Basic Trigonometric Identities Page 427 Check for Understanding 1. Sample answer: x 45° 2. Pythagorean identities are derived by applying the Pythagorean Theorem to a right triangle. The opposite angle identities are so named because A is the opposite of A. 3. tan v co 1 t v , cot v ta 1 n v , c s o in s v v cot v, 1 cot 2 v csc 2 v 4. tan(A) c s o in s ( ( A A ) ) c s o i s n A A c s o in s A A tan A 5. Rosalinda is correct; there may be other values for which the equation is not true. 6. Sample answer: v 0° sin v cos v tan v sin 0° cos 0° tan 0° 0 1 0 1 0 7. Sample answer: x 45° sec 2 x csc 2 x 1 sec 2 45° csc 2 45° 1 (2 ) 2 (2 ) 2 1 2 2 1 4 1 8. sec v co 1 s v 9. tan v co 1 t v sec v tan v 1 2 5 1 2 3 7-1 12. 7 3 23 cos 7 3 cos 23 cos 3 13. 330° 360° 30° csc (330°) sin (1 330°) 205 Chapter 7 Chapter 7 Trigonometric Identities and Equations sec v 3 2 tan v 2 5 tan v 25 5 10. sin 2 v cos 2 v 1 1 5 2 cos 2 v 1 2 1 5 cos 2 v 1 cos 2 v 2 2 4 5 cos v 25 6 Quadrant III, so 25 6 11. tan 2 v 1 sec 2 v 4 7 2 1 sec 2 v 1 4 6 9 1 sec 2 v 6 4 5 9 sec 2 v 7 65 sec v Quadrant IV, so 7 65 1 sin (360° 30°) sin 1 30° csc 30° 14. c c s o c t v v si 1 n v c s o in s v v co 1 s v sec v 15. cos x csc x tan x cos xsin 1 x c s o in s x x 1 16. cos x cot x sin x cos x c s o in s x x sin x c s o i s n 2 x x sin x cos 2 x si n x sin 2 x sin 1 x csc x 17. B F c I s c v BIF csc v F c B sc Iv F BIcs 1 c v F BIsin v Pages 427–430 Exercises 18. Sample answer: 45° sin v cos v cot v sin 45° cos 45° cot 45° 1 2 2 2 2 si 1 n v — c s o in s v v 1 2 1 19. Sample answer: 45° t s a e n c v v sin v t s a e n c4 4 5 5 ° ° sin 45° 1 2 2 2 2 2 2
40
Embed
Chapter 7 Trigonometric Identities and Equations · Basic Trigonometric Identities Page 427 Check for Understanding 1. Sample answer: x 45° 2. Pythagorean identities are derived
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Basic Trigonometric Identities
Page 427 Check for Understanding1. Sample answer: x � 45°2. Pythagorean identities are derived by applying
the Pythagorean Theorem to a right triangle. Theopposite angle identities are so named because �Ais the opposite of A.
3. tan v � �co
1t v�, cot v � �
ta1n v�, �
csoins
v
v� � cot v,
1 � cot2 v � csc2 v
4. tan(�A) � �csoins
((�
�
AA
))
�
� ��
csoisnAA
�
� ��csoins
AA
�
� �tan A5. Rosalinda is correct; there may be other values for
which the equation is not true.6. Sample answer: v � 0°
sin v � cos v � tan vsin 0° � cos 0° � tan 0°
0 � 1 � 01 � 0
7. Sample answer: x � 45°sec2 x � csc2 x � 1
sec2 45° � csc2 45° � 1(�2�)2 � (�2�)2 � 1
2 � 2 � 14 � 1
8. sec v � �co
1s v� 9. tan v � �
co1t v�
sec v � tan v �1
�
���25��
1��23
�
7-1 12. �73�� � 2� � �
�3
�
cos �73�� � cos �2� ��
�3
��� cos �
�3
�
13. �330° � �360° � 30°
csc (�330°) � �sin (�
1330°)�
205 Chapter 7
Chapter 7 Trigonometric Identities and Equations
sec v � �32
� tan v � �2
��5�
tan v � ��2�5
5��
10. sin2 v � cos2 v � 1
���15
��2 � cos2 v � 1
�215� � cos2 v � 1
cos2 v � �2245�
cos v � ��2�5
6��
Quadrant III, so ��2�5
6��
11. tan2 v � 1 � sec2 v
���47
��2 � 1 � sec2 v
�1469� � 1 � sec2 v
�6459� � sec2 v
���765�� � sec v
Quadrant IV, so ��765��
�1
���sin (�360° � 30°)
� �sin
130°�
� csc 30°
14. �ccsoct v
v� �
� �si
1n v� � �
csoins
v
v�
� �co
1s v�
� sec v
15. cos x csc x tan x � cos x��sin1
x����c
soins
xx
��� 1
16. cos x cot x � sin x � cos x ��csoins
xx
�� � sin x
� �csoisn
2
xx
� � sin x
� �cos2 x
si�
n xsin2 x�
� �sin
1x
�
� csc x
17. B � �F c
Is�c v�
BI� � F csc v
F � �cBscI�
v�
F � BI���cs1c v��
F � BI�sin v
Pages 427–430 Exercises18. Sample answer: 45°
sin v cos v � cot vsin 45° cos 45° � cot 45°
� �� � � 1�2��
2�2��
2
�si
1n v�
—�csoins
v
v�
�12
� � 1
19. Sample answer: 45°�tsaenc v
v� � sin v
�tsaenc 4
455°°
� � sin 45°
��12�� � �
�22��
�2� � ��22��
20. Sample answer: 30°
sec2 x � 1 � �ccoscs
xx
�
sec2 30° � 1 � �ccoscs
3300°°
�
� �2� 1 �
��23�
�
�2
2�3��
3
29. 1 � cot2 v � csc2 v
1 � cot2 v � ���311���2
1 � cot2 v � �191�
cot2 v � �29�
cot v � � ��32�
�
Quadrant II, so � ��32�
�
30. tan2 v � 1 � sec2 v
tan2 v � 1 � ���54��2
tan2 v � 1 � �2156�
tan2 v � �196�
tan v � ��34�
Quadrant II, so ��34�
31. sin2 v � cos2 v � 1
���13��2
� cos2 v � 1�19� � cos2 v � 1
cos2 v � �89�
cos v � � �2�
32�
�
Quadrant III, so cos v � ��2�
32�
�
tan v � �csoins
vv
�
tan v ���
13�
—
��2�
32�
�
Chapter 7 206
�192� � 1 �
�3��
4
�13
� �
21. Sample answer: 30°sin x � cos x � 1
sin 30° � cos 30° � 1�12
� � ��23�� � 1
�3��
4
�1 �
2�3�� � 1
22. Sample answer: 0°sin y tan y � cos y
sin 0° tan 0° � cos 0°0 � 0 � 1
0 � 123. Sample answer: 45°
tan2 A � cot2 A � 1tan2 45° � cot2 45° � 1
1 � 1 � 12 � 1
24. Sample answer: 0
cos �v � ��2
�� � cos v � cos ��2
�
cos �0 � ��2
�� � cos 0 � cos ��2
�
cos ��2
� � cos 0 � cos ��2
�
0 � 1 � 00 � 1
25. csc v � �si
1n v� 26. cot v � �
ta1n v�
csc v � cot v �1
���43�
�
1��25
�
csc v � �52
� cot v �4
��3�
cot v �
27. sin2 v � cos2 v � 1
��14
��2� cos2 v � 1
�116� � cos2 v � 1
cos2 v � �1156�
4�3��
3
cos v � ��15��
4
Quadrant I, so
28. sin2 v � cos2 v � 1
�15��
4
sin2 v � ���23
��2� 1
sin2 v � �49� � 1
sin2 v � �59�
sin v � ��5��
3
Quadrant II, so �5��
3
tan v � �2�
12�
� or ��42�
�
32. tan2 v � 1 � sec2 v
��23��2
� 1 � sec2 v
�49� � 1 � sec2 v
�193� � sec2 v
���
313�� � sec v
Quadrant III, so sec v � ���
313��
cos v � �se1c v�
cos v �1
�
���
313��
cos v � � or �
33. cos v � �se1c v�
cos v �1
���
75
�
3�13��
133
��13�
sin2 v � cos2 v � 1
sin2 v � ���57��2
� 1
sin2 v � �2459� � 1
sin2 v � �2449�
sin v � � �2�7
6��
Quadrant III, so � �2�
76�
�
cos v � ��57�
34. sec v � �co1s v�
sec v �
sec v � 8Quadrant IV, so �3�7�
35. 1 � cot2 v � csc2 v sin v � �cs
1c v�
1 � ���43��2
� csc2 v sin v �
1 � �196� � csc2 v sin v � ��
35�
�295� � csc2 v
� �53� � csc v
