods used to solve trigonometric equations. Consider 1 = 2 sin x. a. NUMERICAL Isolate the trigonometric function in the equation so that sin x is the only expression on one side of the equation. b. GRAPHICAL Graph the left and right sides of the equation you found in part a on the same graph over [0, 2 ). Locate any points of intersection and express the values in of radians. c. GEOMETRIC Use the unit circle to the answers you found in part b. d. GRAPHICAL Graph the left and right sides of the equation you found in part a on the same graph over –2 < x < 2 . Locate any points of intersection and express the values in of radians. e. VERBAL Make a conjecture as to the solutions of 1 = 2 sin x. Explain your reasoning.
SOLUTION:
Page 13
5-2
over –2 < x < 2 . Locate any points of intersection and express the values in of radians. e. VERBAL Trigonometric ing Make a conjecture asIdentities to the solutions of 1 = 2 sin x. Explain your reasoning.
The graphs of y = sin x and y =
SOLUTION:
,
,
and
intersect at
over (–2π, 2π).
e. Since sine is a periodic function, with period 2 , the solutions repeat every 2n from and
. Thus, the solutions of sin x =
are x
b.
=
+ 2nπ and x =
+ 2n , where n is an
integer.
60. REASONING Can substitution be used to determine whether an equation is an identity? Explain your reasoning.
SOLUTION: The graphs of y = sin x and y = and
intersect at
Substitution can be used to determine whether an equation is not an identity. However, this method cannot be used to determine whether an equation is an identity, because there is no way to prove that the identity is true for the entire domain.
over [0, 2 ).
61. CHALLENGE that the area A of a triangle
c. On the unit circle, sin x refers to the y-coordinate.
is given by A=
The y coordinate is
at
and
,
.
where a, b, and c represent the sides of the triangle and , , and are the respective opposite angles.
SOLUTION:
Use the Law of Sines to write an equation for b in of side a and angles Β and α.
d.
=
, so b =
.
The graphs of y = sin x and y = eSolutions Manual - Powered by Cognero
,
,
and
intersect at
over (–2π, 2π).
Page 14
5-2
Substitution can be used to determine whether an equation is not an identity. However, this method cannot be used to determine whether an equation is an identity, because there is no wayIdentities to prove that the ing Trigonometric identity is true for the entire domain.
61. CHALLENGE that the area A of a triangle is given by A=
62. Writing in Math Use the properties of logarithms to explain why the sum of the natural logarithm of the six basic trigonometric functions for any angle θ is 0.
,
SOLUTION: where a, b, and c represent the sides of the triangle and , , and are the respective opposite angles. According to the Product Property of Logarithms, the sum of the logarithms of the basic trigonometric functions is equal to the logarithm of the product. Since the product of the absolute values of the functions is 1, the sum of the logarithms is ln 1 or 0.
SOLUTION:
Use the Law of Sines to write an equation for b in of side a and angles Β and α. =
, so b =
.
63. OPEN ENDED Create identities for sec x and csc x in of two or more of the other basic trigonometric functions.
SOLUTION: Start with the identities sec x = sec x and csc x = csc x. Manipulate one side of each equation until you come up with an identity with two or more trigonometric functions.
Two possible identities are: tan x sin x + cos x = sec x and sin x + cot x cos x = csc x.
62. Writing in Math Use the properties of logarithms to explain why the sum of the natural logarithm of the six basic trigonometric functions for any angle θ is 0.
SOLUTION:
According to the Product Property of Logarithms, the sum of the logarithms of the basic trigonometric functions is equal to the logarithm of the product. Since the product of the absolute values of the functions is 1, the sum of the logarithms is ln 1 or 0. eSolutions Manual - Powered by Cognero
Page 15
63. OPEN ENDED Create identities for sec x and csc x in of two or more of the other basic
64. REASONING If two angles
and
are
5-2
According to the Product Property of Logarithms, the sum of the logarithms of the basic trigonometric functions is equal to the logarithm of the product. Since the product of the absolute values of the ing Trigonometric Identities functions is 1, the sum of the logarithms is ln 1 or 0.
63. OPEN ENDED Create identities for sec x and csc
64. REASONING If two angles
and
are
x in of two or more of the other basic trigonometric functions.
complementary, is cos α + cos β = 1? Explain your reasoning. Justify your answers.
SOLUTION:
SOLUTION:
Start with the identities sec x = sec x and csc x = csc x. Manipulate one side of each equation until you come up with an identity with two or more trigonometric functions.
Two possible identities are: tan x sin x + cos x = sec x and sin x + cot x cos x = csc x.
2
2
and are complementary angles, then + = 90° and hence =90° – . Use this to make substitution in the equation. If
2
2
+ cos
cos
2
= cos
2
= cos
2
+ cos (90° – + sin
2
)
= 1.
65. Writing in Math Explain how you would a trigonometric identity in which both sides of the equation are equally complex.
SOLUTION: You can break the identity into two steps. For example consider the identity:
First manipulate one side of the equation until it is simplified reasonably.
