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July 2011
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Trigonometric Equations and Identities Notes These notes belong to_________________
Date Topic Notes Questions 1. Solve by Calculator
2. Exact Solutions
3.
4. Trig Identities introduction
5. Pythagorean Identities
6. Pythagorean Identities
7. Addition Identities
8. Double Angle Identities
9. Double Angle Identities
10. Review
11. Review
12. TEST
Formula sheet is required for this chapter Questions that I find challenging
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Solving by graphing and by Graphing Calculator
Solve.
1
π 2π
1. Graph 5.0sin =x • Y=sinx • Y=0.5
2. What is the period? 3. How many solutions are there between
π20 ≤≤ x ?
-1
1
π 2π
4. Graph 5.02sin =x • Y=sin2x • Y=0.5
5. What is the period? 6. How many solutions are there between
π20 ≤≤ x ?
-1
1
π 2π
7. Graph 5.04sin =x • Y=sin4x • Y=0.5
8. What is the period? 9. How many solutions are there between
π20 ≤≤ x ?
-1
10. How many solutions would
you expect for sin5x = 0.5 over 0 ≤ x ≤ 2π
11. How many solutions would you expect for sinbx = 0.7 over 0 ≤ x ≤ 2π
12. How many solutions would you
expect for sin 1
2x = 0.5 over
0 ≤ x ≤ 2π
Check your answers using your graphing calculator
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Determine the number of solutions for the following between. π20 ≤≤ x
13. 4.02cos =x
14. 4.03cos =x 15. 4.0cos =Bx 16. 4.0tan =x 17. 4.02tan =x
Challenge#1: Solve sinx = 0.5 -π π 2π 3π 4π
18. Graph y = sinx &
y = 0.5 above.
19. How many solutions are there between 0 ≤ x ≤ 2π ?
20. How many solutions are there over the set of real numbers?
21. How often do solutions happen? Find the pattern.
22. Solve sinx = 0.5 between 0 ≤ x ≤ 2π .
23. State the period of
y = sinx .
24. Solve sinx = 0.5 over the set of all real numbers.
This solution is called the general solution.
25. What equations would you graph to see the solutions for cos2x = 0.5?
26. What equations would you graph to see the solutions for sinx = x2 ?
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General Solution Is the solution over the set of real numbers. The general solution can be found by adding multiples of the period.( i.e. 2� or �…).
i.e. sin x =0.5 x = π
6+ 2πn where n is an integer.
Challenge #2: Solve cos(x + π) = 1
2 using your graphing calculator.
What equations would you type into your graphing calcualtor? State the period:
State intelligent window setting: State the solution over 0 ≤ x ≤ 2π . State the general solution.
27. Solve: 21)cos( =+πx
over the set of real numbers.
28. Solve: 21)sin( =+ πx over
the set of real numbers
29. Solve: xTanx log22 =+ over 0 ≤ x ≤ 2π .
A possible solution strategy. Graphing Calculator.
y = cos(x + π)
y = 12
Set your GC Window
X min = 0X max = 2π
Find intersection point(s)
x = 2.094x = 4.189
General Solution*
x = 2.094 + 2πnx = 4.189 + 2πn
Challenge #3: Given
sin2 x = sinx( ) sinx( ) , graph y = sin2 x on your graphing calculator and state
the period.
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Challenge #4: Solve: 2sin2 x + sinx = 2 over the set of real numbers.
Challenge #5: 2sinx = x over the set of real numbers
30. Solve: 2sin2 x + sinx = 2 31. Solve: 2cos2 x + cos x = 2 32. Solve: sin2 x + tanx = 2
over the set of real numbers
Over 0 ≤ x ≤ 2π Over 0 ≤ x ≤ 2π
Graphing Calc
y = 2(sinx)2 + sinxy = 2
Set your GC Window
X min = 0X max = 2π
Find intersection point(s)
x = 0.8959x = 2.2457
General Solution
x = 0.8959 + 2πnx = 2.2457 + 2πn
Solve over the set of real numbers 33. Solve. 2sinx = x
34. Solve. 2cos x = x2 35. Solve.
cos π
2(x + 1) = cos π
2x
Graphing Calc
y = 2sinxy = x
Set your GC Window
X min = −10X max = 10
Since we may not be exactly sure what this one will do, set window to default.
Find intersection point(s)
x = −1.8955x = 0x = 1.8955
There is no general solution since the two functions do not have repeat intersections.