Quadrant IV, so ��53�
36. 1 � cot2 v � csc2 v
1 � (�8)2 � csc2 v
1 � 64 � csc2 v
65 � csc2 v
��65� � csc vQuadrant IV, so ��65�
37. sec v � �co1s v�
sec v �1
���
�43�
�
1���
53�
1��18
�
40. �19
5�� � 2(2�) � �
�5
�
207 Chapter 7
Quadrant II, so ��413��
tan v � �csoins
vv
�
tan v ���
413��
—
���43�
�
tan v � ����
13�3�
� or ���
339��
����
4�3
3���2
� ����
339���2
���
2���
413���2
� 2����43�
��2sec2 A � tan2 A���2sin2 A � 2cos2 A
tan �195�� �
sin �19
5��
—cos �
195��
�
sin �2(2�) � ��5
��——cos �2(2�) � �
�5
��
�
� �tan ��5
�
41. —10
3�— � 3� � �
�3
�
�sin ��5
�
—cos �
�5
�
csc —10
3�— �
1�sin �
103��
�
1��
sin �3� � ��3
���
1��sin �
�3
�
� �csc ��3
�
42. �1290° � �7(180°) � 30°
sec (�1290°) � �cos (�
11290°)�
�
� ��co
1s 30°�
� �sec 30°43. �660° � �2(360°) � 60°
cot (�660°) � �csoins
((�
�
666600°°))
�
�
� �csoins
6600°°
�
� cot 60°
44. —tsaenc x
x— �
� �sin
1x
�
� csc x
45. —ccoost v
v— �
� �si
1n v�
� csc v
46. —csoins
((v
v
�
�
�
�
))
— � —�
�
csoins
v
v—
� tan v47. (sin x � cos x)2 � (sin x � cos x)2
� sin2 x � 2sin x cos x � cos2 x � sin2 x� 2sin x cos x � cos2 x
� 2sin2 x � 2 cos2 x� 2(sin2 x � cos2 x)� 2
�csoins
v
v�
�cos v
�co
1s x�
—�csoins
xx
�
cos (�2(360°) � 60°)���sin (�2(360°) � 60°)
1���cos (�7(180°) � 30°)
tan2 v � 1 � sec2 v
tan2 v � 1 � 82
tan2 v � 1 � 64tan2 v � 63tan v � � 3�7�
sin2 v � cos2 v � 1
sin2 v � ����43�
��2� 1
sin2 v � �136� � 1
sin2 v � �1136�
sin v � ���
413��
�
��99
�
�
�3126�
�498� � �
399�
��
2��1136�� � 2��1
36��
� �12�
38. 390° � 360° � 30°sin 390° � sin (360° � 30°)
� sin 30°
39. �27
8�� � 3� � �
38��
cos �278�� � cos �3� � �
38���
� �cos �38��
sec v � ���43�� or ��
4�3
3��
48. sin x cos x sec x cot x � sin x cos x�—co1s x—��—cs
oins
xx
—�� cos x
49. cos x tan x � sin x cot x � cos x�—csoins
xx
—� � sin x�—csoins
xx
—�� sin x � cos x
50. (1 � cos v)(csc v � cot v) � (1 � cos v)�—si1n v— � —
W � 0.80(0.75)(1000) cos 40°W � 459.6266659459.63 W
57. FN � mg cos v � 0FN � mg cos v
mg sin v � mkFN � 0mg sin v � mk(mg cos v) � 0
mk(mg cos v) � mg sin v
mk � �mm
gg
csoins
v
v�
mk � �csoins
v
v�
mk � tan v
58.
v � �326n0°� � �
18n0°�, tan v � , so h � �
2 taan v� � �
a2
� cot v.�a2
�
�h
Chapter 7 208
� �1
2�
scionsx2 x
�
� �2si
sni2n
xx
�
� �sin
2x
�
� 2csc x53. cos4 a � 2cos2 a sin2 a � sin4 a � (cos2 a � sin2 a)2
� 12 or 154. I � I0 cos2 v
0 � I0 cos2 v
0 � cos2 v
0 � cos vcos�1 0 � v
90° � v
55. Let (x, y) be the point where the terminal side of A intersects the unit circle when A is in standardposition. When A is reflected about the x-axis toobtain �A, the y-coordinate is multiplied by �1,but the x-coordinate is unchanged. So, sin (�A) � �y � � sin A and cos (�A) � x � cos A.
y
xO
(x, y)
(x, �y)
�A
A
ha
�
The area of the isosceles triangle is �12
�(a)��a2
� cot �18
n0°��
� �a4
2� cot ��18
n0°��. There are n such triangles, so
A � �14
�na2 cot ��18n0°��.
59.
sin v � EF and cos v � OF since the circle is a unit
circle. tan v � �OCD
D� � �
C1D� � CD.
sec v � �OCO
D� � �
C1O� � CO. �EOF � �OBA, so
�OEF
F� � �
OBA
A� � �
B1A� � BA. Then cot v � �
csoins
v
v� � �
OEF
F� � BA.
Also by similar triangles, �EE
OF� � �
OO
BA�, or �
E1F� � �
O1B�.
Then csc v � �si
1n v� � �
E1F� � �
O1B� � OB.
60. Cos�1 �� � � 135°�2��
2
y
xO
A C
DF
E
�
B
61.
62. 2(3° 30�) � 7° 7° � 7° � �18
�0°�
� �178�0
�
s � rv
s � 20��178�0
��s � 2.44 cm
63. B � 180° � (90° � 20°) or 70°
sin A � �ac
� cos A � �bc
�
sin 20° � �3a5� cos 20° � �
3b5�
35 sin 20° � a 35 cos 20° � b11.97070502 � a 32.88924173 � ba � 12.0, B � 70°, b � 32.9
64. 2��
2 1 �8 �4
���4����
1�0���
4�
2 5 2 02x2 � 5x � 2 � 0
(2x � 1)(x � 2) � 02x � 1 � 0 or x � 2 � 0
x � ��12
� x � �2
�2, ��12
�, 2
65. 2x2 � 7x � 4 � 0
x2 � �72
�x � 2 � 0
x2 � �72
�x � 2
x2 � �72
�x � �4196� � 2 � �
4196�
�x � �74
��2 � �8116�
x � �74
� � ��94
�
x � ��74
� � �94
�
x � 0.5 or �466. continuous67. 4(x � y � 2z) � 4(3) 4x � 4y � 8z � 12
6b. Sample answer: If a row for sin a were placedabove Exercises 1-5, the entries for Exercise 6acould be obtained by interchanging the first andthird columns and leaving the middle columnalone.
7a. (1) cos, sin, �cos(2) sin, �cos, �sin(3) cot, �tan, cot(4) tan, �cot, tan(5) csc, �sec, �csc(6) sec, csc, �sec
7b. Sample answer: The entries in the rows for cos aand sec a are unchanged. All other entries aremultiplied by �1.
8a. Sample answer: They can be used to reducetrigonometric functions of large positive ornegative angles to those of angles in the firstquadrant.
8b. Sample answer: sum or difference identities
7-3B
5 � 4�3���4 � 5�3�
1 � cot2 v � csc2 v
1 � cot2 v � ��72
��2
1 � cot2 v � �449�
cot2 v � �445�
cot v � �3�
25�
�
Double-Angle and Half-AngleIdentities
Page 453 Check for Understanding1. If you are only given the value of cos v, then
cos 2v � 2 cos2 v � 1 is the best identity to use. If you are only given the value of sin v, then cos 2v � 1 � 2 sin2 v is the best identity to use. Ifyou are given the values of both cos v and sin v,then cos 2v � cos2 v � sin2 v is just as good as theother two.
2. cos 2� � 1 � 2 sin2 v
cos 2v � 1 � �2 sin2 v
�cos
�
2v
2� 1
� � sin2 v
�1 � c
2os 2v� � sin2 v
��1 �c
2o�s 2v�� � sin v
Letting v � �a
2� yields sin �
a
2� � �� ,
or sin �a2� � ��
1 �
2c�os ���.
3a. III or IV 3b. I or II 3c. I, II, III or IV4. sin 2v � 2 sin v
sin 2���2
�� � 2 sin ��2
�
sin � � 2 sin ��2
�
0 � 2(1)0 2
Sample answer: v � ��2
�
5. Both answers are correct. She obtained twodifferent representations of the same number. Oneway to verify this is to evaluate each expressionwith a calculator. To verify it algebraically, squareeach answer and then simplify. The same result isobtained in each case. Since each of the originalanswers is positive, and they have the samesquare, the original answers are the samenumber.