64. REASONING If two angles 2
and
are
2
complementary, is cos α + cos β = 1? Explain your reasoning. Justify your answers.
SOLUTION: and are complementary angles, then + = 90° and hence =90° – . Use this to make substitution in the equation.
Since all steps performed were legitimate operations, we know that the following identity holds:
If
2
+ cos
cos
2
= cos
2
= cos
2
2
+ cos (90° – + sin
2
Now perform operations on the left side to .
)
= 1.
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65. Writing in Math Explain how you would a trigonometric identity in which both sides of the
Page 16
68. sin
cot
SOLUTION:
5-2 ing Trigonometric Identities Now perform operations on the left side to . 69. SOLUTION:
70.
SOLUTION:
We have shown
and by the transitive property
of equality, we have verified that . Simplify each expression. 66. cos csc
71. SOLUTION:
SOLUTION:
72. BALLOONING As a hot-air balloon crosses over
67. tan
cot
SOLUTION:
68. sin
a straight portion of interstate highway, its pilot eyes two consecutive mileposts on the same side of the balloon. When viewing the mileposts, the angles of depression are 64 and 7 . How high is the balloon to the nearest foot?
cot
SOLUTION:
SOLUTION:
69. SOLUTION:
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First, find the measures of and DAE.
CAD,
BCA,
ACD
Page 17
SOLUTION:
5-2 ing Trigonometric Identities 72. BALLOONING As a hot-air balloon crosses over a straight portion of interstate highway, its pilot eyes two consecutive mileposts on the same side of the balloon. When viewing the mileposts, the angles of depression are 64 and 7 . How high is the balloon to the nearest foot?
Locate the vertical asymptotes, and sketch the graph of each function.
73. y =
tan x
SOLUTION: The graph of y =
tan x is the graph of y = tan x or . Find
compressed vertically. The period is
the location of two consecutive vertical asymptotes.
SOLUTION:
and
First, find the measures of and DAE.
CAD,
BCA,
ACD
Create a table listing the coordinates of key points for y =
tan x for one period on
.
Function
Vertical Asymptote Intermediate Point x-int
In ACD, use the law of sines to find the length of .
y= tan x
(0, 0)
tan x
y=
(0, 0)
Intermediate Point Vertical Asymptote
Next, use right triangle ADE and the cosine function to find the length of .
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.
73. y =
74. y = csc 2x
tan x
SOLUTION:
SOLUTION: eSolutions Manual - Powered by Cognero
The graph of y =
tan x is the graph of y = tan x
The graph of y = csc 2x is the graph of y = csc x
Page 18
compressed horizontally. The period is
or .
5-2 ing Trigonometric Identities 74. y = csc 2x
75. y =
sec 3x
SOLUTION: The graph of y = csc 2x is the graph of y = csc x or .
compressed horizontally. The period is
Find the location of two vertical asymptotes.
SOLUTION: is the graph of y = sec x
The graph of
compressed vertically and horizontally. The period is or
and
. Find the location of two vertical
asymptotes. Create a table listing the coordinates of key points for y = csc 2x for one period on
. and
Function
Vertical Asymptote Intermediate Point x-int
Intermediate Point Vertical Asymptote
y = csc x
y = csc 2x
x = −π
Create a table listing the coordinates of key points for one period on
for x =0
x =0
.
Function
Vertical Asymptote Intermediate Point x-int
x=π
Sketch the curve through the indicated key points for the function. Then repeat the pattern.
y = sec x
(0, 1)
Intermediate Point Vertical Asymptote
Sketch the curve through the indicated key points for the function. Then repeat the pattern.
75. y =
sec 3x
SOLUTION: The graph of
is the graph of y = sec x
compressed vertically and horizontally. The period is or
. Find the location of two vertical
asymptotes. eSolutions Manual - Powered by Cognero
Write each degree measure in radians as a multiple of π and each radian measure in degrees. Page 19 76. 660
SOLUTION:
5-2 ing Trigonometric Identities Write each degree measure in radians as a multiple of π and each radian measure in degrees. 76. 660
80. SOLUTION:
SOLUTION:
81. 9 SOLUTION: 77. 570 SOLUTION:
Solve each inequality. 82. x2 – 3x – 18 > 0
78. 158
SOLUTION:
SOLUTION: f(x) has real zeros at x = –3 and x = 6. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.
79. SOLUTION:
80.
SOLUTION:
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81. 9 SOLUTION:
Page 20
2
5-2 ing Trigonometric Identities Solve each inequality. 82. x2 – 3x – 18 > 0
The solutions of x – 3x – 18 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the solution set is .
83. x2 + 3x – 28 < 0 SOLUTION:
SOLUTION:
f(x) has real zeros at x = –3 and x = 6. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.
f(x) has real zeros at x = –7 and x = 4. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.
2
The solutions of x – 3x – 18 > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the solution set is .
2
The solutions of x + 3x – 28 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution set is .
84. x2 – 4x ≤ 5 83. x2 + 3x – 28 < 0
SOLUTION:
SOLUTION:
First, write x – 4x ≤ 5 as x – 4x – 5 ≤ 0.
f(x) has real zeros at x = –7 and x = 4. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.
f(x) has real zeros at x = –1 and x = 5. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.
