1. Trigonometry Identities
ALGEBRA 2 LESSON 14-1
(For help, go to Lessons 1-1, 1-2, and 9-4.)
Determine whether each equation is true for all real numbers x.
Explain your reasoning.
1. 2x + 3x = 5x 2. –(4x – 10) = 10 – 4x
= 4x = x + 1
4x2 x
x2 + 1
x – 1
14-1

2. Trigonometry Identities
ALGEBRA 2 LESSON 14-1
Solutions
1. 2x + 3x = (2 + 3)x Distributive Property
= 5x Addition
So 2x + 3x = 5x is true for all real numbers x.
2. –(4x – 10) = –4x + 10 Distributive Property
= 10 – 4x Commutative Property
So –(4x – 10) = –4x + 10 is true for all real numbers x.
= 4x is not defined for x = 0, and therefore not true for all real numbers x.
= x + 1 is not defined for x = 1, and therefore not true for all real numbers x.
4x2 x
x2 + 1
x – 1
14-1

3. Trigonometry Identities
ALGEBRA 2 LESSON 14-1
Verify the identity (sin + cos )2 = 1 + 2sin cos .
1 + 2 sin cos = (sin + cos )2 Rewrite the equation.
= sin sin cos + cos2 Multiply.
= (sin cos2 ) + 2 sin cos Rewrite the expression.
= sin cos Pythagorean identity
14-1

4. Trigonometry Identities
ALGEBRA 2 LESSON 14-1
Verify the identity = sin + cos .
sin
tan
cos
cot
+ = sin + cos Tangent identity
sin
cos
+ = Simplify.
sin • cos
sin
cos • sin
cos
cos + sin = Simplify.
sin + cos = Rewrite the expression.
14-1

5. Trigonometry Identities
ALGEBRA 2 LESSON 14-1
Simplify the trigonometric expression
(1 + cot2 )(sec2 – 1).
(1 + cot2 )(sec2 – 1) = csc2 • tan2 Pythagorean identities
= •    Reciprocal and
Tangent identities 1
sin2
cos2
= Simplify. 1
cos2
= sec2 Reciprocal identity
14-1

6. Trigonometry Identities
ALGEBRA 2 LESSON 14-1
pages 766–768  Exercises
1. cos cot = cos = = – sin
2. sin cot = sin = cos
3. cos tan = cos = sin
4. sin sec = sin = = tan
5. cos sec = cos = 1
6. tan cot = = 1
7. sin csc = sin = = 1
8. cot = = • = ÷ =
sin
cos 1
csc
sec
1 – sin2
9. 1
10. sin2
11. tan2
12. –cot2
13. csc
14. sin
15. cos
16. 1
17. sin
18. 1
19. 1
20. 1
14-1

7. Trigonometry Identities
ALGEBRA 2 LESSON 14-1
45. sin2 tan2 = sin =
(1 – cos2 ) = =
– =
tan2 – sin2
46. sec – sin tan =
– sin =
– = =
= cos
sin2
cos2
sin2 – sin2 cos2
cos
sin 1
1 – sin2
sin2 cos2
21. sec
22. 1
23. sec2
24. sec2
25. cot
26. csc
27. –tan2
28. tan
29. sec
30. csc
31. sin2
32. sin2
33. sin
34. sec
35. sec csc2
36. 1
37. 1
38. 1
39. ± – cos2
40.
41.
42. ± cot2
43. ± csc2 – 1
44. ± tan2
± – cos2
± – sin2
14-1

8. Trigonometry Identities
ALGEBRA 2 LESSON 14-1
50. (cot )2 = cot cot =
cot cot = csc cot
51.
52. 1 – sin
53. Check students’ work.
54. When checking a root of an equation,
you substitute the solution into the
equation to see if it is true. When
verifying an identity, you substitute
equivalent expressions until both
sides are the same.
47. sin cos (tan + cot ) =
sin cos = =
sin cos2 = 1
= • =
= = =
49. = • =
= = sin
sin
cos +
sin2 cos
cos2 sin
1 – sin
(1 – sin )cos
cos2
1 – sin2
(1 – sin )(1 – sin )
1 + sin
sec
cot + tan 1
sin cos
cos sin2
sin2
14-1

9. Trigonometry Identities
ALGEBRA 2 LESSON 14-1
sec = =
+ 1 =
56. = =
cot = =
= cos sin2 = 1
58. sin2 tan cos2 tan2 =
tan2 (sin cos2 ) 1
cos
tan
1 + cos
1 + tan
58. (continued)
= tan2 (1) = tan2 = sec2 – 1
59. 1
60. csc2
61. If n2 > n1, then 1 > 2; if n2 < n1,
then 1 < 2; if n2 = n1, then 1 = 2.
62. C
63. H
64. C
65. F
66. [2] (sec )(sec – 1) =
sec2 + sec – sec – 1 =
sec2 – 1 = tan2
[1] does not show work
cot sin
sec
tan cos
csc
sin +
14-1

10. Trigonometry Identities
ALGEBRA 2 LESSON 14-1
67. [2] Replace tan and sec with their equivalents.
This results in . Combine the terms in the denominator
by multiplying the first term by , so the terms have a common
denominator. This leaves Invert and multiply: •
Simplify and substitute – sin2 for cos2 – 1, which yields –
The final answer is – csc .
[1] omits explanation OR includes minor error
sin
cos 1
cos –
cos2 – 1
14-1

11. Trigonometry Identities
ALGEBRA 2 LESSON 14-1
68. [4] Using the Pythagorean identities,
1 – sin2 x is replaced by cos2 x.
This leaves = sec x. The next
step is to cancel cos x, leaving =
sec x, and since = sec x, then
sec x = sec x, which proves the identity.
[3] appropriate methods with minor error
[2] correct steps without explanation
[1] steps with a minor error and
without explanation
69.
cos x
cos2x 1
70.
71.
72.
14-1

12. Trigonometry Identities
ALGEBRA 2 LESSON 14-1 ° °
80. 79° ° °
83.
84.
73.
74.
75. 35°
76. 45° °
14-1

13. Trigonometry Identities
ALGEBRA 2 LESSON 14-1
1 + cot2
cot2
Answers may vary. Sample:
= +
= tan
= 1 + tan2
= sec2
1 + cot2
cot2 1
1. Verify the identity = sec2 .
2. Simplify (1 – sec )(1 + sec ).
3. Express in terms of cos .
– tan2
1 + tan2
1 – tan2 1
2 cos2 – 1
14-1

14. Solving Trigonometric Equations Using Inverses
ALGEBRA 2 LESSON 14-2
(For help, go to Lessons 7-7, 13-2, and 13-6.)
For each function ƒ, find ƒ–1.
1. ƒ(x) = x ƒ(x) = 2x – 3 3. ƒ(x) = x2 + 4
Find each value.
4. sin 30° 5. cos 6. cos 135°
7. tan – 8. tan 315° 9. sin – 4 7 3
14-2

