5-1 Trigonometric Identities
Find the value of each expression using the
given information.
1. If cot
= , find tan
.
4. If sin
= , find csc
.
SOLUTION: SOLUTION: 5. If cos x =
2. If cos x =
, find sec x. and sin x =
, find cot x.
SOLUTION: SOLUTION: 3. If tan
= , find cot
.
SOLUTION: 6. If sec
= 2 and tan φ =
, find sin
.
SOLUTION: Use the reciprocal identity
to find .
4. If sin
= , find csc
.
SOLUTION: Then use the quotient identity
to find .
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Page 1
5-1 Trigonometric Identities
6. If sec
= 2 and tan φ =
, find sin
.
7. If csc
= and cot = , find sec
.
SOLUTION: Use the reciprocal identity
to find SOLUTION: Use the reciprocal identity
.
to find sin α.
Then use the quotient identity
to find Use the quotient identity
to find cos α.
.
7. If csc
= and cot = , find sec
.
SOLUTION: Use the reciprocal identity
to find sec Use the reciprocal identity
.
to find sin α.
Use the quotient identity
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to find cos α.
8. If sec
= 8 and tan = 3
SOLUTION: , find csc
.
Page 2
5-1 Trigonometric Identities
8. If sec
= 8 and tan = 3
, find csc
.
SOLUTION: Use the reciprocal identity
to find cos SOLUTION: Use the Pythagorean Identity that involves tan θ to
find sec θ.
.
Use the quotient identity
Find the value of each expression using the
given information.
9. sec and cos ; tan θ = –5, cos > 0
to find sin .
Since we are given that cos θ is positive, the
reciprocal function sec θ must be positive, so
.
Use the reciprocal identity
cos θ.
Use the reciprocal identity
to find csc .
to find
10. cot
and sec ; sin θ =
, tan
< 0
SOLUTION: Use the Pythagorean Identity that involves sin
find cos .
Find the value of each expression using the
given information.
9. sec Manual
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by Cognero
and- cos
; tan
θ = –5, cos > 0
SOLUTION: to Page 3
10. cot
and sec ; sin θ =
, tan
< 0
SOLUTION: 5-1 Use
Trigonometric
Identities
the Pythagorean Identity that involves sin
find cos
to .
11. tan
and sin ; sec
= 4, sin > 0
SOLUTION: Use the reciprocal identity
to find cos
.
Use the Pythagorean Identity that involves sec
find tan .
Since tan θ =
then cos
to is negative, and sin θ is positive,
must be negative. Therefore, to find sec Use the reciprocal identity
.
Since sin θ is positive, and
is positive, tan θ must be positive. Therefore,
to find sin .
Use the quotient identity
.
Use the quotient identity
to find cot .
12. sin
and cot ; cos
= , sin
< 0
SOLUTION: Use the Pythagorean Identity that involves cos
find sin .
11. tan
and sin ; sec
to = 4, sin > 0
SOLUTION: eSolutions Manual - Powered by Cognero
Use the reciprocal identity
Page 4
to find cos
13. cos
= , sin
, tan
> 0
to Use the Pythagorean Identity that involves csc
find cot .
and cot ; cos
; csc = SOLUTION: 5-1 Trigonometric Identities
12. sin
and tan
< 0
SOLUTION: Use the Pythagorean Identity that involves cos
find sin .
to Since we are given that tan θ is positive, cot θ must
also be positive. Therefore,
.
Use the reciprocal identity
to find tan .
Since sin θ is negative,
.
to find cot .
Use the quotient identity
.
Use the reciprocal identity
to find sin .
13. cos
and tan
; csc = , tan
> 0
SOLUTION: Use the Pythagorean Identity that involves csc
find cot .
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to Use the quotient identity
to find cos
.
Page 5
5-1 Trigonometric Identities
Use the quotient identity
.
to find cos
14. sin
and cos ; cot
= 8, csc < 0
SOLUTION: Use the Pythagorean Identity that involves cot
find csc .
14. sin
and cos ; cot
to Since we are given that csc θ is negative, csc θ =
.
= 8, csc < 0
SOLUTION: to find sin Use the reciprocal identity
Use the Pythagorean Identity that involves cot
find csc .
to .