Did you get a straight line? See the answer key.
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Finding the Exact Answer for Trig Equations Put down your calculator! All you need is what you already know about trigonometry.
Use the sine or cosine waves or the unit circle Special Triangles Sine Wave
The Unit Circle
Cosine Wave
30º,60º,90º
45º,45º,90º
0° 90° 180° 270° 30° 60° 45°
45°
Sin 0 1 0 -1 Sin 21
23
Sin 2
1
2
1
Cos 1 0 -1 0 Cos 23 2
1 Cos 2
1
1
2
Tan 0 Ø 0 Ø Tan 31 3 Tan 1 1
Put down you calculator and find the exact answer. Challenge #6: Find the exact answer to sinx = 0.5 for 0 ≤ x ≤ 2π .
Challenge #7: Find the exact answer to 5sinx + 8 = 3sinx + 9 for 0 ≤ x ≤ 2π .
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36. Find the exact answer to
sinx = 1
2 for 0 ≤ x ≤ 2π .
Below is one of many ways of solving trig equations. a) Special triangle
sin x =
1
2=
O
H
b) Reference angle
30°
37. Find the exact answer to
cos x = 3
2 for 0 ≤ x ≤ 2π .
38. Find the exact answer to
tanx = 3 for 0 ≤ x ≤ 2π .
c) Quadrants
Sine is (+) in Q1 and Q2.
d) Solution °° 150&30
The answer is asked in radians so the answer is written in radians.
65&
6ππ
611&
6ππ
39. Find the exact answer to
secx = 2 for 0 ≤ x ≤ 2π .
40. Find the exact answer to 3cot −=x for
0 ≤ x ≤ 2π .
41. Find the exact answer to 3csc37csc5 +=+ xx for
0 ≤ x ≤ 2π .
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Reminders Solve. 2x2 = −2x2 + 1
Solve. 2x2 + x = 0
Solve. 2x2 − 5x − 3 = 0
x=±0.5 x=0 & x=-0.5 x=-0.5 & x=3
Challenge #8: Find the exact answer to
2sin2 x = −2sin2 x + 1 for π20 ≤≤ x .
Challenge #9: Find the exact answer to 2cos2 x − 5cos x − 3 = 0 for 0 ≤ x ≤ 2π .
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Solve. 42. Find the exact answer to
2sin2 x = −2sin2 x + 1 for π20 ≤≤ x
Rearrange 4sin2 x = 1
Isolate sin2 x = 1
4
Square root both sides
sinx = ± 1
4= ± 1
2
Use special triangle
Reference Angle
°= 30x6
π=x
Determine quadrants by signs
Sine is (+) in Q1,Q2
& (-) in Q3,A4
x = 30°,150°,210°,330°
6
11,
6
7,
6
5,
6
ππππ=x
43. Find the exact answer. 0sinsin2 2 =+ xx for
0 ≤ x ≤ 2π .
44. Find the exact answer.
6tan2 x = 3tan2 x + 1 for 0 ≤ x ≤ 2π .
45. Find the exact answer to
2cos2 x − 5cos x − 3 = 0 for 0 ≤ x ≤ 2π .
Factor 0)3)(cos1cos2( =−+ xx
Solve for x
2cos x + 1 = 0 or cos x − 3 = 0
cos x = − 1
2 or cos x = 3
Use Special Triangle
Reference Angle
x = 60° x = π
3
Determine Quadrants by Signs
Cosine is (-) in Q2,Q3
NO SOLUTION CosX=3 means that the adjacent side is bigger than the hypotenuse. Impossible! You could also check on the unit circle or use your graphing calculator to check.
x = 120°,240°
Remember to answer in Radians
x = 2π
3, 4π
3
46. Find the exact answer to
2sin2 x − 11sinx + 5 = 0 for 0 ≤ x ≤ 2π .
47. Find the exact answer to
4cos3 x + 8cos2 x − 5cos x = 0 for 0 ≤ x ≤ 2π .
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Challenge #10: Find the two smallest exact answers to212sin =x for 0 ≤ x ≤ 2π .