6. sin ��8
� � sin ��
4�
�2
1 � cos 2 ���
2��
��2
7-4 8. sin2 v � cos2 v � 1 tan v � �csoins
v
v�
��25
��2 � cos2 v � 1 �
�25
�
—
��
521��
Chapter 7 220
� �� (Quadrant I)1 � cos �
�
4�
��2
cos2 v � �2215� � or �
2�21
21��
cos v � ��
521�� (Quadrant I)
sin 2v � 2 sin v cos v
� 2��25
�����521���
� �4�
2521�
�
cos 2v � cos2 v � sin2 v
� ���521���2 � ��
25
��2
� �1275�
tan 2v � �1
2�
ttaannv2 v
�
�
2��2�21
21���
——
1 � ��2�21
21���
2
2��21�
� or �4�
1721�
�
9. tan2 v � 1 � sec2 v sin2 v � cos2 v � 1
��43
��2 � 1 � sec2 v sin2 v � ���35
��2 � 1
�295� � sec2 v sin2 v � �
1265�
��53
� � sec v (Quadrant III) sin v � ��45
�
cos v � �se
1c v� (Quadrant III)
� or ��35
�
sin 2v � 2 sin v cos v
� 2���45
�����35
��� �
2245�
cos 2v � cos2 v � sin2 v
� ���35
��2 � ���45
��2
� ��275�
tan 2v � �1
2�
ttaannv2 v
�
�
2��43
��—1 � ��
43
��2
1���
35�
�4�
2121�
�
—�1271�
� or ��274�
10. tan 2v � �cot v �
2tan v�
tan 2v � �cot v �
2tan v� � �
ttaann
v
v�
tan 2v �
tan 2v � �1
2�
ttaannv2 v
�
tan 2v � tan 2v
2 tan v���cot v tan v � tan2 v
�83
�
—
��79
�
� ��� �
�2 �
2��2��
�
7. tan 165° � tan �33
20°�
� ��11 �
�
ccoo�ss
333300°°
�� (Quadrant II)
� ���� �(2 � �3�)� �3� � 2
1 � ��33�
�
—1 � �
�33�
�
1 � ��22��
��2
11. 1 � �12
� sin 2A � �sec A
se�
c Asin A�
1 � �12
� sin 2A �
1 � �12
� sin 2A � � �ccooss
AA
�
1 � �12
� sin 2A � 1 � sin A cos A
1 � �12
� sin 2A � 1 � �12
� � 2 sin A cos A
1 � �12
� sin 2A � 1 � �12
� sin 2A
12. sin �2x
� cos �2x
� � �sin
2x
�
� �sin
2x
�
2 sin �2x
� cos �2x
�
——2
�co
1s A� � sin A
——�co
1s A�
�co
1s A� � sin A
——�co
1s A�
16. tan �51�2� � tan
�56��
�2
221 Chapter 7
� �sin
2x
�
�sin
2x
� � �sin
2x
�
13. cos 2v � 2 cos2 v � 1cos 2v � 1 � 2 cos2 v
�12
� cos 2v � �12
� � cos2 v
P � I02 R sin2 qt
P � I02 R (1 � cos2 qt)
P � I02 R�1 ���
12
� cos 2qt � �12
���P � I0
2 R��12
� � �12
� cos 2qt�P � �
12
� I02 R � �
12
� I02 R cos 2qt
Pages 454–455 Exercises14. cos 15° � cos �
320°�
� ��1 � c2�os 30°�� (Quadrant I)
� ��1 � ��23�
�
—2
sin 2��2x
��—
2
� ��15. sin 75° � sin �
1520°�
� ��1 � c�o2s 150�°�� (Quadrant I)
� ��1 � ����23�
����
2
2 � �3��
2
� ��2 �
2��3��
�
� �� (Quadrant I)1 � cos �
56��
——1 � cos �
56��
� ��1 � ����23�
����
1 � ����23�
��
� ���2 �
2�3��
—
�2 �
2�3��
� ��(2 � �3�)(2 � �3�)���(2 � �3�)(2 � �3�)
� �� (Quadrant I)1 � cos �
34��
——2
� ��� 2 � �3�
17. sin �38�� � sin
�34��
�2
(2 � �3�)2��
4 � 3
� ��1 � ����22�
��—
2
� ��18. cos �
71�2� � cos
�76��
�2
�2� � �2���
2
� ��� (Quadrant II)1 � cos �
76π�
——2
� ���2 �
2�2��
—
�2 �
2�2��
� ��(2 � �2�)(2 � �2�)���(2 � �2�)(2 � �2�)
� �2�2�
2� 2�
� �2� � 1
� ��(2 � �2�)2��
4 � 2
� ��2 � �2��
�2�
� ��1 � ��22�
�
—1 � �
�22�
�
� ���1 � ����23���
��2
� � ��2 �
2��3��
�
19. tan 22.5° � tan �425°�
� ��11
�
�
ccoo�ss
4455°°
�� (Quadrant I)
20. tan �2v
� � ��11
�
�
ccoo�ss
v
v�� 23. tan2 v � 1 � sec2 v
(�2)2 � 1 � sec2 v
5 � sec2 v
��5� � sec v (Quadrant II)cos v � sec v
� or ���55�
�
sin2 v � cos2 v � 1
sin2 v � ����55�
��2 � 1
sin2 v � �2205�
sin v � �2�
55�
� (Quadrant II)
sin 2v � 2 sin v cos v
� 2��2�5
5������
�55�
��� ��
45
�
cos 2v � cos2 v � sin2 v
� ����55�
��2 � ��2�5
5���2
� ��35
�
tan 2v � �1
2�
ttaannv2 v
�
� �1 �
2(�(�
22))2
�
� ��
�
43� or �
43
�
24. cos v � �se
1c v�
1���5�
Chapter 7 222
� ��1 � �14
�
�
1 � �14
�
� ���34
�
—�54
�
� ��35
�� or ��
515��
sin 2v � 2 sin v cos v
� 2��35
����45
��� �
2245�
cos 2v � cos2 v � sin 2 v
� ��45
��2 � ��35
��2
� �275�
tan 2v � �1
2�
ttaannv2 v
�
�
2��34
���1 � ��
34
��2
� or �274�
�32
�
—�176�
22. sin2 v � cos2 v � 1
��13
��2 � cos2 v � 1
cos2 v � �89
�
cos v � �2�
32�
�
(Quadrant I)
sin 2v � 2 sin v cos v
� 2��13
����2�3
2���
� �4�
92�
�
cos 2v � cos2 v � sin2 v
� ��2�3
2���2 � ��
13
��2
� �79
�
tan 2v � �1
2�
ttaannv2 v
�
�2 ���4
2���
��
1 � ���42�
��2
�1
�
��43
�
� ��34
�
sin2 v � cos2 v � 1
sin2 v � ���34
��2 � 1
sin2 v � �176�
sin v � ��47�
� (Quadrant II)
tan v � �csoins
v
v�
���47�
�
—
��34
�
� ���37�
�
sin 2v � 2 sin v cos v
� 2���47�
�����34
��� ��
3�8
7��
cos 2v � cos2 v � sin2 v
� ���34
��2 � ���47�
��2
� �126� or �
18
�
tan 2v � �1
2�
ttaannv2 v
�
�2 ���
�37�
����
1 � ����37�
��2
� � or �3�7��2�
37�
�
—�29
�
21. sin2 v � cos2 v � 1
sin2 v � ��45
��2 � 1
sin2 v � �295�
sin v � �35
�
(Quadrant I)
tan v � �csoins
v
v�
�
� �34
�
�35
�
—�45
�
tan v � �csoins
v
v�
��13
�
—
�2�
32�
�
� or ��42�
�1
�2�2�
� or �4�
72�
�
��22�
�
—�1146�
25. 1 � cot2 v � csc2 v tan v � �co
1t v�
1 � ��32
��2 � csc2 v �
�143� � csc2 v � �
23
�
���
213�� � csc v (Quadrant III)
sin v � �cs
1c v� sin2 v � cos2 v � 1
� ���2�
1313�
��2 � cos2 v � 11
�
���
213��
1�
�32
�
tan 2v � �1
2�
tatannv2 v
�
223 Chapter 7
(Quadrant III)sin 2v � 2 sin v cos v
� 2���2�
1313�
�����3�
1313�
��� �
1123�
cos 2v � cos2 v � sin2 v
� ���3�
1313���2 � ���
2�13
13���2
� �153�
tan 2v � �1
2�
tatannv2 v
�
�
� or �152�
�43
�
�
�59
�
2��23
���
1 � ��23
��2
26. sin v � �cs
1c v�
�
� ��25
�
1—��
52
�
(Quadrant IV)
tan v � �csoins
v
v�
�
��25
�
—
��
521��
�
2���2�
2121�
����
1 � ���2�
2121�
��2
� or ��4�
1721�
�
��4�
2121�
�
——�1271�
27. sin2 a � cos2 a � 1 tan a � �csoins
a
a�
sin2 a � ����32�
��2 � 1 �
sin2 a � �79
� � � or ���
214��
�7���2�
��37��
�
���32��
� or �2�
514�
�
28. csc 2v � �12
� sec v csc v�sin
12v� � �
12
� sec v csc v
�2 sin
1v cos v� � �
12
� sec v csc v
�12
� � �si
1n v� � �
co1s v� � �
12
� sec v csc v
�12
� csc v sec v � �12
� sec v csc v
�12
� sec v csc v � �12