Since we are given that csc θ is negative, csc θ =
.
to find sin Use the reciprocal identity
.
to find cos .
Use the quotient identity
to find cos .
Use the quotient identity
15. cot
and sin ; sec
= , sin
> 0
SOLUTION: Use the Pythagorean Identity that involves sec θ to
find tan .
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15. cot
and sin ; sec
= Page 6
, sin
> 0
15. cot
and sin ; sec
= , sin
> 0
SOLUTION: 5-1 Use
Trigonometric
Identities
the Pythagorean Identity that involves sec θ to
find tan
.
Since we are given that sec is negative, we know that cos is negative. We are also given that sin θ
is positive. Since cos is negative and sin is positive, tan is negative. Therefore, .
Then use the Pythagorean Identity that involves cos
θ to find sin θ. Since we are given that sin θ is positive, sin θ =
.
16. tan
and csc ; cos
=
, sin
< 0
SOLUTION: Use the reciprocal identity
to find cot Use the Pythagorean Identity that involves cos θ to
find sin θ.
.
.
Since we are given that sin θ is negative, we know
Use the reciprocal identity
to find cos .
Then use the Pythagorean Identity that involves cos
θ to find sin θ. eSolutions Manual - Powered by Cognero
that sin θ = . Use the reciprocal identity
to find csc θ.
.
Page 7
Since we are given that sin θ is positive, sin θ =
5-1 Trigonometric
Identities
.
16. tan
and csc ; cos
=
, sin
< 0
Find the value of each expression using the
given information. 17. If csc
SOLUTION: Use the Pythagorean Identity that involves cos θ to
find sin θ.
= –1.24, find
.
SOLUTION: 18. If cos x = 0.61, find
.
SOLUTION: Since we are given that sin θ is negative, we know
that sin θ = . Use the reciprocal identity
to find csc θ.
19. If tan
= –1.52, find
.
SOLUTION: .
to find tan θ.
Use the quotient identity
20. If sin
= 0.18, find
.
SOLUTION: Find the value of each expression using the
given information. 17. If csc
= –1.24, find
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SOLUTION: .
21. If cot x = 1.35, find
SOLUTION: .
Page 8
5-1 Trigonometric Identities
21. If cot x = 1.35, find
.
24. sec x cot x – sin x
SOLUTION: SOLUTION: Simplify each expression.
22. csc x sec x – tan x
SOLUTION: 25. SOLUTION: 23. csc x – cos x cot x
SOLUTION: 26. SOLUTION: 24. sec x cot x – sin x
SOLUTION: eSolutions Manual - Powered by Cognero
Page 9
27. SOLUTION: 5-1 Trigonometric Identities
27. 29. SOLUTION: SOLUTION: 30. cot x – csc2 x cot x
SOLUTION: 28. SOLUTION: 31. cot x – cos3 x csc x
SOLUTION: Simplify each expression.
32. 29. SOLUTION: SOLUTION: eSolutions Manual - Powered by Cognero
Page 10
5-1 Trigonometric Identities
Simplify each expression.
34. 32. SOLUTION: SOLUTION: 33. 35. SOLUTION: SOLUTION: 36. SOLUTION: 34. SOLUTION: eSolutions Manual - Powered by Cognero
Page 11
5-1 Trigonometric Identities
37. SUNGLASSES Many sunglasses are made with 36. polarized lenses, which reduce the intensity of light.
The intensity of light emerging from a system of two
polarizing lenses I can be calculated by I = I0 –
SOLUTION: , where I0 is the intensity of light entering the
system of lenses and θ is the angle of the axis of the
second lens in relation to that of the first lens.
a. Simplify the formula for the intensity of light
emerging from a system of two polarized lenses.
b. If a pair of sunglasses contains a system of two
polarizing lenses with axes at 30º to one another,
what proportion of the intensity of light entering the
sunglasses emerges?
SOLUTION: 37. SUNGLASSES Many sunglasses are made with a.
polarized lenses, which reduce the intensity of light.