15. Solving Trigonometric Equations Using Inverses
ALGEBRA 2 LESSON 14-2
Solutions
1. ƒ(x) = x + 1; rewrite as y = x + 1
Inverse: x = y + 1
x – 1 = y
ƒ–1(x) = x – 1
3. ƒ(x) = x2 + 4; rewrite as
y = x2 + 4
Inverse: x = y 2 + 4
x – 4 = y 2
± x – 4 = y
ƒ–1(x) = ± x – 4, x 4
4. sin 30° = cos = =
6. cos 135° = – 7. tan – = 0
8. tan 315° = –1 9. sin – = –
2. ƒ(x) = 2x – 3; rewrite as
y = 2x – 3
Inverse: x = 2y – 3
x + 3 = 2y
= y
y = +
ƒ–1(x) = + 7 3 4 2 x
x + 3
> –
14-2

16. Solving Trigonometric Equations Using Inverses
ALGEBRA 2 LESSON 14-2
Refer to the graph of the inverse of y = cos in Example 1.
a. Find the radian measures of the angles whose cosine is 0. 2
The line x = 0 intersects the graph at (0, ) and (0, – ). 2
So the measures of two angles whose cosine is 0 are and – .
Other points of intersection are (0, ), (0, ), and so on. 3 2 5
The measures of all the angles whose cosine is 0 can be written as n,
where n is an integer. 2
b. Find the radian measures of the angles whose cosine is –1.5.
The line x = –1.5 does not intersect the graph. –1.5 is not in the domain
of the inverse of y = cos .
There is no angle whose cosine is –1.5.
14-2

17. Solving Trigonometric Equations Using Inverses
ALGEBRA 2 LESSON 14-2
Use a unit circle to find the degree measure of the angles
whose cosine is 2
These points and the origin form 45°–45°–90° triangles.
458 and 3158 are the measures of two angles whose cosine is 2
Draw a unit circle and mark the points on
the circle that have x-coordinates of 2
All their coterminal angles also have a cosine 2
The measure of all the angles whose cosine is can be written as
45° + n • 360° and 315° + n • 360°. 2
14-2

18. Solving Trigonometric Equations Using Inverses
ALGEBRA 2 LESSON 14-2
Use a calculator and an inverse function to find the radian measures of all the angles whose sine is 0.98.
The angle is in Quadrant I.
The sine function is also positive in Quadrant II,
as shown in the figure at the right.
sin–1 (0.98) Use a calculator.
So – 1.37 ≈ 1.77 is another solution.
The radian measures of all the angles whose sine is 0.98
can be written as n and n.
14-2

19. Solving Trigonometric Equations Using Inverses
ALGEBRA 2 LESSON 14-2
Use a calculator and an inverse function to find the measures in radians of all the angles whose tangent is 1.34.
The tangent function is also positive in Quadrant III,
as shown in the figure at the right.
So is another solution.
tan– Use a calculator.
The radian measures of all angles whose tangent is 0.93
can be written as 0.93 = n.
14-2

20. Solving Trigonometric Equations Using Inverses
ALGEBRA 2 LESSON 14-2
Solve 7 cos – 3 = 2 cos for 0 < < 2 .
7 cos – 3 = 2 cos
5 cos = 3 Add 3 – 2 cos to each side.
cos = Divide each side by 5. 3 5
cos– Use the inverse function to find one value for . 3 5
The cosine function is also positive in Quadrant IV.
So another value of is 2 –
The two solutions between 0 and 2 are approximately 0.93 and 5.36.
14-2

21. Solving Trigonometric Equations Using Inverses
ALGEBRA 2 LESSON 14-2
Solve 2 sin cos – cos = 0 for 0 < < 2 .
2 sin cos – cos = 0
cos (2 sin – 1) = 0 Factor.
cos = 0 or 2 sin – 1 = 0 Use the Zero-Product Property.
cos = 0 sin = Solve for sin . 1 2
= and = and Use the unit circle. 2 3 6 5
The four values of are , , , and 2 3 6 5
14-2

22. Solving Trigonometric Equations Using Inverses
ALGEBRA 2 LESSON 14-2
The noise level d in decibels close to an electronic alarm
device is modeled by d = 40 sin + 40, where t is the number of seconds after the device is activated. How many seconds does it
take for the sound to reach a noise level of 60decibels?
(t – 6) 12
d = 40 sin + 40
(t – 6) 12
Substitute 60 for d.
60 = 40 sin + 40
Subtract 40 from each side.
20 = 40 sin
(t – 6) 12
Divide each side by 40 and simplify.
= sin
(t – 6) 12 1 2
14-2

23. Solving Trigonometric Equations Using Inverses
ALGEBRA 2 LESSON 14-2
(continued)
sin–1 =
(t – 6) 12 1 2
Use the inverse of the sine to solve for t.
= Evaluate the inverse.
(t – 6) 12
= t – 6 Multiply each side by . 12
= t Add 6 to each side. 12
8 = t Simplify.
The device will reach 60 decibels in 8 seconds.
14-2

24. Solving Trigonometric Equations Using Inverses
ALGEBRA 2 LESSON 14-2 2 6 4 5 11 3
n and n,
or n
n and n
13. no solution
n and n
n and n , , , ,
, 5.44
, 3.61
pages 773–776  Exercises
1. – n n n
4. 30° + n • 360° and 150° + n • 360°
5. 30° + n • 360° and 210° + n • 360°,
or 30° + n • 180°
6. 210° + n • 360° and 330° + n • 360°
7. 120° + n • 360° and 300° + n • 360°,
or 120° + n • 180°
n and n
n and n 10. – n and n
14-2

25. Solving Trigonometric Equations Using Inverses
ALGEBRA 2 LESSON 14-2
22. 0
, 5.25
24. no solution ,
, , ,
28. 0, , ,
, , ,
30. 0, , ,
31. 0,
32. 0, 2 4 3 5 7 ,
34. a. t = • cos–1
b s, 1.33 s, 1.54 s
c s, 3.33 s, 3.54 s
35. 30° + n • 360° and 150° + n • 360°
36. 60° + n • 360° and 300° + n • 360°
° + n • 360° and 330° + n • 360° ,
39.
, 3.24
, 2.80
, 6.18 6 11 h –4
14-2

26. Solving Trigonometric Equations Using Inverses
ALGEBRA 2 LESSON 14-2 2 3 4 6 5
< – 1
cos 0.5 11 7
n, n, n
n, n
n, n
n, n
56. The student misinterpreted the
meaning of cos–1 0.5 as being
equal to
57. The student divided both sides
of the equation by sin , which
in the given interval can be equal
to zero. Since division by zero is
not possible, this is where the
error was.
43. 0
s, s
45. a. 30° + n • 360° x 150° + n • 360°
b. 150° + n • 360° x ° + n • 360°
c. Find the values of x where the graphs
intersect, and then choose the
appropriate interval.
n, n, n
n, n
n, n, n n
n, n, n
n, n, n, n
14-2