� sec v csc v
29. cos A � sin A � �cos
cAos
�
2sAin A
�
cos A � sin A � �ccooss
2 AA
�
�
ssiinn
2
AA
�
cos A � sin A �
cos A � sin A � cos A � sin A30. (sin v � cos v)2 � 1 � sin 2v
sin2 v � 2 sin v cos v � cos2 v � 1 � sin 2v
2 sin v cos v � 1 � 1 � sin 2v
2 sin v cos v � sin 2v
sin 2v � sin 2v
31. cos x � 1 � �2c(ocsos
2xx
�
�
11)
�
cos x � 1 � �2 c
2o(sc
2
osx
x�
�
11�
)1
�
cos x � 1 � �22(
ccooss
2
xx�
�
12)
�
cos x � 1 � �22((ccooss
2
xx
�
�
11))
�
cos x � 1 �
cos x � 1 � cos x � 1
32. sec 2v � �ccooss
2
2v
v
�
�
ssiinn
2
2v
v�
sec 2v � �cos
12v�
sec 2v � sec 2v
33. tan �A2
� � �1 �
sinco
As A
�
2(cos x � 1)(cos x � 1)���
2(cos x � 1)
(cos A � sin A)(cos A � sin A)����
cos A � sin A
��14��
��52
�
� � or ��2�
1313�
�2
��13�
cos2 v � �11617
9�
cos v � ��3�
1313�
�
sin2 v � cos2 v � 1
���25
��2 � cos2 v � 1
cos2 v � �2215�
cos v � ��
521��
� � or ��2�
2121�
�
sin 2v � 2 sin v cos v
� 2���25
�����521���
� ��4�
2521�
�
cos 2v � cos2 v � sin2 v
� ���521���2 � ���
25
��2
� �1275�
2��21�
sin a � ��37�
� (Quadrant II)
tan 2a � �1
2�
tatanna2 a
�
�
2����
214���
��
1 � ����
214���2
39a. tan�45° � �12
�� �tan 45° � tan �
L2
�
��
1 � tan 45° tan �L2
�
Chapter 7 224
tan �A2
� �sin 2��
A2
����
1 � cos 2��A2
��
tan �A2
� �2 sin �
A2
� cos �A2
�
��1 � 2 cos2 �
A2
� � 1
tan �A2
� �2 sin �
A2
� cos �A2
�
��2 cos2 �
A2
�
tan �A2
� �
tan �A2
� � tan �A2
�
34. sin 3x � 3 sin x � 4 sin3 xsin(2x � x) � 3 sin x � 4 sin3 x
sin 2x cos x � cos 2x sin x � 3 sin x � 4 sin3 x2 sin x cos2 x � (1 � 2 sin2 x) sin x � 3 sin x � 4 sin3 x
2 sin x(1 � sin2 x) � (1 � 2 sin2 x) sin x � 3 sin x � 4 sin3 x2 sin x � 2 sin3 x � sin x � 2 sin3 x � 3 sin x � 4 sin3 x
3 sin x � 4 sin3 x � 3 sin x � 4 sin3 x
35. cos 3x � 4 cos3 x � 3 cos xcos(2x � x) � 4 cos3 x � 3 cos x
(2 cos2 x � 1)cos x � 2 sin2 x cos x � 4 cos3 x � 3 cos x(2 cos2 x � 1)cos x � 2(1 � cos2 x)cos x � 4 cos3 x � 3 cos x2 cos3 x � cos x � 2 cos x � 2 cos3 x � 4 cos3 x � 3 cos x
4 cos3 x � 3 cos x � 4 cos3 x � 3 cos x
36. � �ssiinn
2
22v
v�
�2vg
2� sin2 2v
��2vg
2� sin2 v
sin �A2
�
�cos �
A2
�
R �2v2 cos v sin(v � 45°)���
g cos2 45°
� �(2 sin
sinv
2cov
s v)2�
� �4 sin
s
2
inv2cv
os2 v�
� 4 cos2 v
37. ∠ PBD is an inscribed angle that subtends thesame arc as the central angle ∠ POD, so m∠ PBD
� �12
�v. By right triangle trigonometry, tan�12
�v � �BP
AA�
� �1 �
PAOA� � �
1 �
sinco
v
s v�.
38. R �2v2 cos v sin(v � a)���
g cos2 a
R �2v2 cos v(sin v cos 45° � cos v sin 45°)�����
g cos2 45°
R �2v2 cos v�(sin v)���2
2��� � (cos v)���2
2����
����
g���22�
��2
R �2�
�22�
� v2 cos v(sin v � cos v)���
g � �12
�
R �
R � �v2 �
g2�
�(2 cos v sin v � (2 cos2 v � 1)� 1)
R � �v2 �
g2�
�(sin 2v � cos 2v � 1)
�2�v2 (2 cos v sin v � 2 cos2 v)����
g
�1 � tan �
L2
�
��
1 � 1 � tan �L2
�
�1 ��
11
�
�
ccoo�ss
LL
����
1 � ��11
�
�
ccoo�ss
LL
��
39b. �1 � ��11
�
�
ccoo�ss
6600˚˚
�����
1 � ��11�
�
ccoo�ss
6600˚˚
��1 ��
11
�
�
ccoo�ss
LL
����
1 � ��11
�
�
ccoo�ss
LL
��
� 1 ���1 � �12
�
�
1 � �12
�
1 ���1 � �12
�
�
1 � �12
�
�1 � ��
13
���
1 � ��13
��
��3 �
3�3��
��
�3 �
3�3��
� �12 �
66 �3�� or 2 � �3�
40. tan (a � 30°) � �271�
tan (a � 30°) � 3
� 3
tan a � ��33�
� � 3 � �3� tan a
tan a � �3� tan a � 3 � ��33�
�
(1 � �3�) tan a � 3 � ��33�
�
tan a �3 � �
�33��
��1 � �3�
tan a � tan 30°���1 � tan a tan 30°
tan a �9 � �3���3 �3�3�
tan a � ��6 �
35�3��
41. cos �1�2� � cos ��
�3
� � ��4
��� cos �
�3
� cos ��4
� � sin ��3
� sin ��4
�
� �12
� � ��22�
� � ��23�
� � ��22�
�
� ��2� �
4�6�
�
sec �1�2� �
1�
cos �1�2�
�1
��
��2� �
4�6�
�
� �4�2�
�
�
44�6�
� or �6� � �2�
42. Sample answer: sin(���)2 � cos(���)2 � sin � � cos �
� 0 � (�1)� �1 1
43. s � rv �1170� � �
1170� � �
18�0°�
17 � 10 � v 97.4°�1170� � v
44. Let x � the distance from A to the point beneaththe mountain peak.tan 21°10� � �
570h� x�
h � (570 � x) tan 21°10�
tan 36°40� � �hx
�
h � x tan 36°40�
(570 � x) tan 21°10� � x tan 36°40�570 tan 21°10� � x tan 36°40� � x tan 21°10�570 tan 21°10� � x(tan 36°40� � tan 21°10�)
4. The x-coordinates are the solutions of theequations. Substitute the x-coordinates and seethat the two sides of the equation are equal.
5.
[0, 2�] sc1���4
� by [�3, 3] sc1�15a. The x-intercepts of the graph are the solutions of
the equation sin x � 2 cos x. They are the same.5b. y � tan 0.5x � cos x or y � cos x � tan 0.5x
Page 459 Check for Understanding1. A trigonometric identity is an equation that is
true for all values of the variable for which eachside of the equation is defined. A trigonometricequation that is not an identity is only true forcertain values of the variable.
2. All trigonometric functions are periodic. Addingthe least common multiple of the periods of thefunctions that appear to any solution to theequation will always produce another solution.
3. 45° � 360x° and 135° � 360x°, where x is anyinteger
7-5
4. Each type of equation may require adding,subtracting, multiplying, or dividing each side bythe same number. Quadratic and trigonometricequations can often be solved by factoring. Linearand quadratic equations do not require identities.All linear and quadratic equations can be solvedalgebraically, whereas some trigonometricequations require a graphing calculator. A linearequation has at most one solution. A quadraticequation has at most two solutions. Atrigonometric equation usually has infinitelymany solutions unless the values of the variableare restricted.