The intensity of light emerging from a system of two
polarizing lenses I can be calculated by I = I0 –
, where I0 is the intensity of light entering the
system of lenses and θ is the angle of the axis of the
second lens in relation to that of the first lens.
b. Evaluate I = I0 cos2 θ for θ = 30°.
a. Simplify the formula for the intensity of light
emerging from a system of two polarized lenses.
b. If a pair of sunglasses contains a system of two
polarizing lenses with axes at 30º to one another,
what proportion of the intensity of light entering the
sunglasses emerges?
SOLUTION: a.
Therefore,
of the intensity of light entering the sunglasses emerges.
Rewrite as an expression that does not involve
a fraction. eSolutions Manual - Powered by Cognero
38. Page 12
SOLUTION: Therefore,
of the intensity of light entering the 5-1 Trigonometric Identities
sunglasses emerges.
Rewrite as an expression that does not involve
a fraction. 40. 38. SOLUTION: SOLUTION: 41. SOLUTION: 39. SOLUTION: 40. SOLUTION: 42. SOLUTION: eSolutions Manual - Powered by Cognero
Page 13
5-1 Trigonometric Identities
42. 44. SOLUTION: SOLUTION: 43. SOLUTION: 45. SOLUTION: 44. SOLUTION: 46. SOLUTION: eSolutions Manual - Powered by Cognero
Page 14
5-1 Trigonometric Identities
Determine whether each parent trigonometric
function shown is odd or even. Explain your
reasoning.
46. SOLUTION: 48. SOLUTION: Functions that are symmetric with respect to the yaxis are even function. Functions that are symmetric
with respect to the origin are odd functions. The
graph shown is an odd function. 47. The graph is f (x) = cot x. f (–x) = cot (–x) = –cot x =
–f (x). Since f (–x) = –f (x), f (x) = cot x is an odd
function.
SOLUTION: 49. SOLUTION: Functions that are symmetric with respect to the yaxis are even function. The graph shown is an even
function.
The graph is f (x) = sec x. f (–x) = sec (–x) = sec x =
f(x). Since f (–x) = f (x), f (x) = sec x is an even
function.
50. SOCCER When a soccer ball is kicked from the ground, its height y and horizontal displacement x are
related by
Determine whether each parent trigonometric
function shown is odd or even. Explain your
reasoning.
, where v0 is
the initial velocity of the ball, is the angle at which it was kicked, and g is the acceleration due to
gravity. Rewrite this equation so that tan is the only trigonometric function that appears in the
equation.
SOLUTION: eSolutions Manual - Powered by Cognero
Page 15
5-1
function.
The graph is f (x) = sec x. f (–x) = sec (–x) = sec x =
f(x).
Since f (–x) = f (x),
f (x) = sec x is an even
Trigonometric
Identities
function.
50. SOCCER When a soccer ball is kicked from the ground, its height y and horizontal displacement x are
related by
52. cos x + tan x sin x SOLUTION: , where v0 is
the initial velocity of the ball, is the angle at which it was kicked, and g is the acceleration due to
gravity. Rewrite this equation so that tan is the only trigonometric function that appears in the
equation.
SOLUTION: 53. csc x tan2 x – sec2 x csc x
SOLUTION: Write each expression in terms of a single
trigonometric function.
51. tan x – csc x sec x
SOLUTION: 54. sec x csc x – cos x csc x
SOLUTION: 52. cos x + tan x sin x SOLUTION: eSolutions Manual - Powered by Cognero
Page 16
made in part c valid for the entire domain of each
function? Explain your reasoning.
SOLUTION: 5-1 Trigonometric Identities
a. Evaluate
.
for ,
, 0,
, and
54. sec x csc x – cos x csc x
SOLUTION: Evaluate
for .
Evaluate
and
.
for ,
, 0,
,
55. MULTIPLE REPRESENTATIONS In this
problem, you will investigate the verification of
trigonometric identities. Consider the functions
shown.
i.
Evaluate
.
ii.
a. TABULAR Copy and complete the table below,
without graphing the functions.
for ,
, 0,
, and
Complete the table.
b. GRAPHICAL Graph each function on a graphing calculator.
c. VERBAL Make a conjecture about the relationship between y 1 and y 2. Repeat for y 3 and
b. To graph y 1, enter the following into Y1 and then
press Zoom Trig.
y 4.
d. ANALYTICAL Are the conjectures that you made in part c valid for the entire domain of each
function? Explain your reasoning.