27. Solving Trigonometric Equations Using Inverses
ALGEBRA 2 LESSON 14-2
n, n
n, n n
n, n
n, n, n
n, n
64. Answers may vary. Sample:
In the first equation, you isolate
the variable x to get the solution.
In the trigonometric equation,
you first isolate the trig. part,
sin , but then you must
continue to solve for . n
66. a. Answers may vary. Sample:
cos = –1, 2 cos = –2,
3 cos = –3
b. Start with cos = –1, and then
multiply both sides of the equation
by any nonzero number.
67. sin–1
cos–1(y)
69. sin– – 2
cos–1 –
71. cos–1(y – 1) 1 2 y 3 4
14-2

28. Solving Trigonometric Equations Using Inverses
ALGEBRA 2 LESSON 14-2 11 6
cos–1
73. a. K = 1250( – sin ) ft2 b
74. a. 1:55 A.M., 11:05 A.M.,
and 2:55 P.M.
b. 12:00 midnight to 1:55 A.M.,
11:05 A.M. to 2:55 P.M.
75. D
76. H
77. C
78. G
79. D 1
y – 1 2 4 7
80. [2] 2 cos = 2
cos =
cos– =
= and
[1] finds solution in Quadrant I only
81. [4] 2 sin2 = –sin
2 sin sin = 0
sin (2 sin ) = 0
sin = 0 2 sin = 0
sin = –
sin–1 0 = 0, sin–1 – = ,
[3] appropriate methods, with minor error
[2] answer only, without work shown
[1] finds only one solution for each instance
where it might equal 0
14-2

29. Solving Trigonometric Equations Using Inverses
ALGEBRA 2 LESSON 14-2
82. cot
83. tan2
84. 1
85. 1
86. sin
87. tan
88. y = 4 cos
89. y = 3 cos
90. y = 3 cos 2
91. y = cos
92. 4 2 3 91
100
14-2

30. Solving Trigonometric Equations Using Inverses
ALGEBRA 2 LESSON 14-2
1. Use a unit circle and 30°–60°–90° triangles to find the degree measures
of the angles that have a sine of –
2. Use a calculator and inverse functions to find the radian measures of the
angles whose cosine is 0.75.
Solve each equation for < 2 .
3. 2 cos2 – 1 = 0
sin cos + cos = 0 3 2
240° + n • 360°, 300° + n • 360° n, n
< –
, , , 4 5 3 7
, , , 2 3 5 4 7
14-2

31. Right Triangles and Trigonometric Ratios
ALGEBRA 2 LESSON 14-3
(For help, go to Skills Handbook page 844.)
ABC is similar to RST. Complete the following proportions.
1. = 2. =
3. = 4. = a b c t r s
14-3

32. Right Triangles and Trigonometric Ratios
ALGEBRA 2 LESSON 14-3
Solutions
1. = 2. =
3. = 4. = a b c t r s
14-3

33. Right Triangles and Trigonometric Ratios
ALGEBRA 2 LESSON 14-3
A tourist visiting Washington D.C. is seated on
the grass at point A and is looking up at the top of the Washington Monument. The angle of her line of sight with the ground is 27°.
Given that sin 27° , cos 27° , and tan 27° , find her approximate distance AC from the base of the monument.
tan 27° = Definition of tan
height of monument
distance from monument
tan 27° = Substitute.
555 AC
0.51 = Use a calculator in degree mode.
555 AC
AC =
555
0.51
The distance of the visitor from the monument is approximately 1088 feet.
14-3

34. Right Triangles and Trigonometric Ratios
ALGEBRA 2 LESSON 14-3
In PQR, R is a right angle and cos P = .
Find sin P, tan P, and cos Q in fraction and in decimal form. 7 25
Step 1: Draw a diagram.
Step 2:  Use the Pythagorean
Theorem to find p.
r2 = p2 + q2
252 = p2 + 72
625 = p2 + 49
576 = p2
24 = p
Step 3: Calculate the ratios.
sin P = = = = 0.96 24 25
length of leg opposite P
length of the hypotenuse p r
tan P = = = 3.43
length of leg opposite P
length of leg adjacent P p q 24 7
cos Q = = = = 0.96 24 25
length of leg adjacent Q
length of the hypotenuse p r
14-3

35. Right Triangles and Trigonometric Ratios
ALGEBRA 2 LESSON 14-3
A man 6 feet tall is standing 50 feet from a tree. When he looks at the top of the tree, the angle of elevation is 42°. Find the height of the tree to the nearest foot.
In the right triangle, the length of the leg
adjacent to the 42° angle is 50 ft.
The length of the leg opposite the 42° angle is unknown.
You need to find the length of the leg opposite the 42° angle.
Use the tangent ratio.
x = 50 tan 42° Solve for x.
tan 42° = Definition of tan x 50
45 Use a calculator in degree mode.
The height of the tree is approximately or 51 ft.
14-3

36. Right Triangles and Trigonometric Ratios
ALGEBRA 2 LESSON 14-3
In KMN, N is a right angle, m = 7, and n = 25. Find m K to the nearest tenth of a degree.
Step 1: Draw a diagram.
Step 2:  Use a cosine ratio.
cos K =
= 0.28
m K = cos–1 0.28
73.74° Use a calculator. 7 25
Since K is acute, the other solutions of cos– do not apply.
To the nearest tenth of a degree, m K is 73.7°.
14-3

37. Right Triangles and Trigonometric Ratios
ALGEBRA 2 LESSON 14-3
A straight road that goes up a hill is 800 feet higher at the top than at the bottom. The horizontal distance covered is 6515 feet. To the nearest degree, what angle does the road make with level ground?
Let = the measure of the angle of the slope of the road.
You know the length of the leg opposite of the angle you need to find.
You know the length of the hypotenuse. So, use the tangent ratio.
tan =
800
6515
= tan–1 Use the inverse of the tangent function.
800
6515
7.0 Use a calculator.
The angle the road makes with level ground is approximately 7°.
14-3

38. Right Triangles and Trigonometric Ratios
ALGEBRA 2 LESSON 14-3 3. a b c d e
f. not defined 40 41 9
pages 782–785  Exercises
1. a ft
b ft
2. a b c d e f 17 8 15 4.
sin P = ,
cos P = ,
tan P = = 2.4,
csc P = ,
sec P = = 2.6 12 13 5
14-3

39. Right Triangles and Trigonometric Ratios
ALGEBRA 2 LESSON 14-3
8. a. 300 ft
b. 445 ft
c. Answers may vary. Sample:
The flagpole must be straight,
the ground must be flat, and
the flagpole and the ground
must be perpendicular. You
assume these things so that
form a right triangle. By having
a right triangle you can use
its properties to find the
missing parts. ° ° ° ° ° ° ° ° °
18. a , m A °, m B °
19. c , m A °, m B °
20. a = 9, m A °, m B °
14-3