Solve for x where 0 ≤ x ≤ 2π . 48. Find the two smallest
exact answers
to212sin =x
Let 2x=A
21sin2sin == Ax
By calculator of special triangle
A = 30°,150° Since 2x=A
x = 15°,75°
x = π
12, 5π12
49. Find the two smallest exact answers to
213sin =x
50. Find the two smallest exact answers to
219sin =x
51. Find the two smallest exact answers to
215cos =x
Solve for π20 ≤≤ x . 52. Solve xx cossin =
53. Solve 0sin =x
54. Solve 0cos =x 55. Solve 2sin =x
Challenge #11: Find the exact answer to sin2 x = sinx cos x for 0 ≤ x ≤ 2π .
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Solve for x where 0 ≤ x ≤ 2π 56. Find the exact answer to
sin2 x = sinx cos x for 0 ≤ x ≤ 2π .
Rearrange
sin2 x − sinx cos x = 0 Factor
sinx(sinx − cos x) = 0
Solve for x
sinx = 0 or sinx − cos x = 0
sinx = 0 or sinx = cos x Use Sine wave
Sinx=0
Use Special Triangle
Sinx=cosx
Reference Angle
x = 0°,180°
x = 0,π
Reference Angle
x = 45°
x = π
4
Determine Quadrants by Signs
57. Find the exact answer to xxxx cossincoscos2 22 +=
for 0 ≤ x ≤ 2π .
58. Find the exact answer to 0sincos 22 =− xx for
0 ≤ x ≤ 2π .
x = 0°, 45°,225°,180° Remember to answer in Radians
x = 0, π
4, 5π
4,π
Challenge #12: Find secx if the terminal arm of angle x passes through (-2,5).
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Find the trigonometric ratio. 59. Find secx if the
terminal arm of angle x passes through (-2,5).
60. Find sinx if the terminal arm of angle x passes through (-4,5).
61. Find secx if the terminal arm of angle x passes through (-4,-2).
62. Find cos x if the terminal arm of angle x passes through (m,n). m>0 & n>0
Solution: Adjacent side=-2 Opposite side =5 Hypotenuse=
29
5)2( 22
=
+−=
h
h
Remember cos x = ADJ
HYP
So
secx = HYP
ADJ= − 29
2
The next section requires that you are a master of manipulating fractions. 63. Simplify
72+ 6
7
64. Simplify
72− 6
7 65. Simplify
72× 6
7 66. Simplify
72÷ 6
7
Challenge #13: Show that
a + 12
a − 12
=
2a + 12a − 1
. There is no answer in the back. Check with a
peer.
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Simplifying Trigonometric Fractions Simplify.
67. Simplify. c × a
b× 2
a× b
c=
68. Simplify. =+cb
ab
69. Simplify. =⎟⎠⎞⎜
⎝⎛
cab
70. Simplify. =⎟⎠⎞⎜
⎝⎛
acab
71. Simplify. =−
+
a
a11
11 72. Simplify. =
−
+
baba
1
1
Complete the fractions using sinx , cos x and one.
73. tanx =
74. secx =
75. cot x =
76. cscx =
77. sinx × cos x
sinx× 2sinx
cos x=
78. =xx
coscos2
79. =⎟⎠⎞⎜
⎝⎛
xxx
cossincos
80. =⎟⎠⎞⎜
⎝⎛
x
xx
cos2
cossin
81.
1 + 1sinx
1 − 1sinx
=
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Simplify by writing each expression as a single trigonometric ratio. 82. tanx cos x =
83. secx cos x = 84. tanx cscx =
85. cot x secx =
86. cos2 x secx = 87.
1 + cos xsinx
1 + sinxcos x
=
88.
sinxtanx
+ cos x
89.
cot x sec2 xcsc2 x
90.
tanx csc2 xsec2 x
Challenge #14: Show that cos x cscx tanx = 1 .
cos x cscx tanx = 1 is a trigonometric identity.
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What is an identity? • An equation is an identity if the left side exactly equals the right side • An equation that is satisfied for all values of the variable for which both sides of the equation are
defined.