SOLUTION: a. Evaluate
.
for eSolutions Manual - Powered by Cognero
,
, 0,
, and
Page 17
5-1 Trigonometric Identities
To graph y 2, enter the following into Y1 and then
press Zoom Trig.
To graph y 4, enter the following into Y1 and then
press Zoom Trig.
To graph y 3, enter the following into Y1 and then
press Zoom Trig.
c. The graphs for y 1 and y 2 are not the same, so y 1
y 2. For the window shown, the graphs of y 3 and
y4 appear identical. Therefore it appears that y 3 =
y4.
d. yes
Since y 2 =
the entire domain of each function, y 1
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To graph y 4, enter the following into Y1 and then
for , and
y 2 .
Page 18
5-1
, and
Since y
Trigonometric
Identities
2=
the entire domain of each function, y 1
for y 2 .
58. ln (cot2 x + 1) + ln |sec x|
SOLUTION: 59. ln (sec2 x – tan2 x) – ln (1 – cos2 x)
Since y 4 =
SOLUTION: = y 3, y 3 = y 4 for the
entire domain of each function.
Rewrite each expression as a single logarithm
and simplify the answer.
56. ln |sin x| – ln |cos x|
SOLUTION: Alternate solution:
57. ln |sec x| – ln |cos x|
SOLUTION: 60. ELECTRICITY A current in a wire in a magnetic
field causes a force to act on the wire. The strength
of the magnetic field can be determined using the
formula
, where F is the force on the
wire, I is the current in the wire, is the length of
the wire, and θ is the angle the wire makes with the
magnetic field. Some physics books give the formula
as F = I Bsin θ. Show that the two formulas are
equivalent.
SOLUTION: Start with one equation and use algebraic and
trigonometric identities to form the other.
58. ln (cot2 x + 1) + ln |sec x|
SOLUTION: 61. LIGHT WAVES When light shines through two
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2
2
2
59. ln (sec x – tan x) – ln (1 – cos x)
SOLUTION: narrow slits, a series of light and dark fringes appear.
Page 19
The angle θ, in radians, locating the mth fringe can
be calculated by sin θ =
, where d is the
5-1 Trigonometric Identities
62. PROOF Prove that the area of the triangle is A =
where
61. LIGHT WAVES When light shines through two
narrow slits, a series of light and dark fringes appear.
The angle θ, in radians, locating the mth fringe can
be calculated by sin θ =
, where d is the
s=
distance between the two slits, and λ is the
wavelength of light.
sin A.)
(a + b + c). (Hint: The area of an oblique triangle is A =
bc
SOLUTION: a. Rewrite the formula in terms of csc θ.
b. Determine the angle locating the 100th fringe
when light having a wavelength of 550 nanometers is
shined through double slits spaced 0.5 millimeters
apart.
SOLUTION: By the Law of Cosines,
, so
. Rewrite each factor under the radical using this
value for cos A.
a.
b. Solve the equation
when m = 100,
= 550 nanometers, and d = 0.5 mm. To do this,
first convert 0.5 mm to nanometers. One nanometer
–9
is 1 × 10 m.
62. PROOF Prove that the area of the triangle is A =
where
s=
(a + b + c). eSolutions Manual - Powered by Cognero
(Hint: The area of an oblique triangle is A =
sin A.)
Page 20
bc
Rewrite each of these expression in terms of s if s =
.
5-1 Trigonometric Identities
63. ERROR ANALYSIS Jenelle and Chloe are
simplifying
. Jenelle thinks that the
expression simplifies to
, and Chloe
2
2
thinks that it simplifies to csc x – tan x. Is either of
them correct? Explain your reasoning.
SOLUTION: Jenelle is correct.
Rewrite each of these expression in terms of s if s =
.
as
Chloe incorrectly simplified
.
CHALLENGE Write each of the basic
trigonometric functions in terms of the
following functions.