40. Right Triangles and Trigonometric Ratios
ALGEBRA 2 LESSON 14-3
21. c , m A °, m B °
22. a , m A °, m B °
23. b , m A °, m B °
24. a. m A = cos–1
b. 37°
c. 53°
25. cos = ,
tan = ,
csc = ,
sec = ,
cot = 55 8
1200 d
26. sin = ,
tan = ,
csc = ,
sec = ,
27. sin = ,
tan = ,
csc = ,
sec = 5, 3 20 7
117
2 6 5
5 6 12 6
14-3

41. Right Triangles and Trigonometric Ratios
ALGEBRA 2 LESSON 14-3
30. sin = ,
cos = ,
tan = ,
csc = ,
cot =
31. sin = ,
cos = ,
tan = ,
sec =
9 7 35
28. sin = ,
csc = ,
sec = ,
29. sin = ,
sec = , 24 25 7 33 4
5 7 16 9 41 5
32. cos = 0.937, tan = 0.374,
csc = 2.857, sec = 1.068,
cot = 2.676
33. sin = 0.192, cos = 0.981,
tan = 0.196, sec = 1.019,
cot = 5.103
34. a. d =
b ft, ft
35. a = 15, m A °,
m B °
36. c , m A °,
m B °
37. a , b , m B = 38°
38. a , c , m B = 55.8°
100
sin
14-3

42. Right Triangles and Trigonometric Ratios
ALGEBRA 2 LESSON 14-3
39. a , c , m A = 72.8°
40. a , b , m A = 81.7° ° ft m2
44. a ft
b ft
45. a. 12
b. Answers may vary. Sample:
cos E =
sin E =
m E = sin– °
EF = 13 cos 22.6° 12 EF 13 5
46. Using inverse sine, you can find that
= 30°. Since sine is positive in the
first and second quadrants, another
solution is 150°. All the solutions would
be 30° + n • 360° and 150° + n • 360°.
47.
Since APQ and ABC are
similar triangles, =
So, cos = AQ = = = = cos A. AQ AP AC AB 1
14-3

43. Right Triangles and Trigonometric Ratios
ALGEBRA 2 LESSON 14-3
48. a ft
b °
49. sec A =
50. tan A 1
cos A
51. cos2 A + sin2 A 1
b2 + a2 c2
c2 = c2
52. a. 5.9 units
b units2
c. 6.2 units2
53. a. 72°
b cm
54. A
55. I
56. A
57. G
sin A 2 c b a
58. C
59. [2] tan A =
m A tan–
m A °
tan B
m B tan–
m B °
[1] answer only, without
work shown
° + n • 360°
61. no solution
62. 45° + n • 180°
63. 0° + n • 360° and 180° + n • 360°
135 95
14-3

44. Right Triangles and Trigonometric Ratios
ALGEBRA 2 LESSON 14-3
64.
65.
66.
67.
68. P(0)
P(1)
P(2)
P(3)
P(4)
P(5)
P(6)
P(7)
P(8)
14-3

45. Right Triangles and Trigonometric Ratios
ALGEBRA 2 LESSON 14-3
69. P(0)
P(1)
P(2)
P(3)
P(4)
P(5)
P(6)
P(7)
P(8)
70. P(0)
P(1)
P(2)
P(3)
P(4)
P(5)
P(6)
P(7)
P(8)
14-3

46. Right Triangles and Trigonometric Ratios
ALGEBRA 2 LESSON 14-3
1. Find sin A, tan A, and cos B for
the triangle shown to the right.
2. Find csc P, sec Q, and cot Q for
3. In KST, S is a right angle, s = 17, and t = 15. Find m T and
m K to the nearest tenth of a degree.
4. A flagpole that is 35 feet tall casts a shadow 22 feet long at a certain time
of the morning. What is the angle of elevation of the sun to the nearest degree? 4 3
, , 5 2 3 13
, ,
61.9°, 28.1°
58°
14-3

47. Area and the Law of Sines
ALGEBRA 2 LESSON 14-4
(For help, go to Lesson 13-4.)
Simplify each expression.
Find the area of a triangle with the given base b and height h.
7. b = 3 cm, h = 4 cm b = 6 in., h = 15 in.
9. b = 5.2 mm, h = 12.6 mm 10. b = 6.17 ft, h = 3.25 ft
sin 30° 6 12
sin 45° 4 8
sin 60° 10 9
14-4

48. Area and the Law of Sines
ALGEBRA 2 LESSON 14-4
Solutions
sin 30° 6 12
sin 45° 4 8
sin 60° 10 9
0.5 50 1 2 3 20 5
120 24 16 18
1. = = =
2. = = ÷ 4 = • =
3. = = ÷ 10 = • =
4. = = =
5. = = ÷ 8 = • =
6. = = ÷ 9 = • =
14-4

49. Area and the Law of Sines
ALGEBRA 2 LESSON 14-4
Solutions (continued)
7. A = bh = (3 cm)(4 cm) = 6 cm2
8. A = bh = (6 in.)(15 in.) = 45 in.2
9. A = bh = (5.2 mm)(12.6 mm) = mm2
10. A = bh = (6.17 ft)(3.25 ft) ft2 1 2
14-4

50. Area and the Law of Sines
ALGEBRA 2 LESSON 14-4
Find the area of the triangle shown below.
The area K = bc sin A. 1 2
b = 23, and since sin 25° = , h = 40 sin 25°. h 40
So K = (23)(40 sin 25°) cm2. 1 2
14-4

51. Area and the Law of Sines
ALGEBRA 2 LESSON 14-4
In ABC, m A = 33°, m C = 64°, and BC = 8 cm.
Find AC.
Step 1: Draw and label a diagram.
Step 2:  Find the measure of the angle opposite AC.
m B = 180° – 33° – 64° = 83°
Step 3:  Find AC.
= Law of Sines.
AC = Solve for AC.
AC cm Use a calculator.
sin 83° AC
sin 33° 8
8 sin 83°
14-4

52. Area and the Law of Sines
ALGEBRA 2 LESSON 14-4
In PQR, p = 7 in., q = 10 in., and m Q = 98°. Find m R.
Step 1:  Draw and label a diagram.
Step 2:  Find the measure of the angle opposite p.
7 sin 98° 10
= Law of Sines.
sin P = Solve for sin P.
m P = sin–1 Solve for m P.
m P °
sin P 7
sin 98°
Step 3:  Find the measurement of R.
m R ° – 98° – 44°
= 38°
14-4