5. 2 sin x � 1 � 0 6. 2 cos x � �3� � 0
2 sin x � �1 2 cos x � �3�
sin x � ��12� cos x � �
�23�
�
x � �30° x � 30°
7. sin x cot x � ��23�
�
sin x ��csoins
xx
�� � ��23�
�
cos x � ��23�
�
x � 30° or x � 330°8. cos 2x � sin2 x � 2
2 cos2 x � 1 � (1 � cos2 x) � 22 cos2 x � 1 � �cos2 x � 1
3 cos2 x � 0cos2 x � 0cos x � 0
x � 90° or x � 270°9. 3 tan2 x � 1 � 0
3 tan2 x � 1tan2 x � �
13
�
tan x � ��33�
�
x � ��6
� or x � �56�� or x � �
76�� or x � �
116��
10. 2 sin2 x � 5 sin x � 32 sin2 x � 5 sin x � 3 � 0
(2 sin x � 1)(sin x � 3) � 02 sin x � 1 � 0 or sin x � 3 � 0
sin x � ��12� sin x � 3
x � �76�� or x � �
116�� no solutions
11. sin2 2x � cos2 x � 01 � cos2 2x � cos2 x � 0
1 � (2 cos2 x � 1)2 � cos2 x � 01 � (4 cos4 x � 4 cos2 x � 1) � cos2 x � 0
�4 cos4 x � 5 cos2 x � 0cos2 x(�4 cos2 x � 5) � 0
cos2 x � 0 or �4 cos2 x � 5 � 0cos x � 0 cos2 x � �
54
�
x � ��2
� � �k cos x � ��25�
�
no solutions
12. tan2 x � 2 tan x � 1 � 0(tan x � 1)(tan x � 1) � 0
tan x � 1 � 0tan x � �1
x � �34�� � �k
13. cos2 x � 3 cos x � �2cos2 x � 3 cos x � 2 � 0
(cos x � 1)(cos x � 2) � 0cos x � 1 � 0 or cos x � 2 � 0
cos x � �1 cos x � �2x � (2k � 1)� no solutions
14. sin 2x � cos x � 02 sin x cos x � cos x � 0
cos x (2 sin x � 1) � 0cos x � 0 or 2 sin x � 1 � 0
x � ��2
� � k� sin x � �12
�
x � ��6
� � 2�k
or x � �56�� � 2�k
15. 2 cos v � 1 02 cos v �1
cos v ��12
�
cos v � ��12
� at �23�� and �
43��
�23�� v �
43��
16. W � Fd cos v1500 � 100 � 20 cos v0.75 � cos v
v 41.41°
Pages 459–461 Exercises17. �2� sin x � 1 � 0 18. 2 cos x � 1 � 0
�2� sin x � 1 2 cos x � �1
sin x � ��12�
� cos x � ��12
�
sin x � ��22�
� x � 120°
x � 45°19. sin 2x � 1 � 0
2 sin x cos x � 1 � 0
sin2 x cos2 x � �14�
sin2 x (1 � sin2 x) � �14
�
sin2 x � sin4 x � �14
� � 0
sin4 x � sin2 x � �14
� � 0
�sin2 x � �12
���sin2 x � �12
�� � 0
sin2 x � �12
� � 0
sin2 x � �12
�
sin x � ��12�
� or ��22�
�
x � 45°
Chapter 7 226
20. tan 2x � �3� � 0tan 2x � �3�
�1
2�
ttaannx2 x
� � �3�
2 tan x � �3� (1 � tan2 x)2 tan x � �3� � �3� tan2 x
�3� tan2 x � 2 tan x � �3� � 0(�3� tan x � 1)(tan x � �3�) � 0�3� tan x � 1 � 0 tan x � �3� � 0
tan x � ��13�� tan x � ��3�
tan x � ��33�� x � �60°
x � 30°21. cos2 x � cos x
cos2 x � cos x � 0cos x(cos x � 1) � 0cos x � 0 or cos x � 1 � 0
x � 90° cos x � 1x � 0°
22. sin x � 1 � cos2 xsin x � 1 � 1 � sin2 x
sin2 x � sin x � 2 � 0(sin x � 1)(sin x � 2) � 0sin x � 1 � 0 or sin x � 2 � 0
sin x � 1 sin x � �2x � 90° no solution
23. �2� cos x � 1 � 0�2� cos x � �1
cos x � ���22��
x � 135° or x � 225°24. cos x tan x � �
12
�
cos x �csoins
xx
� � �12
�
sin x � �12
�
x � 30° or x � 150°25. sin x tan x � sin x � 0
sin x (tan x � 1) � 0sin x � 0 or tan x � 1 � 0
x � 0° or x � 180° tan x � 1x � 45° or x � 225°
26. 2 cos2 x � 3 cos x � 2 � 0(2 cos x � 1)(cos x � 2) � 02 cos x � 1 � 0 or cos x � 2 � 0
2 cos x � 1 cos x � �2cos x � �
12
� no solution
x � 60° or x � 300°27. sin 2x � �sin x
2 sin x cos x � –sin x2 sin x cos x � sin x � 0
sin x (2 cos x � 1) � 0sin x � 0 or 2 cos x � 1 � 0
x � 0° or x � 180° 2 cos x � �1
cos x � ��12
�
x � 120°or x � 240°
28. cos (x � 45°) � cos (x � 45°) � �2�cos x cos 45° � sin x sin 45°� cos x cos 45° � sin x sin 45° � �2�
cos x � ��22�
� � sin x � ��22�
�
� cos x � ��22�
� � sin x � ��22�
� � �2�
�2� cos x � �2�cos x � 1
x � 0°29. 2 sin v cos v � �3� sin v � 0
sin v (2 cos v � �3�) � 0sin v � 0 or 2 cos v + �3� � 0
v � 0° or v � 180° 2 cos v � ��3�
cos v � ���23�
�
v � 150°or v � 210°
30. (2 sin x � 1)(2 cos2 x � 1) � 02 sin x � 1 � 0 or 2 cos2 x � 1 � 0
2 sin x � 1 2 cos2 x � 1
sin x � �12
� cos2 x � �12
�
x � ��6
� cos x � ��22�
�
or x � �56�� x � �
�4
� or x � �34��
or x � �54�� or x � �
74��
31. 4 sin2 x � 1 � �4 sin x4 sin2 x � 4 sin x � 1 � 0
(2 sin x � 1)(2 sin x � 1) � 02 sin x � 1 � 0
2 sin x � �1
sin x � ��12
�
x � �76�� or x � �
116��
32. �2� tan x � 2 sin x
�2� �csoins
xx
� � 2 sin x
�2� � 2 cos x��22�
� � cos x
x � ��4
� or x � �74��
�2� tan x � 2 sin x would also be true if both tan xand sin x equal 0. Since tan x � �
csoins
xx
�, tan x equals0 when sin x � 0. Therefore x can also equal 0 and�.
0, ��4
�, �, �74��
33. sin x � cos 2x � 1sin x � 1 � 2 sin2 x � 1
2 sin2 x � sin x � 0sin x(2 sin x � 1) � 0sin x � 0 or 2 sin x � 1 � 0
x � 0 or x � � 2 sin x � �1
sin x � ��12
�
x � �76�� or
x � �11
6��
227 Chapter 7
34. cot2 x � csc x � 1csc2 x � 1 � csc x � 1csc2 x � csc x � 2 � 0
(csc x � 2)(csc x � 1) � 0csc x � 2 � 0 or csc x � 1 � 0
csc x � 2 or csc x � �1
sin x � �12
� sin x � �1
x � ��6
� or x � �56�� x � �
32��
35. sin x � cos x � 0sin x � �cos x
sin2 x � cos2 xsin2 x � cos2 x � 0
sin2 x � 1 � sin2 x � 02 sin2 x � 1 � 0
sin2 x � �12
�
sin x � ��12�
� or ��22�
�
sin x and cos x must be opposites, so x � �34��
or x � �74��.