64. sin x
SOLUTION: csc x =
cos x:
Substitute these value into the equation
.
sec x:
63. ERROR ANALYSIS Jenelle and Chloe are
simplifying
. Jenelle thinks that the
eSolutions
Manual -simplifies
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expression
, and Chloe
2
2
thinks that it simplifies to csc x – tan x. Is either of
Page 21
5-1 sec
Trigonometric
Identities
x:
65. cos x
SOLUTION: sin x:
tan x:
cot x:
csc x:
sec x:
sec x =
tan x: cot x:
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65. cos x
SOLUTION: Page 22
5-1 cot
Trigonometric
Identities
x:
csc x:
sin x:
66. tan x
SOLUTION: cot x =
sec x:
cos x:
REASONING Determine whether each
statement is true or false . Explain your
reasoning.
67. csc2 x tan x = csc x sec x is true for all real
numbers.
SOLUTION: 2
csc x:
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False; The expressions csc x tan x and csc x sec x
are equivalent for all values of x for which both
expressions are defined, as shown below.
However, these expressions are not defined for all
2
real numbers. For example, when x = , csc x tan
x and csc x sec x are undefined.
Page 23
68. The odd-even identities can be used to prove that the
2
= 5-1 Trigonometric Identities
2
tan
70. cot2
REASONING Determine whether each
statement is true or false . Explain your
reasoning.
67. csc2 x tan x = csc x sec x is true for all real
numbers.
is equivalent to 1 + tan
2
+ 1 = sec
2
= sec
or .
2
+ 1 = csc
SOLUTION: 2
2
2
According to the Pythagorean Theorem, x + y = r .
2
When each side of the equation is divided by y ,
SOLUTION: 2
False; The expressions csc x tan x and csc x sec x
are equivalent for all values of x for which both
expressions are defined, as shown below.
+ 1 = . Since cot
+ 1 = is equivalent to cot
2
or 1 + cot
2
= csc
= and csc =
2
,,
+ 1 = csc
2
.
71. PREWRITE Use a chart or a table to help you
organize the major trigonometric identities found in
Lesson 5-1.
However, these expressions are not defined for all
2
real numbers. For example, when x = , csc x tan
x and csc x sec x are undefined.
SOLUTION: Reciprocal
Identities
sin θ = cos θ =
68. The odd-even identities can be used to prove that the
tan θ =
graphs of y = cos x and y = sec x are symmetric
with respect to the y-axis.
csc θ =
sec θ =
SOLUTION: True; sample answer: The graphs of even functions
are symmetric with respect to the y-axis. Since cos x
and sec x are both even functions, the graphs of
these functions are therefore symmetric with respect
to the y-axis.
cot θ =
Quotient
Identities
tan θ =
cot θ =
PROOF Prove each Pythagorean identity.
69. tan2
+ 1 = sec
2
Pythagorean
Identities
SOLUTION: 2
2
2
= = 2
tan
70. cot2
. Since tan
=
and sec
2
is equivalent to 1 + tan
2
+ 1 = sec
=
,1+
2
= sec
Cofunction
Identities
or .
2
+ 1 = csc
Odd-Even
Identities
SOLUTION: 2
2
2
According to the Pythagorean Theorem, x + y = r .
eSolutions
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When
each- Powered
side of the
equation
+ 1 = 2
cot θ + 1 = csc θ
r . When each side of the equation is divided by x ,
1+
2
tan2 θ + 1 = sec2 θ
2
According to the Pythagorean Theorem, x + y =
2
2
sin θ + cos θ = 1 . Since cot
= 2
is divided by y ,
and csc =
,,
sin (–θ) = –sin θ csc (–θ) = –csc θ
cos (–θ) = cos θ sec (–θ) = sec θ
tan (–θ) = –tan θ cot (–θ) = –cot θ
Page 24
+ 1 = is equivalent to cot
2
+ 1 = csc
cos (–θ) = cos θ sec (–θ) = sec θ
tan (–θ) = –tan θ cot (–θ) = –cot θ
2
5-1 Trigonometric
Identities
2
2
or 1 + cot
= csc
.
Solve each triangle. Round to the nearest
tenth, if necessary.
71. PREWRITE Use a chart or a table to help you
organize the major trigonometric identities found in
Lesson 5-1.