53. Area and the Law of Sines
ALGEBRA 2 LESSON 14-4
Two observers view the same mountain peak from two points on level ground and 2 miles apart, as shown in the diagram. The angle of elevation for the peak for the observer most distant from the mountain, T, is 31°. For the other observer, S, the angle of elevation for the peak is 58°.
a. Find TK, the distance from T to the summit.
First find m KST and m SKT.
m KST = 180° – 58° = 122°
m SKT = 180° – 31° – 122° = 27°
Use the Law of Sines in SKT. Write a proportion that includes the side
you know, ST, and the side you want to know KT.
14-4

54. Area and the Law of Sines
ALGEBRA 2 LESSON 14-4
(continued)
= Law of Sines
= Substitute.
sin S KT
sin K
sin 122°
sin 27° 2
KT = Solve for KT.
KT Use a calculator.
2 sin 122°
sin 27°
The distance from T to the summit is about 3.74 mi.
b. Find RK, the height of the mountain.
In right KRT, you know KT and m T. Use the sine ratio.
sin 31° = Definition of Sine RK
3.74
RK = 3.74 sin 31° Solve for RK.
RK Use a calculator.
The summit is about 1.93 mi high.
14-4

55. Area and the Law of Sines
ALGEBRA 2 LESSON 14-4
pages 789–792  Exercises
cm2
in.2 m2
cm2
6. 9.2
7. 7.4
in. m
in. ° ° ° ° °
16. a. m 1 = 48°, m 2 = 58°
b. 227 ft
17. m D = 100°, e = 22.3, f = 34.2
18. m A = 38.7°, m B = 33.3°, b = 21.9
19. m T = 29.3°, m R = 26.7°, r = 8.3
20. m B = 85°, b = 11.3, c = 8.7
21. m C = 40°, b = 10, c = 6.8
22. m E = 62.8°, m F = 82.2°, e = 17.1
14-4

56. Area and the Law of Sines
ALGEBRA 2 LESSON 14-4
23. m F = 72°, e = 20 in., f = 23.5 in.
24. m E = 40.3°, m F = 85.7°, f = 12.3 m
25. a. when m C = 50°
b. when m C = 50°
c. 90°; a right triangle has
the greatest height.
26. Check students’ work.
mi, 7.9 mi cm
29. 65° m ft yd cm ft
miles
14-4

57. Area and the Law of Sines
ALGEBRA 2 LESSON 14-4
44. No; you need at least one
side in order to set up a
proportion you can solve.
45. a. 56.4°
b. 93.6°, 26.4°
c. No; EFG could be congruent
to ABC instead of ABD.
46. A
47. B
48. B
49. C ° ° ° °
54. a = 3, m A = 36.9°, m C = 53.1°
55. b = , m A = 45°, m C = 45°
56. c = 11.2, m A = 41.8°, m C = 48.2°
57. b = 8.3, m A = 11.4°, m C = 78.6°
58. c = 8.5, m A = 25.9°, m C = 64.1°
59. a = 89.3, m A = 63.3°, m C = 26.7°
60.
14-4

58. Area and the Law of Sines
ALGEBRA 2 LESSON 14-4
61.
62.
63. 7 5
14-4

59. Area and the Law of Sines
ALGEBRA 2 LESSON 14-4
1. Find the area of EFG.
2. Find KM in KLM.
3. In ABC, m A = 56°, AB = 11 ft, and BC = 13 ft. Find m C.
4. In PQR, m P = 49°, m R = 37°, and PR = 26 cm.
Find QR and the area of PQR.
88.5 cm2
3.7 in.
44.5°
19.7 cm; cm2
14-4

60. 1. In ABC, m C = 33°, a = 17, and c = 21. Find m A.
The Law of Cosines
ALGEBRA 2 LESSON 14-5
(For help, go to Lesson 14-4.)
Use the Law of Sines.
1. In ABC, m C = 33°, a = 17, and c = 21. Find m A.
2. In RST, m R = 59°, m S = 45°, and r = 16. Find t.
3. In DEF, m D = 55°, d = 18, and e = 21. Find m F.
14-5

61. The Law of Cosines Solutions 2. In RST, m R = 59°, m S = 45°,
ALGEBRA 2 LESSON 14-5
Solutions
1. In ABC, m C = 33°, a = 17, and
c = 21. Find m A. =
21 sin A = 17 sin 33°
sin A =
m A = sin–1
m A °
2. In RST, m R = 59°, m S = 45°,
and r = 16. Find t.
m T = 180° – (59° + 45°) = 76°
t sin 59° = 16 sin 76°
t = t
sin A a
sin C c
sin R r
sin T t 17
sin 33° 21
17 sin 33°
sin 59° 16
sin 76°
16 sin 76°
14-5

62. Solutions (continued)
The Law of Cosines
ALGEBRA 2 LESSON 14-5
3. In DEF, m D = 55°, d = 18,
and e = 21. Find m F. =
18 sin E = 21 sin 55°
sin E =
m E = sin–1
m E °
m F ° – (55° °) = 52.1°
sin D d
sin E e
sin 55° 18 21
21 sin 55°
Solutions (continued)
14-5

63. Choose the form of the Law of Cosines that has a2.
ALGEBRA 2 LESSON 14-5
Suppose two stars are 9.5 and 4.6 light years from Earth. When an astronomer observes the stars with the Earth as vertex, the angle between the stars is about 43°. What is the approximate distance between the stars?
Choose the form of the Law of Cosines that has a2.
a2 = b2 + c2 – 2bc cos A
a2 = (9.5)2 + (4.6)2 – 2(9.5)(4.6) cos 43° Substitute.
Use a calculator.
a Use a calculator.
The distance between the two stars is about 6.9 light years.
14-5

64. Choose the form of the Law of Cosines that contains C.
ALGEBRA 2 LESSON 14-5
A triangle has sides of lengths 9 cm, 11 cm, and 14 cm. Find the measure of the angle opposite the longest side. Round your answer to the nearest tenth of a degree.
c2 = a2 + b2 – 2ab cos C
Choose the form of the Law of Cosines that contains C.
142 = – 2(9)(11) cos C Substitute.
196 = – 198 cos C Simplify.
–6 = –198 cos C Combine like terms.
= cos C Solve for cos C. 6
198
cos–1 = m C Solve for m C. 6
198
m C ° Use a calculator.
14-5

65. In PQR, r = 10, q = 12, and m P = 32°. Find m Q.
The Law of Cosines
ALGEBRA 2 LESSON 14-5
In PQR, r = 10, q = 12, and m P = 32°. Find m Q.
Step 1: Draw a diagram.
Step 2:  Find p. Since you cannot find m Q directly,
use the Law of Cosines to find p.
p2 = q2 + r2 – 2qr cos 32°
p2 = – 2(12)(10) cos 32° Substitute.
p Simplify.
p Solve for p.
p Find the principal square root.
14-5