36. �1 � 3 sin v � cos 2v
�1 � 3 sin v � 1 � 2 sin2 v
2 sin2 v � 3 sin v � 2 � 0(2 sin v � 1)(sin v � 2) � 0
2 sin v � 1 � 0 or sin v � 2 � 02 sin v � �1 sin v � 2
sin v � ��12
� no solution
v � �76�� or v � �
116��
37. sin x � ��12
�
x � �76�� � 2 �k or x � �
116�� � 2�k
38. cos x tan x � 2 cos2 x � �1
cos x �csoins
xx
� � 2 cos2 x � �1
sin x � 2(1 � sin2 x) � �12 sin2 x � sin x � 1 � 0
(2 sin x � 1)(sin x � 1) � 02 sin x � 1 � 0 or sin x � 1 � 0
2 sin x � 1 sin x � �1
sin x � �12
� x � �32�� � 2�k
x � ��6
� � 2�k or x � �56�� � 2�k
39. 3 tan2 x � �3� tan x3 tan2 x � �3� tan x � 0tan x(3 tan x � �3�) � 0tan x � 0 or 3 tan x � �3� � 0
x � �k 3 tan x � �3�
tan x � ��33�
�
x � ��6
� � �k40. 2(1 � sin2 x) � 3 sin x
2 � 2 sin2 x � 3 sin x2 sin2 x � 3 sin x � 2 � 0
(2 sin x � 1)(sin x � 2) � 02 sin x � 1 � 0 or sin x � 2 � 0
2 sin x � 1 sin x � �2sin x � �
12
� no solution
x � ��6
� � 2�k or
x � �56�� � 2�k
41. �cos x �
1sin x� � cos x � sin x
(cos x � sin x)(cos x � sin x) � 1cos2 x � sin2 x � 1
cos2 x � (1 � cos2 x) � 12 cos2 x � 1 � 1
2 cos2 x � 2cos2 x � 1cos x � 1
x � �k42. 2 tan2 x � 3 sec x � 0
2(sec2 x � 1) � 3 sec x � 0(2 sec x � 1)(sec x � 2) � 0
2 sec2 x � 3 sec x � 2 � 02 sec x � 1 � 0 or sec x � 2 � 0
2 sec x � �1 sec x � 2
sec x � ��12
� cos x � �12
�
cos x � �2 x � ��3
� � 2�k or
no solution x � �53�� � 2�k
43. sin x cos x � �12
�
sin2 x cos2 x � �14
�
sin2 x(1 � sin2 x) � �14
�
sin2 x � sin4 x) � �14
�
sin4 x � sin2 x � �14
� � 0
�sin2 x � �12
���sin2 x � �12
�� � 0
sin2 x � �12
� � 0
sin2 x � �12
�
sin x � ��22�
�
x � ��4
� � �k
44. cos2 x � sin2 x � ��23�
�
cos2 x � (1 � cos2 x) � ��23�
�
2 cos2 x � 1 � ��23�
�
2 cos2 x � �2 �
2�3��
cos2 x � �2 �
4�3��
cos x �
x � �1�2� � �k or x � �
1112�� � �k
45. sin4 x � 1 � 0(sin2 x � 1)(sin2 x � 1) � 0sin2 x � 1 � 0 or sin2 x � 1 � 0
sin2 x � 1 sin2 x � �1sin x � 1 no solutions
x � ��2
� � �k46. sec2 x � 2 sec x � 0
sec x(sec x � 2) � 0sec x � 0 or sec x � 2 � 0
cos x � �10
� sec x � �2
no solution cos x � ��12
�
x � �23�� � 2�k or
x � �43�� � 2�k
�2 � ��3����
2
Chapter 7 228
47. sin x � cos x � 1sin2 x � 2 sin x cos x � cos2 x � 1
sin2 x � 2 sin x cos x � 1 � sin2 x � 12 sin x cos x � 0
sin x cos x � 0sin2 x cos2 x � 0
sin2 x (1 � sin2 x) � 0sin2 x � 0 or 1 � sin2 x � 0sin x � 0 sin2 x � 1
x � 2�k sin x � 1x � �
�2
� � 2�k
48. 2 sin x � csc x � 32 sin2 x � 1 � 3 sin x
2 sin2 x � 3 sin x � 1 � 0(2 sin x � 1)(sin x � 1) � 02 sin x � 1 � 0 or sin x � 1 � 0
2 sin x � 1 sin x � 1
sin x � �12
� x � ��2
� � 2�k
x � ��6
� � 2�k or
x � �56�� � 2�k
49. cos v � ���23�
�
cos v � ���23�
� at �56�� and �
76��
�56�� � v � �
76��
50. cos v � �12
� � 0
cos v � �12
�
cos v � �12
� at ��3
� and �53��
0 � v ��3
� or �53�� v 2�
51. �2� sin v � 1 0�2� sin v 1
sin v ��22�
�
sin v � ��22�
� at ��4
� and �34��
0 � v ��4
� or �34�� v 2�
52. 0.4636, 3.6052 53. 0, 1.895554. 0.3218, 3.463355. sin v � �
D�
�
sin v � �5.5
0�
.00130�7
�
sin v 0.0001833333333v 0.01°
56. sin 2x sin x2 sin x cos x sin x
2 sin x cos x – sin x 0sin x(2 cos x � 1) 0
The product on the left side of the inequality isequal to 0 when x is 0, �
�3
�, �, or �53��. For the product
to be negative, one factor must be positive and theother negative. This occurs if �
�3
� x � or �53�� x
2�.
57. R � �vg
2� sin 2v
20 � �195.8
2� sin 2v
0.8711111111 sin 2v
2v 60.5880156 or 2v 119.4119844v 30.29° v 59.71°
58a. n1 sin i � n2 sin r1.00 sin 35° � 2.42 sin r
sin r � �1.00
2s.4in2
35°�
sin r 0.2370150563r 13.71°
58b. Measure the angles of incidence and refractionto determine the index of refraction. If the indexis 2.42, the diamond is genuine.
59. D � 0.5 sin (6.5 x) sin (2500t)0.01 � 0.5 sin (6.5(0.5)) sin (2500t)0.02 � sin 3.25 sin 2500t
�0.1848511958 sin 2500t�0.1859549654 2500tThe first positive angle with sine equivalent to sin (�0.1859549654) is � � 0.1859549654 or3.326477773.
t �3.32
26540707773�
t 0.0013 s
60. a sin(bx � c) � d � d � �a2
�
a sin(bx � c) � �a2
�
sin(bx � c) � �12
�
The period of the function sin(bx � c) is �36
b0°�, so
the given interval consists of � b periods. 360°��36
b0°�
229 Chapter 7
The equation sin (bx � c) � �12
� has two solutionsper period, so the total number of solutions is 2b.
Slope-Intercept Form: y � mx � b, displaysslope and y-interceptPoint-Slope Form: y � y1 � m(x � x1),displays slope and a point on the lineStandard Form: Ax � by � C � 0, displaysno informationNormal Form: x cos f � y sin f � p � 0,displays length of the normal and the anglethe normal makes with the x-axis
33b. p � 1.25, f � 45°x cos (�45°) � y sin (�45°) � 1.25 � 0
��22�
�x � ��22�
�y � 1.25 � 0
�2�x � �2�y � 2.5 � 034a.
f and the supplement of v are complementaryangles of a right triangle, so f � 180° � v � 90°.Simplifying this equation gives v � f � 90°.
34b. tan v. The slope of a line is the tangent of theangle the line makes with the positive x-axis
34c. Since the normal line is perpendicular to �, theslope of the normal line is the negativereciprocal of the slope of �. That is, ��
ta1n v� �
�cot v.34d. The slope of � is the negative reciprocal of the
slope of the normal, or ��ta
1n f� � �cot f.
35a. �A2 ��B2� � �52 � 1�22� or 13�153�x � �
1123�y � �
3193� � 0
�153�x � �
1123�y � 3 � 0
35b. sin f � �1123�, cos f � �
153�; Quadrant I
tan f � or �152�
�1123�
—
�153�
36a.
The angles of the quadrilateral are 180° � a,90°, f2 � f1, and 90°. Then 180° � a � 90° �f2 � f1 � 90° � 360°, which simplifies to f2 � f1 � a. If the lines intersect so that a is aninterior angle of the quadrilateral, the equationworks out to be f2 � 180° � f1 � a.
36b. tan f2 � tan(f1 � a)
� �1ta�
ntf
a1n
�
f1
tatanna
a�
If the lines intersect so that a is an interiorangle of the quadrilateral, the equation works out to be tan f2 � �
1ta�
ntf
a1n
�
f1
tatanna
a�.
37. 5x � y � 155x � y � 15 � 0�A2 ��B2� � �52 � (��1)2� or �26�
13.85564879 � 500 6927.824395; $6927.8238. 2 cos2 x � 7 cos x � 4 � 0
(2 cos x � 1)(cos x � 4) � 02 cos x � 1 � 0 or cos x � 4 � 0
2 cos x � 1 cos x � �4
cos x � �12
� no solution
x � ��3
� or x � �53��
39. sin x � �1 � co�s2� x sin y � �1 � co�s2 y�
Chapter 7 234
y
xO� �
�
f 67°f � 90° � 67° � 90° or 157°x cos 157° � y sin 157° � 3 � 0
��1123�x � �
153�y � 3 � 0
35c. See students’ work.35d. The line with normal form x cos f � y sin f �
p � 0 makes an angle of f with the positive x-axis and has a normal of length p. The graphof Armando’s equation is a line whose normalmakes an angle of f � d with the x-axis and alsohas length p. Therefore, the graph of Armando’sequation is the graph of the original line rotatedd° counterclockwise about the origin. Armando iscorrect. See students’ graphs.
y
xO
� x � y � 3 � 01213
513
y
xO
�
�1
�2
� �1 � ��16���2� � �1 � ��
23���2�
� ��3366�� or �
�635�� � ��
59
�� or ��35�
�
sin(x � y) � sin x cos y � cos x sin y
� ���635�����
23
�� � ��16
�����35�
��� �
2�35�18
� �5��
O
y
x
(2, 6)
(2, 3)
(5, 3)
x � y � 8
y � 3
x � 2
40. A � 1, �24�� � �
�2
� or 90°
41. r � �d2
�
r � �13
2.4� or 6.7
x2 � 6.72 � 6.72 � 2(6.7)(6.7) cos 26°20�x2 9.316604344x 3.05 cm
V(0.5) is closer to zero, so x � 0.5.4 � x � 4 � 0.5 or 4.56 � x � 6 � 0.5 or 6.52 � x � 2 � 0.5 or 2.54.5 in. by 6.5 in. by 2.5 in.