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2
2
Page 21
2
5-2
The solutions of x + 3x – 28 < 0 are x-values such that f (x) is negative. From the sign chart, you can ing Trigonometric. Identities see that the solution set is
84. x2 – 4x ≤ 5
2
The solutions of x – 4x –5 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see that the solution set is .
85. x2 + 2x ≥ 24
SOLUTION:
SOLUTION:
2
2
2
2
First, write x – 4x ≤ 5 as x – 4x – 5 ≤ 0.
First, write x + 2x ≥ 24 as x + 2x – 24 ≥ 0.
f(x) has real zeros at x = –1 and x = 5. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.
f(x) has real zeros at x = –6 and x = 4. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.
2
The solutions of x – 4x –5 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see that the solution set is .
85. x2 + 2x ≥ 24 SOLUTION: 2
2
First, write x + 2x ≥ 24 as x + 2x – 24 ≥ 0.
2
The solutions of x + 2x – 24 ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see that the solution set is .
86. −x2 − x + 12 ≥ 0 SOLUTION: 2
2
First, write – x – x + 12 ≥ 0 as x + x – 12 ≤ 0. f(x) has real zeros at x = –6 and x = 4. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or eSolutions Manualat- that Powered by Cognero negative point.
f(x) has real zeros at x = –4 and x = 3. Set up aPage sign22 chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.
2
5-2
The solutions of x + 2x – 24 ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see that the ing Trigonometric. Identities solution set is
86. −x2 − x + 12 ≥ 0
2
The solutions of x + x – 12 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see that the solution set is .
87. −x2 – 6x + 7 ≤ 0
SOLUTION:
SOLUTION:
2
2
2
2
First, write – x – x + 12 ≥ 0 as x + x – 12 ≤ 0.
First, write – x – 6x +7 ≤ 0 as x + 6x – 7 ≥ 0.
f(x) has real zeros at x = –4 and x = 3. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.
f(x) has real zeros at x = –7 and x = 1. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.
2
The solutions of x + x – 12 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see that the solution set is .
87. −x2 – 6x + 7 ≤ 0
88. FOOD The manager of a bakery is randomly
SOLUTION: 2
2
First, write – x – 6x +7 ≤ 0 as x + 6x – 7 ≥ 0.
f(x) has real zeros at x = –7 and x = 1. Set up a sign chart. Substitute anbyx-value eSolutions Manual - Powered Cogneroin each test interval into the polynomial to determine if f (x) is positive or negative at that point.
2
The solutions of x + 6x – 7 ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see that the solution set is .
checking slices of cake prepared by employees to ensure that the correct amount of flavor is in each slice. Each 12-ounce slice should contain half chocolate and half vanilla flavored cream. The amount of chocolate by which each slice varies can be represented by g(x) =
| x – 12|. Describe the Page 23
transformations in the function. Then graph the function.
2
5-2
The solutions of x + 6x – 7 ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see that the ing Trigonometric. Identities solution set is
88. FOOD The manager of a bakery is randomly
89. SAT/ACT
checking slices of cake prepared by employees to ensure that the correct amount of flavor is in each slice. Each 12-ounce slice should contain half chocolate and half vanilla flavored cream. The amount of chocolate by which each slice varies can be represented by g(x) =
| x – 12|. Describe the
transformations in the function. Then graph the function.
SOLUTION: The parent function of g(x) is f (x) = |x|. The factor of
will cause the graph to be compressed since and the subtraction of 12 will translate the
graph 12 units to the right. Make a table of values for x and g(x). x 4 8 12 16 g(x) 4 2 0 2 Plot the points and draw the graph of g(x).
a, b, a, b, b, a, b, b, b, a, b, b, b, b, a, … If the sequence continues in this manner, how many bs are there between the 44th and 47th appearances of the letter a? A 91 B 135 C 138 D 182 E 230
SOLUTION: The number of bs after each a is the same as the number of a in the list (i.e., after the 44th a there are 44 bs). Between the 44th and 47th appearances of a the number of bs will be 44 + 45 + 46 or 135. Therefore, the correct answer choice is B.
90. Which expression can be used to form an identity 20 4
, when
with tan −1? F sin G cos H tan J csc
SOLUTION:
89. SAT/ACT a, b, a, b, b, a, b, b, b, a, b, b, b, b, a, … If the sequence continues in this manner, how many
Therefore, the correct answer choice is J.
91. REVIEW Which of the following is not equivalent to cos
, when 0 < θ <
?
A
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B C cot D tan
Page 24
sin csc
5-2 ing Trigonometric Identities Therefore, the correct answer choice is J.
91. REVIEW Which of the following is not equivalent to cos
, when 0 < θ <
?
A
B C cot D tan
sin csc
SOLUTION:
Therefore, the correct answer choice is D.
92. REVIEW Which of the following is equivalent to sin + cot cos F 2 sin G
?
H cos2 J
SOLUTION:
Therefore, the correct answer choice is G.
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