66. Step 3: Now you can use the Law of Sines or
The Law of Cosines
ALGEBRA 2 LESSON 14-5
(continued)
Step 3:  Now you can use the Law of Sines or
the Law of Cosines to find m Q.
= Law of Sines.
sin Q 12
sin 32°
6.36
sin Q = Solve for sin Q.
12 sin 32°
6.36
m Q = sin–1 Solve for m Q.
12 sin 32°
6.36
m Q Use a calculator.
14-5

67. 30. a-b. Check students’ work
The Law of Cosines
ALGEBRA 2 LESSON 14-5
pages 796–799  Exercises ft cm
in. ° ° ° ° ° ° ° °
18. b2 = a2 + c2 – 2ac cos B
19. c2 = a2 + b2 – 2ab cos C = = =
sin B b
sin C c
sin A a
23. a2 = b2 + c2 – 2bc cos A
24. b = 34.7, m A = 26.7°,
m B = 33.3°
25. m A = 50.1°, m B = 56.3°,
m C = 73.6°
26. f = 6.2, m D = 34.6°,
m E = 83.4°
27. m A = 90°, m B = 36.9°,
m C = 53.1°
28. g = 53.3, m F = 31.9°,
m H = 38.1°
29. m A = 56.1°, m B = 70°,
m C = 53.9°
30. a-b. Check students’ work
14-5

68. The Law of Cosines 39. 117.3° 40. 13.0 cm 41. 32.6° 42. 21.5°
ALGEBRA 2 LESSON 14-5 ° cm ° ° ° ft ° ° cm cm
49. Yes; since cos 90° = 0,
c2 = a2 + b2 – 2ab cos C
reduces to c2 = a2 + b2. ft
32. Assume one side is 1.
Then use that side and its corresponding angle in the Law of Sines to find the ratio of the lengths of the other two sides. °
34. a mi
b. 14.4° left;
4.4° west of north ° cm cm °
in.
51. a. 2.1 m
b. 9.8 m2
in. m
14-5

69. The Law of Cosines 61. 26.3 in. 62. 54.0° 63. 2 , x = ± 64. , x = ±
ALGEBRA 2 LESSON 14-5
in. °
, x = ±
, x = ±
, x = ±
, x = ±
67. 3 6 2 1 5 4
14-5

70. 1. In ABC, m A = 127°, c = 9 cm, and b = 8.5 cm. Find a.
The Law of Cosines
ALGEBRA 2 LESSON 14-5
1. In ABC, m A = 127°, c = 9 cm, and b = 8.5 cm. Find a.
2. A triangle has side lengths 12 ft, 13 ft, and 14 ft.
Find the measure of the angle opposite the side of length 13 ft.
Round your answer to the nearest tenth of a degree.
3. In STU, m U = 87°, t = 17, and u = 19. Find s and m S.
15.7 cm
59.4°
9.4, °
14-5

71. Complete the three reciprocal identities. 1. csc = 2. sec = 3. cot =
Angle Identities
ALGEBRA 2 LESSON 14-6
(For help, go to Lesson 14-1.)
Complete the three reciprocal identities.
1. csc = 2. sec = 3. cot =
Complete the three Pythagorean identities.
4. cos sin2 =
tan2 =
cot2 = 1
14-6

72. Angle Identities Solutions 1. csc = 2. sec = 3. cot =
ALGEBRA 2 LESSON 14-6
Solutions
1. csc =
2. sec =
3. cot =
4. cos sin2 = 1
tan2 = sec2
cot2 = csc2 1
sin
cos
tan
14-6

73. Use the fact that – = – – to verify the identity sin ( – ) = sin .
Angle Identities
ALGEBRA 2 LESSON 14-6 2
Use the fact that – = – – to verify the
identity sin ( – ) = sin .
sin ( – ) = sin – – Take the Sine of each side. 2
= cos – Cofunction identity 2
= cos – Negative angle identity 2
= sin Cofunction identity
14-6

74. Solve cos – = sin – for 0 < < 2 .
Angle Identities
ALGEBRA 2 LESSON 14-6 2
Solve cos – = sin – for 0 < < 2 .
cos – = sin – 2
= 1 Divide by sin – .
cos –
sin – 2
= 1 Cofunction identities
sin
cos
tan = Tangent identity = 4
= tan–1 1 Solve for .
Another solution is = 4 5
14-6

75. In a right triangle, the acute angles are complimentary
Angle Identities
ALGEBRA 2 LESSON 14-6
Find a cofunction identity for tan (90° – A), where A is an acute angle of a right triangle.
In a right triangle, the acute angles are complimentary
(the sum of their measures is 90°).
So A + B = 90° and B = 90° – A, where A and B are the
measure of the acute angles.
tan (90° – A) = Tangent identity
sin (90° – A)
cos (90° – A)
= Cofunction identities
cos A
sin A
= cot A Cotangent identity
14-6

76. Find the exact value of cos 165°.
Angle Identities
ALGEBRA 2 LESSON 14-6
Find the exact value of cos 165°.
You know the exact values for 180°, 60°, and 45°.
Use the fact that 165° = 180° – (60° – 45°).
cos (A – B) = cos A cos B + sin A sin B Cosine Angle Difference Identity
cos (180° – (60° – 45°)) =   Substitute 180° for A, and (60° – 45°) for B.
cos 180° cos (60° – 45°) + sin 180° sin (60° – 45°)
= (–1)cos (60° – 45°) + (0)sin (60° – 45°) Use the exact values for
cos 180° and sin 180°.
= (–1)cos (60° – 45°) Simplify.
= (–1)(cos 60° cos 45° + sin 60° sin 45°) Substitute 60° for A, and
45° for B.
14-6

77. = (–1) + Replace with exact values.
Angle Identities
ALGEBRA 2 LESSON 14-6
(continued)
= (–1) Replace with exact values. 2 3 1
= – Simplify.
= – 2 4 6
Since the cosine is negative in Quadrant II, cos 165° = –
( ) 4
14-6

78. Find the exact value of sin 195°.
Angle Identities
ALGEBRA 2 LESSON 14-6
Find the exact value of sin 195°.
Use the fact that 195° = 135° + 60°.
sin (A + B) = sin A cos B + cos A sin B Sine Angle Sum Identity
sin (135° + 60°) = sin 135° cos 60° + cos 135° sin 60° Substitute 135° for A,
and 60° for B.
= Replace with exact values. 2 3 1 –
= – Simplify. = 2 4 6
( – )
So, sin 195° =
( – ) 4
14-6

79. Angle Identities pages 804–805 Exercises 1. csc – = = = = = –sec
ALGEBRA 2 LESSON 14-6
pages 804–805  Exercises
1. csc – = =
= =
= –sec
2. sec – = =
= csc
3. cot – = =
= tan
4. csc – = =
= sec
5. tan – = tan – – =
– tan – = –cot
6. sec – = =
= csc 2
sin – 1
sin – –
– sin –
– cos
cos –
cos – –
cos –
sin
sin –
cos
14-6