44.
f(x, y) � 3x � y � 4f(2, 3) � 3(2) � 3 � 4 or 7f(2, 6) � 3(2) � 6 � 4 or 4f(5, 3) � 3(5) � 3 � 4 or 1616, 4
45. � � � �15
�� ��2�1
34
�2�1
34
1
�1 24 3
235 Chapter 7
y
90˚45˚
�1
1
O
y � sin 4
x V(x)0.4 �4.4160.5 1.125
� � � � � � � �015
xy
23
�1�4
�15
�� � � � � � � � � �15
�� � � � �015
�2�1
34
xy
23
�1�4
�2�1
34
� � � �15
�� ��30�15
xy
� � � � �(�6, �3)
46. The value of 2a � b cannot be determined fromthe given information. The correct choice is E.
Distance From a Point to a Line
Page 474 Check for Understanding1. The distance from a point to a line is the distance
from that point to the closest point on the line.2. The sign should be chosen opposite the sign of C
where Ax � By � C � 0 is the standard form ofthe equation of the line.
3. In the figure, P and Q are any points on the lines.The right triangles are congruent by AAS. Thecorresponding congruent sides of the trianglesshow that the same distance is always obtainedbetween the two lines.
4. The formula is valid in either case. Examples willvary. For a vertical line, x � a, the formulasubtracts a from the x-coordinate of the point. Fora horizontal line, y � b, the formula subtracts bfrom the y-coordinate of the point.
5. 2x � 3y � �2 → 2x � 3y � 2 � 0
d �Ax1 � By1 � C��
��A2 ��B2�
7-7
�6�3
xy
d �2(1) � (�3)(2) � 2���
��22 � (��3)2�
d � ���
�213�
� or �2�
1313�
�
6. 6x � y � �3 → 6x � y � 3 � 0
d �Ax1 � By1 � C��
��A2 ��B2�
P
Q
d �6(�2) � (�1)(3) � 3���
��62 � (��1)2�
d � ��
�
�1327�
� or �12�
3737�
�
7. 3x � 5y � 1 When x � 2, y � 1. Use (2, 1).3x � 5y � �3 → 3x � 5y � 3 � 0
30. The radius of the circle is �[(�5) �� (�2)]2�� (6 �� 2)2�or 5. Now find the distance from the center of thecircle to the line.
d �Ax1 � By1 � C��
�A2 ��B2�
d �
d � ��
�
6153
�
d � 5Since the distance from the center of the circle tothe line is the same as the radius of the circle, theline can only intersect the circle in one point. Thatis, the line is tangent to the circle.
39. 2x � y � z � �9 2x � y � z � �92(�x � 3y � 2z) � 2(10)
→�2x � 6y � 4z � 20
7y � 5z � 11x � 2y � z � �7
�x � 3y � 2z � 10y � 2z � 3
�5(y � z) � �5(3)→
�5y � 5z � �157y � 5z � 11 7y � 5z � 11
2y � �4y � �2
y � z � 3 x � 2y � z � �7�2 � z � 3 x � 2(�2) � (�5) � �7
�5 � z x � �6(�6, �2, �5)
40. square: A � s2 triangle: A � �12
�bh
16 � s2 6 � �12
�(4)h
4 � s 3 � hAE � s � hAE � 4 � 3 or 7EF � AEEF � 7The correct choice is C.
Chapter 7 238
a1 �
a1 � ��354�
m2 � ��
�
13�
�
(�43)
� or ��72
�
y � 4 � ��72
�(x � (�3))
7x � 2y � 13 � 0
a2 �Ax1 � By1 � C��
��A2 ��B2�
3(�1) � (�4)(�3) � 25���
��32 � (��4)2�
a2 �
a2 � ���
3453�
� or ��34�
5353�
�
m3 � �71
�
�
((�
�
31))
� or 5
y � 7 � 5(x � 1)5x � y � 2 � 0
a3 �Ax1 � By1 � C��
��A2 ��B2�
7(1) � 2(7) � 13��
��72 � 2�2�
a3 �
a3 � ��
�
�1276�
� or �17�
2626�
�
�354�, �
34�53
53��, �
17�26
26��
32.
The standard form of the equation of the linethrough (0, 0) and (4, 12) is 3x � y � 0. Thestandard form of the equation of the line through(4, 12) and (10, 0) is 2x � y � 20 � 0. Thestandard form for the x-axis is y � 0. To find thebisector of the angle at the origin, set �3
�x �
10�y
� � yand solve to obtain y � �
1 � �3
10��x. To find the
bisector of the angle of the triangle at (10, 0), set
�2x �
�y
5�� 20� � �y and solve to obtain 2x � (1 � �5�)y
� 20 � 0. The intersection of these two bisectorsis the center of the inscribed circle. To solve the
system of equations, substitute y � �1 � �
310�
�x into
the equation of the other bisector and solve for x to
get x � . Then y � �20(1 � �10�)
���5 � 3 �5� � 2�10�
20(1 � �10�)���5 � 3 �5� � 2�10�
5(�3) � (�1)(4) � 2���
��52 � (��1)2�
y
xO
2468
10
2 4 6 8
�1 � �
310�
� � . This y-coordinate is the
inradius of the triangle. The approximate value is3.33.
60���5 � 3�5� � 2�10�
y
120˚ 300˚ 480˚�1
1
Oy � csc ( � 60˚)
Chapter 7 Study Guide and Assessment
Page 477 Understanding and Using theVocabulary
1. b 2. g 3. d 4. a5. i 6. j 7. h 8. f9. e 10. c
Pages 478–480 Skills and Concepts11. csc v � �
si1n v�
�
� 2
12. tan2 v � 1 � sec2 v
42 � 1 � sec2 v
17 � sec2 v
�17� � sec v13. sin v � �
cs1c v� sin2 v � cos2 v � 1
� ��35
��2 � cos2 v � 11��53
�
1��12
�
19. �sin4
sxin�
2cxos4 x
� � 1 � cot2 x
� 1 � cot2 x
�sin2
sx
in�
2cx
os2 x� � 1 � cot2 x
1 � �csoins
2
2
x
x� � 1 � cot2 x
1 � cot2 x � 1 � cot2 x20. cos 195° � cos (150° � 45°)
� cos 150° cos 45° � sin 150° sin 45°
� ����23�
�����22�
�� � �12
� � ��22�
�
� ���6�
4� �2��
21. cos 15° � cos (45° � 30°)� cos 45° cos 30° � sin 45° sin 30°
� ��22�
� � ��23�
� � ��22�
� � �12
�
� ��6� �
4�2�
�
22. sin ���1172��� � �sin �
1172��
� �sin ���4
� � �76���
� ��sin ��4
� cos �76�� � cos �
�4
� sin �76���
� �����22�
������23�
�� � ���22�
�����12
���� �����6�
4� �2���
� ��6� �
4�2�
�
23. tan �1112�� � tan ��
23�� � �
�4
��
�tan �2
3�� � tan �
�4
�
���
1 � tan �23�� tan �
�4
�
(sin2 x � cos2 x)(sin2 x � cos2 x)����
sin2 x
239 Chapter 7
� �35
�
14. sec v � �co
1s v� tan2 v � 1 � sec2 v
� tan2 v � 1 � ��54
��21��45
�
� �1 �
��(�
3��
�
3�)1(1)
�
� �11�
�
��
3�3�
�
� �4 �
�
22�3�
� or �2 � �3�
24. cos x � �1 � si�n2 x� sin y � �1 � co�s2 x�
� �1 � ��2�75��2� � �1 � ��
23���2�
� ��567265
�� or �2245� � ��
59
�� or ��35�
�
cos (x � y) � cos x cos y � sin x sin y
� ��2245����
23
�� � ��275�����3
5���
� �48 �
775
�5��
cos2 v � �1265�
cos v � �45
�
� �54
� tan2 v � �196�
tan v � �34
�
15. csc x � cos2 x csc x � �sin
1x
� � (1 � sin2 x)��sin1
x��
� �sin
1x
� � �sin
1x
� � sin x
� sin x
16. cos2 x � tan2 x cos2 x � 1
cos2 x � �csoins
2
2x
x� cos2 x � 1
cos2 x � sin2 x � 11 � 1
17. �11
�
�
ccooss
v
v� � (csc v � cot v)2
�11
�
�
ccooss
v
v� � ��si
1n v� � �
csoins
v
v��2
�11
�
�
ccooss
v
v� � �
(1 �
sinc2os
v
v)2�
�11
�
�
ccooss
v
v� � �
(11
�
�
ccooss2
v
v
)2�
�11
�
�
ccooss
v
v� �
�11
�
�
ccooss
v
v� � �
11
�
�
ccooss
v
v�
18. �se
tcav
n�
v
1� � �
setcav
n�
v
1�
�se
tcav
n�
v
1� � �
tansev
c(s2ev
c�
v �
11)
�
�se
tcav
n�
v
1� � �
tan v
t(asenc2v
v
� 1)�
�se
tcav
n�
v
1� � �
setcav
n�
v
1�
(1 � cos v)2���(1 � cos v)(1 � cos v)
25. cos y � �se
1c y� sin y � �1 � co�s2 y�
� � �1 � ��23���2�
� �23
� � ��59
�� or ��35�
�
tan y � �csoins
yy
�
� or ��25�
�
