Download 5-2 Verifying Trigonometric Identities

SOLUTION: 5-2 Verifying Trigonometric Identities
Verify 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: eSolutions Manual - Powered by Cognero
Page 1
5-2 Verifying 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
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SOLUTION: 2
sin
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5-2 Verifying 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. Verify 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.
Verify each identity.
20. (csc + cot )(1 – cos
) = sin
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a. Verify that
= Page 3
.
maximumTrigonometric
height of the rocket Identities
is about 598.7
5-2 The
Verifying
meters.
Verify 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
) =
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Page 4
5-2 Verifying 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: eSolutions Manual - Powered by Cognero
31. = (sin
SOLUTION: + cos )
2
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5-2 Verifying Trigonometric Identities
31. = (sin
+ cos )
2
GRAPHING CALCULATOR Test whether
each equation is an identity by graphing. If it
appears to be an identity, verify 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. Verify that 2p cos θ = 2p (1 – sin
)sec
.
SOLUTION: 33. PHOTOGRAPHY The amount of light passing
through a polarization filter can be modeled using I =
2
Im cos , where I is the amount of light passing
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. Verify 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, verify it. If not, find
an x-value for which both sides are defined but
not equal.
34. =
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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
Verifying
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. Verify 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
Verifying
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. Verify 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
Verifying
Trigonometric
Identities
not an identity.
39. cos2 x – sin2 x =
Verify 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. Verify this
algebraically.
41. =
SOLUTION: 42. ln |csc x + cot x| + ln |csc x – cot x| = 0
SOLUTION: Verify each identity.
40. SOLUTION: 43. ln |cot x| + ln |tan x cos x| = ln |cos x|
SOLUTION: Verify each identity.
44. sec2 + tan2 = sec4
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4
– tan
SOLUTION: Start with the right side of the identity.
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5-2 Verifying Trigonometric Identities
Verify 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 terms of tangent α that
can be used to find the distance between the
stations.
b. Verify 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.
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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: Verifying 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
Verify 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 =
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coth x =
, x 0
Verify each identity using the functions shown
above.
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54. cosh (–x) = cosh x
5-2 Verifying Trigonometric Identities
53. sech2 x = 1 – tanh2 x
GRAPHING CALCULATOR Graph each side
of each equation. If the equation appears to be
an identity, verify 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, verify it algebraically.
55. SOLUTION: Graph Y1 =
The graphs appear to be the same, so the equation
appears to be an identity. Verify 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. Verify this algebraically.
2
Graphs are not the same so sec x – cos x csc x
tan x secx.
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57. (tan x + sec x)(1 – sin x) = cos x
5-2 Verifying 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. –
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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 terms of radians.
c. GEOMETRIC Use the unit circle to verify 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 terms 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 terms of
radians.
e.
VERBAL Trigonometric
Verifying
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 Verify 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
terms of side a and angles Β and α.
d.
= , so b =
.
The graphs of y = sin x and y =
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,
,
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
Verifying
Trigonometric
identity is true for the entire domain.
61. CHALLENGE Verify 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
terms of side a and angles Β and α.
= , so b =
.
63. OPEN ENDED Create identities for sec x and csc
x in terms 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.
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Page 15
63. OPEN ENDED Create identities for sec x and csc
x in terms 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
Verifying
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 terms 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 verify 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 verify.
)
= 1.
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65. Writing in Math Explain how you would verify a
trigonometric identity in which both sides of the
Page 16
68. sin
cot
SOLUTION: 5-2 Verifying Trigonometric Identities
Now perform operations on the left side to verify.
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: eSolutions Manual - Powered by Cognero
First, find the measures of
and DAE.
CAD,
BCA,
ACD
Page 17
SOLUTION: 5-2 Verifying 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 Verifying 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.
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Write each degree measure in radians as a
multiple of π and each radian measure in
degrees.
Page 19
76. 660
SOLUTION: 5-2 Verifying 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: eSolutions Manual - Powered by Cognero
81. 9
SOLUTION: Page 20
2
5-2 Verifying 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
Verifying 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
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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
Verifying 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
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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
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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
Verifying 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 Verifying 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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