SOLUTION: Reciprocal
Identities
72. sin θ = SOLUTION: cos θ =
tan θ =
csc θ =
Because two angles are given, B = 180 – (62 +
17 ) or 101 . Use the Law of Sines to find b and c.
sec θ =
cot θ =
Quotient
Identities
tan θ =
cot θ =
Therefore, B = 101 , c
3.0, and b
3.4.
Pythagorean
Identities
2
2
sin θ + cos θ = 1 tan2 θ + 1 = sec2 θ
2
2
cot θ + 1 = csc θ
Cofunction
Identities
73. SOLUTION: Because two angles are given, C = 180 – (59 +4
or 73 . Use the Law of Sines to find a and b.
Odd-Even
Identities
sin (–θ) = –sin θ csc (–θ) = –csc θ
cos (–θ) = cos θ sec (–θ) = sec θ
tan (–θ) = –tan θ cot (–θ) = –cot θ
Therefore, C = 73 , a
55.6, and b
48.2.
Solve each triangle. Round to the nearest
tenth, if necessary.
74. SOLUTION: Use the Law of Sines to find b,and C.
72. SOLUTION: Because two angles are given, B = 180 – (62 +
Sines to find b and c.
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17 Manual
) or 101
. Usebythe
Law of
Page 25
5-1 Trigonometric Identities
Therefore, C = 73 , a
55.6, and b
48.2.
Therefore, B
21 , C
37 , and b
13.1.
74. SOLUTION: 75. Use the Law of Sines to find b,and C.
SOLUTION: Use the Law of Sines to find c,and B.
Find B, B = 180º – (122º +37º) or 21º.
Find C, C = 180º – (65º +46º) or 69º.
Therefore, B
21 , C
37 , and b
13.1.
Therefore, B
46 , C
69 , and c
5.2.
75. SOLUTION: Use the Law of Sines to find c,and B.
76. SOLUTION: Because two angles are given, C = 180º – (63º + 20º
97º. Use the Law of Sines to find a and b.
Find C, C = 180º – (65º +46º) or 69º.
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Therefore, C = 97 , a
5.5, and b
14.4.
Page 26
5-1 Trigonometric Identities
Therefore, C = 97 , a
5.5, and b
14.4.
Therefore, A
40 , B
65 , and b
2.8.
Find the exact value of each expression, if it
exists. 78. cot
SOLUTION: 77. SOLUTION: Let u =
Use the Law of Sines to find a,and A.
, so
u=
.
Because the domain of the inverse sine function is
restricted to Quadrants I and IV, u must line in one
of these quadrants. The solution is similar for each
quadrant, so we will solve for Quadrant I.
Draw a diagram of a right triangle with an acute
angle u, an opposite side length 7 and a hypotenuse
length 9.
Find B,B = 180º – (40º +75º) or 65º.
The length of the side adjacent to u is
or
Therefore, A
40 , B
65 , and b
2.8.
. Now, solve
for cot u.
Find the exact value of each expression, if it
exists. 78. cot
SOLUTION: Let u =
, so
u=
.
Because the domain of the inverse sine function is
restricted to Quadrants I and IV, u must line in one
of these quadrants. The solution is similar for each
quadrant, so we will solve for Quadrant I.
Draw a diagram of a right triangle with an acute
angle u, an opposite side length 7 and a hypotenuse
length 9.
So, cot
=
.
79. tan (arctan 3)
SOLUTION: The inverse property applies, because 3 lies on the
interval
. Therefore,
= 3.
80. cos
SOLUTION: eSolutions Manual - Powered by Cognero
The inverse property applies, because
Page 27
lies on 79. tan (arctan 3)
we find that sin u =
SOLUTION: . Therefore,
The inverse property applies, because 3 lies on the
5-1 interval
Trigonometric
Identities
. Therefore,
= 3.
82. cos – 1
80. cos
SOLUTION: SOLUTION: The inverse property applies, because
Let u =
lies on the interval [–1, 1]. Therefore,
=
.
81. . Then
u=
.
Because the domain of the inverse sine function is
restricted to Quadrants I and IV, u must line in one
of these quadrants. The solution is similar for each
quadrant, so we will solve for Quadrant I.