��35��
��23
�
1��32
�
� ��274�
33. sin 4v � sin 2(2v)� 2 sin 2v cos 2v
� 2��2245�����
275��
� ��363265
�
34. tan x � 1 � sec x(tan x � 1)2 � sec2 x
tan2 x � 2 tan x � 1 � tan2 x � 12 tan x � 0
tan x � 0x � 0°
35. sin2 x � cos 2x � cos x � 01 � cos2 x � 2 cos2 x � 1 � cos x � 0
cos2 x � cos x � 0cos x (cos x � 1) � 0
cos x � 0 or cos x � 1 � 0x � 90° or x � 270° cos x � 1
x � 0°
Chapter 7 240
tan (x � y) � �1ta�
ntxan
�
xtatannyy
�
��54
� � ��25�
�
��
1 � ��54
�����25�
��
��5 � 2
4�5�
�
��
�8 � 5
8�5�
�
� �180
�
�
54��
5�5�
�
� �180
�
�
6812�5�
� or ��180 �
6125�5��
26. cos 75° � cos �15
20°�
� ��1 � c�o2s 150�°�� (Quadrant I)
� ��1 � ����23�
����2
��2 � ��3����
2
27. sin �78�� � sin
�74��
�2
� �� (Quadrant II)1 � cos �
74��
��2
� ��� �
�2 �
2��2���
28. sin 22.5° � sin �425°�
� ��1 � c2�os 45°�� (Quadrant I)
1 � ��22�
�
�2
� ��� �
�2 �
2��2��
�
1 � ��22�
�
�2
29. tan �1�2� � tan
��
6�
�2
� �� (Quadrant I)1 � cos �
�
6�
��1 � cos �
�
6�
���1 � ��23�
�
�
1 � ��23�
�
� ��22
�
�
���3�
3���
� ���(2 � �3�)(2 � �3�)���(2 � �3�)(2 � �3�)
� ��� 2 � �3�
30. sin2 v � cos2 v � 1
sin2 v � ��35
��2 � 1
sin2 v � �1265�
sin v � �45
�
sin 2v � 2 sin v cos v
� 2��45
����35
��� �
2245�
31. cos 2v � 2 cos2 v � 1
� 2��35
��2 � 1
� ��275�
32. tan v � �csoins
v
v� tan 2v � �
12�
ttaannv2 v
�
� or �43
� �2��
43
���
1 � ��43
��2�45
�
�
�35
�
(2 � �3�)2��
4 � 3
36. cos 2x � sin x � 11 � 2 sin2 x � sin x � 1
2 sin2 x � sin x � 0sin x (2 sin x � 1) � 0
sin x � 0 or 2 sin x � 1 � 0x � 0° or x � 180° sin x � �
12
�
x � 30° or x � 150°
37. sin x tan x � ��22�� tan x � 0
tan x �sin x � ��22��� � 0
tan x � 0 or sin x � ��22�� � 0
x � �ksin x � �
�22��
x � ��4
� � 2�k or �34�� � 2�k
38. sin 2x � sin x � 02 sin x cos x � sin x � 0
sin x (2 cosx � 1) � 0sin x � 0 or 2 cos x � 1 � 0
x � �k cos x � ��12
�
x � �23�� � 2�k
or x � �43�� � 2�k
39. cos2 x � 2 � cos xcos2 x � cos x � 2 � 0
(cos x � 1)(cos x � 2) � 0cos x � 1 � 0 or cos x � 2 � 0
3. One way to solve this problem is to label the threeinterior angles of the triangle, a, b, and c. Thenwrite equations using these angles and theexterior angles.a � b � c � 180x � a � 180y � b � 180z � c � 180Add the last three equations.x � a � y � b � z � c � 180 � 180 � 180x � y � z � a � b � c � 180 � 180 � 180Replace a � b � c with 180.x � y � z � 180 � 180 � 180 � 180
x � y � z � 180 � 180 or 360The correct choice is D.
4. Since x � y � 90°, x � 90° � y.Then sin x � sin (90° � y).sin (90° � y) � cos y�csoins
xy
� � �sin(
c9o0s°y� y)
� � �ccooss
yy
� � 1
The correct choice is D.Another solution is to draw a diagram and noticethat sin x � �
bc
� and cos y � �bc
�.
�csoins
xy
� � � 1
5. In order to represent the slopes, you need thecoordinates of point A. Since A lies on the y-axis,let its coordinates be (0, y). Then calculate the twoslopes. The slope of A�B� is �0
y�
�
(�03)
� � �3y
�. The slope of A�D�is �0
y �
�
03
� � ��3y
�. The sum of the slopes is�3y
� � �3y
� � 0.The correct choice is B.
6. Since PQRS is a rectangle, its angles measure 90°.The triangles that include the marked angles areright triangles. Write an equation for the measureof ∠ PSR, using expressions for the unmarkedangles on either side of the angle of x°.
90 � (90 � a) � x � (90 � b)0 � 90 � a � b � x
a � b � 90 � xThe correct choice is A.
�bc
�
—
�bc
�
243 Chapter 7
� ���2 �
2�3��
��
�2 �
2�3��
� ���� ��
(24
�
�
3�3)2
��
(2 � �3�)(2 � �3�)���(2 � �3�)(2 � �3�)
� 2 � �3�
2. Sample answer: sin x tan x � �1 �
cocsoxs2 x
�
sin x tan x � �1 �
cocsoxs2 x
�
sin x �csoins
xx
� � �scions
2
xx
�
�scions
2
xx
� � �scions
2
xx
�
SAT & ACT Preparation
Page 483 SAT and ACT Practice1. The problem states that the measure of ∠ A is 80°.
Since the measure of ∠ B is half the measure of∠ A, the measure of ∠ B must be 40°. Because ∠ A,∠ B, and ∠ C are interior angles of a triangle, thesum of their measures must equal 180°.m∠ A � m∠ B � m∠ C � 180
80 � 40 � m∠ C � 180120 � m∠ C � 180
m∠ C � 60The correct choice is B.
2. To find the point of intersection, you need to solvea system of two linear equations. Substitution orelimination by addition or subtraction can be usedto solve a system of equations. To solve thissystem of equations, use substitution. Substitute2x � 2 for y in the second equation.
7x � 3y � 117x � 3(2x � 2) � 11
7x � 6x � 6 � 11x � 5
Then use this value for x to calculate the value for y.y � 2x � 2y � 2(5) � 2 or 8The point of intersection is (5, 8). The correctchoice is A.
a
b
cx˚
y˚
7. Simplify the fraction. One method is to multiply
both numerator and denominator by �y
y
2
2�.
� �y
y
2
2� � �y � �
1y
�
��1 � �
2y
� � �y
12�
y � �1y
�
��1 � �
2y
� � �y
12�
9. Since the volume V varies directly with thetemperature T, the volume and temperaturesatisfy the equation V � kT, where k is a constant.When V � 12, T � 60. So 12 � 60k, or k � �
15
�. The relationship is V � �
15
�T.To find the volume when the temperature is 70°,substitute 70 for T in the equation V � �
15
�T.V � �
15
�(70) or 14. The volume of the balloon is 14 in3.The correct choice is C.
10. Two sides have the same length. The lengths of allsides are integers. The third side is 13. FromTriangle Inequality, the sum of the lengths of anytwo sides must be greater than the length of thethird side. Let s be the length of the other twosides. Write and solve an inequality.2s � 13s � 6.5
The length of the sides must be greater than 6.5.But the length of the sides must be an integer.The smallest integer greater than 6.5 is 7. Theanswer is 7. If you answered 6.5, you did not findan integer. If you answered 6, you found a numberthat is less than 6.5.
Chapter 7 244
� �y2 �
y3
2�
y
y
� 1�
� �(y
y�
(y1
2
)(�
y �
1)1)
�
� �y((yy
�
�
11))((yy
�
�
11))
�
� �yy
2
�
�
1y
�
Another method is to write both the numeratorand denominator as fractions, and then simplify.
�
y����
2
y���� 1�
——
�y2 �
y22y � 1�
y � �1y
�
��1 � �
2y
� � �y
12�
� �y2
y� 1���y2 �
y2
2
y � 1��
� �y((yy
�
�
11))((yy
�
�
11))
�
� �yy
2
�
�
1y
�
The correct choice is A.8. Since the triangles are similar, use a proportion
with corresponding sides of the two triangles.�BAC