80. 37. sin (A – B) = sin (A + (–B)) = sin A cos (–B) + cos A sin (–B)
Angle Identities
ALGEBRA 2 LESSON 14-6
7. ,
8. ,
9. , , ,
10.
11.
, 5.176 ,
14.
15. sec A
16. tan A
17. 0 2 3 4 5 7
18.
19. –1
20. 0
21.
22. – – 2
23. 2 – 3
24.
25.
27. –
28. –1
2 – 6
6 – 2
29.
30.
31. –
32. –
33. –
34.
35.
36.
37. sin (A – B) = sin (A + (–B))
= sin A cos (–B) + cos A sin (–B)
= sin A cos B – cos A sin B 1
14-6

81. 38. tan (A – B) = tan (A + (–B)) =
Angle Identities
ALGEBRA 2 LESSON 14-6
38. tan (A – B) = tan (A + (–B)) =
39. cos (A + B) = cos (A – (–B))
= cos A cos (–B) + sin A sin (–B)
= cos A cos B – sin A sin B
40. tan (A + B) = tan (A – (–B))
tan A + tan(–B)
1 – tan A tan(–B)
tan A – tan B
1 + tan A tan B
tan A – tan(–B)
1 + tan A tan(–B)
41. sin x sin x –
= sin x cos cos x sin +
[sin x cos – cos x sin ]
sin x cos – cos x sin
= 2 sin x cos
= 2 sin x •
= sin x
42. sin – x
= sin cos x – cos sin x
= –1 cos x – 0 • sin x
= –cos x x 3 1 2
tan A + tan B
1 – tan A tan B
14-6

82. 43. Answers may vary. Sample: sin (30° + 60°) sin 30° + sin 60°
Angle Identities
ALGEBRA 2 LESSON 14-6
43. Answers may vary. Sample:
sin (30° + 60°) sin 30° + sin 60°
sin (90°) sin 30° + sin 60°
1 = +
44. sin 3
45. sin 5
46. cos 7
47. cos 5
48. tan 11
49. tan 2 1 2 3
50. a.
b. No; it is not an identity since it is
true for only a few specific x-values.
c. yes; n, n, or n
d. Answers may vary. Sample:
sin x = 2 sin (0.5x)
51. (5 cos – sin , 5 sin +
cos ) /
14-6

83. 57. Given a parallelogram with adjacent sides x1 and x2 and
Angle Identities
ALGEBRA 2 LESSON 14-6
57. Given a parallelogram with
adjacent sides x1 and x2 and
diagonals of d1 and d2,
then by the Law of Cosines,
d12 + d22 = x12 + x22 – 2x1x2 cos +
[x12 + x22 – 2x1x2 cos( – )]
= 2x12 + 2x22 – 2x1x2 cos – 2x1x2
[cos cos + sin sin ]
= 2x12 + 2x22 – 2x1x2 cos –
2x1x2 [–cos ]
= 2x12 + 2x22 – 2x1x2 cos +
2x1x2 cos
= 2x12 + 2x22 = 2(x12 + x22)
58. D
59. I
60. D
52. a. cosine, secant; sine, cosecant,
tangent, and cotangent
b. No; answers may vary. Sample:
y = sin (x + 0.5) is not odd
because y = sin (–x + 0.5) =
sin (–(x – 0.5)) = –sin (x – 0.5) =
–sin (x + 0.5).
53. cos ( – ) = cos • cos + sin sin =
cos cos = –cos
54. sin ( – ) = sin • cos – cos sin =
–cos sin = sin
55. sin ( ) = sin • cos + cos sin =
–sin
56. cos ( ) = cos • cos – sin sin =
–cos /
14-6

84. [1] incorrect property used OR minor error
Angle Identities
ALGEBRA 2 LESSON 14-6 ft cm mm m
69. and 1.40
70. – and –0.87
71. – and –0.26
and 1.22
73. and 3.32
74. and 3.49 2 3 1 6 4
6 – 2 9 5 18 12 7 19 10
61. G
62. B
63. [2] sin 165° = sin 15°
= sin (45° – 30°)
= sin 45° • cos 30° – cos 45° sin 30°
= • – • = – =
[1] incorrect property used OR minor error
64. [2] cos = cos – = cos cos
sin sin = 0 • • =
14-6

85. Angle Identities
ALGEBRA 2 LESSON 14-6
14-6

86. 2. Solve the equation sec – tan = 0 for 0 < < 2 .
Angle Identities
ALGEBRA 2 LESSON 14-6 2
1. Show that sin – – cos = 0.
2. Solve the equation sec – tan = 0 for 0 < < 2 .
3. Find the value of cos 85° cos 55° + sin 85° sin 55°.
4. Write an expression for sin 17° in terms of sines or cosines of 50° and 33°.
sin – = cos , so sin – – cos = cos – cos = 0 2 2 3 2
sin 50° cos 33° – cos 50° sin 33°
14-6

87. Double-Angle and Half-Angle Identities
ALGEBRA 2 LESSON 14-7
(For help, go to Lesson 14-6.)
Complete the following angle identities.
1. cos (A – B) = 2. cos (A + B) =
3. tan (A – B) = 4. sin (A – B) =
5. sin (A + B) = 6. tan (A + B) =
14-7

88. Double-Angle and Half-Angle Identities
ALGEBRA 2 LESSON 14-7
Solutions
1. cos (A – B) = cos A cos B + sin A sin B
2. cos (A + B) = cos A cos B – sin A sin B
3. tan (A – B) =
4. sin (A – B) = sin A cos B – cos A sin B
5. sin (A + B) = sin A cos B + cos A sin B
6. tan (A + B) =
tan A – tan B
1 + tan A tan B
tan A + tan B
1 – tan A tan B
14-7

89. Double-Angle and Half-Angle Identities
ALGEBRA 2 LESSON 14-7
Use a double-angle identity to find the exact value of sin 600°.
sin 600° = sin 2(300°) Rewrite 600 as 2(300).
= 2 sin 300° cos 300° Use a sine double-angle identity.
= 2 – Replace with exact values. 3 2 1
= – Simplify. 3 2
14-7

90. Double-Angle and Half-Angle Identities
ALGEBRA 2 LESSON 14-7
sin 2
(1 – sin2 )
Verify the identity = 2 tan .
sin 2
(1 – sin2 )
= 2 tan .
= double angle and Pythagorean identities
2 sin • cos
cos2
= Simplify.
2 sin
cos
= 2 tan Tangent Identity
14-7

91. Double-Angle and Half-Angle Identities
ALGEBRA 2 LESSON 14-7
Use the half-angle identities to find each exact value.
a. sin 75°
sin 75° = sin Rewrite 75° as
150° 2
= Use the principal square root, since
sin 75° is positive.
(1 – cos 150°) 2
= Substitute the exact value for cos 150°.
1 – 2 3 –
= Simplify. 4
= Simplify. 2
14-7