Draw a diagram of a right triangle with an acute
angle u, an opposite side length and a hypotenuse of length 2.
SOLUTION: Let u = cos
–1
and substitute.
Now find
. Draw a diagram of a right
triangle with an acute angle u, an adjacent side
length
and a hypotenuse of length 2.
The length of the side adjacent to u is
. Since
is not a real
number, u does not exist. Therefore,
exist, so cos
–1
does not
does not exist.
83. sin
SOLUTION: The length of the side opposite u is
. Using this triangle,
we find that sin u =
. Therefore,
82. cos – 1
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SOLUTION: Let u =
. Then
u=
Let u =
. Then
u= .
Because the domain of the inverse cosine function is
restricted to Quadrants I and II, u must line in one of
these quadrants. The solution is similar for each
quadrant, so we will solve for Quadrant I.
Draw a diagram of a right triangle with an acute
angle u, an adjacent side length 3 and a hypotenuse
length 5.
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.
number, u does not exist. Therefore,
does not
–1
5-1 exist,
Trigonometric
Identities
so cos
does not exist.
linearizing the data and finding the linear regression
equation.
b. Use the model for the linearized data to find a
model for the original data.
c. Use your model to predict the height an American
male whose head circumference is 24 inches.
83. sin
SOLUTION: Let u =
. Then
u=
.
Because the domain of the inverse cosine function is
restricted to Quadrants I and II, u must line in one of
these quadrants. The solution is similar for each
quadrant, so we will solve for Quadrant I.
Draw a diagram of a right triangle with an acute
angle u, an adjacent side length 3 and a hypotenuse
length 5.
SOLUTION: a. For a quadratic model, replace (x, y) with (x, ln y).
x
y
ln y
14.1
19.5
2.97
18.0
26.4
3.27
18.3
29.7
3.39
18.7
32.3
3.48
19.1
34.4
3.54
19.4
36.2
3.59
19.6
37.7
3.63
Find the regression equation of the linearized data.
The length of the side opposite to u is
or 4.
So, sin
= sin u =
.
84. ANTHROPOLOGY Allometry is the study of the
relationship between the size of an organism and the
size of any of its parts. A researcher decided to test
for an allometry between the size of the human head
compared to the human body as a person ages. The
data in the table represent the average American
male.
a. Find a quadratic model relating these data by
linearizing the data and finding the linear regression
equation.
b. Use the model for the linearized data to find a
model for the original data.
c. Use your model to predict the height an American
male whose head circumference is 24 inches.
The rounded equation is
.
Graph the linearized data.
b. Replace with ln y and simplify.
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86. D ⊂ U
SOLUTION: False; all of the elements of D are not elements of
U; 11 is an element of D but not an element of U.
5-1 Trigonometric Identities
b. Replace with ln y and simplify.
87. SAT/ACT If x > 0 then
=
2
A (x + 1)
2
B (x – 1)
C 3x – 1
D 3x
E 3(x – 1)2
c. Substitute 24 for x.
SOLUTION: The height is approximately 59.7 inches.
Let U = {0, 1, 2, 3, 4, 5}, A = {6, 9}, B = {6, 9,
10}, C = {0, 1, 6, 9, 11}, D = {2, 5, 11}.
Determine whether each statement is true or
false . Explain your reasoning.
85. A ⊂ B
SOLUTION: True; all of the elements of A are also elements of B.
88. REVIEW If sin x = m and 0 < x < 90°, then tan x
=
86. D ⊂ U
SOLUTION: F
False; all of the elements of D are not elements of
U; 11 is an element of D but not an element of U.
87. SAT/ACT If x > 0 then
G
=
H
2
2
J
A (x + 1)
B (x – 1)
C 3x – 1
D 3x
E 3(x – 1)2
SOLUTION: SOLUTION: 89. Which of the following is equivalent to
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⋅ tan θ ?
A tan θ
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5-1 Trigonometric Identities
89. Which of the following is equivalent to
⋅ tan θ ?
A tan θ
B cot θ
C sin θ
D cos θ
SOLUTION: The correct answer is choice B.
90. REVIEW Refer to the figure. If cos D = 0.8, what
is the length of
?
F5
G4
H 3.2
J
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