92. Double-Angle and Half-Angle Identities
ALGEBRA 2 LESSON 14-7
(continued)
b. cos 67.5°
135° 2
cos 67.5° = cos Rewrite 65.7° as
= Use the principal square root, since
sin 75° is positive.
(1 + cos 135°) 2
= Substitute the exact value for cos 135°.
1 + 2 –
= Simplify.
2 – 2 4
= Simplify.
2 – 2 2
14-7

93. Double-Angle and Half-Angle Identities
ALGEBRA 2 LESSON 14-7 12 13 2
Given cos = – and 90° < < 180°, find sin .
Since 90° < < 180°, 45° < < 90° and is in Quadrant I. 2
1 – cos 2
sin = ± half-angle identity
1 – 2 – 12 13
Substitute. Choose the positive
square root since is in Quadrant I. =
= Simplify. 25 26
= Simplify. 26
14-7

94. Double-Angle and Half-Angle Identities
ALGEBRA 2 LESSON 14-7 17
19.
20.
21. 3
22.
23.
24. –
25. – 4
26. – 17
10. tan 2 = tan ( ) = =
11. –
13.
14.
15. –
17. 0
18.
2 – 2
pages 810–811  Exercises
1. –
2. –
3. – 3
4. 1
5. –
7. –
8. –
9. sin 2 = sin ( )
= sin cos + cos • sin
= 2 sin cos 3 1
tan + tan
1 – tan tan
2 tan
1 – tan2
2 – 3
2 – 2 10
14-7

95. Double-Angle and Half-Angle Identities
ALGEBRA 2 LESSON 14-7
27. sin 2R = 2 sin R cos R = 2 • =
28. cos 2R = cos2 R – sin2 R =
– = – =
29. sin 2S = 2 sin S cos S = 2 • • =
= 2 sin R cos R = sin 2R
30. sin2 = sin = ± =
= = – =
31. tan = = =
• = = = r t s
2rs
t 2 2
s 2
r 2
s 2 – r 2
2sr S
1 – cos S
1 – 1 2t
t – s
t + s R
1 – cos R
1 + cos R
1 +
t 2 – s 2
(t + s )2
32. tan2 = (tan )2 = ±
= =
33. No; since the sine function is
periodic, A and B can have many
different values.
34. –
35. –
36.
37. –
1 + cos S
t – r 24 25 7 =
14-7

96. Double-Angle and Half-Angle Identities
ALGEBRA 2 LESSON 14-7
45. 4 cos2 – 1 = 0
, , ,
46. 1
47. –cos
48. cos – sin
49. Answers may vary. Sample:
a. sin 60° = cos 60° =
b. sin 120° =
c. cos 30° =
50. Answers may vary. Sample: No; the graphs
of y = and y = tan are only equal at
certain finite values. 3 2
38.
39.
40. –
41. –2
42. cos (8 sin – 3) = 0
, , 0.384, 2.757
43. sin (4 cos – 3) = 0
0, , 0.723, 5.560
44. cos (2 sin2 – 1) = 0
, , , , , 1 5
2 5 4 7
tan
14-7

97. Double-Angle and Half-Angle Identities
ALGEBRA 2 LESSON 14-7
1± cos
51–56. Answers may vary.
51. 4 sin cos (cos2 – sin2 )
52. 8 cos4 – 8 cos
53.
54. ±
55. ± ± cos
56. ± 1 2
4 tan (1 – tan2 )
tan4 – 6 tan
cos
57. a. tan = ± = ± •
= ± =
± =
Since tan and sin A always have the
same sign, only the positive sign occurs.
b. tan = ± = ± •
= ± =
± =
Since tan and sin A always have the
1 + cos A
1 – cos A
1 – cos 2 A
(1 + cos A)2
sin 2 A
sin A A 2
(1 – cos A)2
1 + cos2 A
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98. Double-Angle and Half-Angle Identities
ALGEBRA 2 LESSON 14-7
58. D
59. G
60. [2] sin = = =
[1] appropriate methods
with minor error
61. [4] tan = = 0.5 =
tan 2 = =
= = =
[3] appropriate method with minor error
[2] correct answer without work shown
[1] recognized using double-angle
identities but did not apply them properly
62.
63.
65. 12, about 4.1
66. 3, 2 3 2
2 2
135°
1 – cos 135°
– 2
1 –
0.5 1
2 tan
1 – tan2 4
14-7

99. Double-Angle and Half-Angle Identities
ALGEBRA 2 LESSON 14-7
1. Use a double-angle identity to find the exact value of sin 660°.
2. Use an angle sum identity and a double-angle identity to
verify the identity sin 3 = 3 sin – 4 sin3 .
3. Use a half-angle identity to find the exact value of sin 67.5°.
4. Given cos = – and 180° < < 270°, find the exact value of cos . 3 2 –
sin 3 = sin ( )
= sin 2 cos + cos 2 sin
= (2 sin cos )cos + (cos2 – sin2 )sin
= 3 sin cos2 – sin3
= 3 sin (1 – sin2 ) – sin3
= 3 sin – 4 sin3 2 20 29 2 2 –
14-7

100. Trigonometric Identities and Equations
ALGEBRA 2 CHAPTER 14
1. csc
2. 1
3. csc2
4. csc cos tan = • • =
= 1
5. csc2 – cot2 = 1 + cot2 – cot2 = 1
6. sec cot = • = = csc
7. sec2 – 1= 1 + tan2 – 1 = tan2
8. 60° + 360° • n, 120° + 360° • n
9. 30° + 360° • n, 330° + 360° • n
° + 360° • n
cos
sin 1
cos sin
sin cos
11. 60° + 360° • n, 240° + 360° • n , , ,
, 0.63
, 1.28
, 1.25
, 1.28
, 1.60
, 1.25
21. 8, 36.9°, 53.1° 3 4 5
6.4
Page 816
14-A

101. Trigonometric Identities and Equations
ALGEBRA 2 CHAPTER 14
, 50.2°, 39.8°
, 24.3°, 65.7°
, 25.3°, 64.7° m2
in. ft ° °
30. No; the Law of Sines requires
at least one angle in order to
set up a ratio of side-to-angle. cm ° °
34. –sin ( – ) = –sin (–( – ))
= sin ( – ) = cos
35. csc ( ) = =
= = sec
36. csc ( – ) =
= = –sec 2 1
sin (0) + cos (1)
sin cos cos sin
cos
sin (– ( – ))
– sin ( – )
– cos
14-A

102. Trigonometric Identities and Equations
ALGEBRA 2 CHAPTER 14
37. cos (– – ) = cos (– )
cos sin (– ) sin =
cos ( )(0) + sin (– )(1) =
sin (– )
38. 0,
39. ,
40. 0;
41.
42.
44.
45. 1
46. –1
47. Check students’ work. 2 3 1
14-A