• SIMPLE VERSION RIGHTsideLEFTside =
Prove each identity. 91. 1tancsccos =xxx
xx
xx
cossin
sin1cos
Reduce
=1
1=
Left =Right Therefore they are identities
92. xxx seccotcsc = 93. x
xx sin
sectan =
There are many ways to prove a trigonometric identity. Follow the directions below to see how factoring, distributing, combining fractions and multiplying by the conjugate may be helpful in your solution. 94. Factoring is sometimes helpful. 95. Distributing is sometimes helpful.
sec2 x − sec2 x sin2 x
Try the following: • Factor
• Pythagorean identity
• Multiply
• Simplify
1sin2 x
cos2 x + sin2 x( ) Try the following: • Distribute
• Simplify
• Pythagorean identity
96. Combining Fractions is sometimes helpful. 97. Multiplying by the conjugate is sometimes helpful.
1sin2 x
− 1 Try the following: • Common
denominators
• Combine
• Pythagorean identity
• Simplify
sin2 x1 − cos x
Try the following: • Multiply N & D by
the conjugate (1+cosx)
• Simplify the bottom
• Pythagorean identity
• Simplify
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Proving Identities can be done in many ways. • The following is a list of steps that may be helpful. Try changing xtan
xx
cossin , xcot
xx
sincos , xsec
xcos1 , xcsc
xsin1
Use the formula sheet. Use your formula sheet. One of these may be helpful.
Factoring sin2 x − sinx cos x = sinx sinx − cos x( )
Distribution
1cos x
cos xsinx
+ 1⎛
⎝⎜
⎞
⎠⎟ = cos x
cos x sinx+ 1
cos x
Combine Fractions
1cos x
+ 1sinx
= sinx + cos xcos x sinx
Conjugate 1 − cos x( ) 1 + cos x( ) = 1 − cos2 x = sin2 x
Prove each identity. 98. xxx csc1)sin1(csc +=+
99. xx
xx tan1cos
cossin +=+ 100.
secx(1 + cot x) = sinx + cos x
sinx cos x
Challenge #15:
The Pythagorean Theorem. Complete the following x
2 + y2 = ____
The Pythagorean Identity. Determine the following trig ratios:
sinθ = cosθ = Fill in the blanks.
x2 + y2 = ________+ ________ = ____
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The Pythagorean Identities Notice the relationship between cosθ, sinθ and 1 in the unit circle.
Remember: xxhypadj
===1
cosθ
x = cosθ
yy
hypopp
===1
sinθ
y = sinθ
Remember Pythagoras c2 = a2 + b2
1 = x2 + y2 or 1 = cos2θ + sin2θ
The Pythagorean identity is more commonly written
sin2θ +cos2θ = 1
There are 2 other Pythagorean Identities that can be created from 1cossin 22 =+ xx 1cossin 22 =+ xx
Divide both sides by x2sin .
xxx
xx
22
2
2
2
sin1
sincos
sinsin =+
Simply using Reciprocal identities
1+cot2x = csc2x
1cossin 22 =+ xx Divide both sides by x2cos .
xxx
xx
22
2
2
2
cos1
coscos
cossin =+
Simply using Reciprocal identities
tan2 x +1= sec2 x
Each of the following simplify to a whole number or a single trigonometric term.
101.
sinxcos x
=
102. =− x2sin1 103. sin2 x − 1 = 104. 2 − 2sin2 x =
105. =− x2csc1
106. =++ 3cossin 22 xx 107. =++ 32cos52sin5 xx 108. sin2 x + cos2 x + tan2 x =
109. 5 − 5cos2 x =
110. 14 + 14cot2 x = 111. 1 − sinx( ) 1 + sinx( ) = 112.
1 − cos x( ) 1 + cos x( ) =
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Pythagorean Identities 1cossin 22 =+ xx xx 22 sec1tan =+ xx 22 csccot1 =+
Prove the identities. (The answer is the process. It needs to neat and easy to follow.)
113. xxx 222 cotsinsin1 =−
114. secx sin2 x + cos x = secx
115. tanx + cot x = cscx secx
116. (cscx + 1)(cscx − 1) = cot2 x
117. x
xx 2
22
tan1tansin+
=
118.
cos3 xsinx
= cot x − cos x sinx
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Prove the identities. (The answer is the process. It needs to neat and easy to follow.) 119. x
xx sin
csc1sin1 =
++
120. sin4 x − cos4 x = sin2 x − cos2 x
121. xxx
xx csccossin1cotcos =
++
122.
1 + tanx1 + cot x
= 1 − tanxcot x − 1
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Prove the identities. (The answer is the process. It needs to neat and easy to follow.)
123. xx
xx
sincos1
cos1sin −=+
124. xxx
x tansecsin1
cos −=+
125. xxx
2sec2sin11
sin11 =
−+
+
126.
11 + cos x
+ 11 − cos x
= 2csc2 x
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Prove the identities. (The answer is the process. It needs to neat and easy to follow.)
127.
cos x1 − cos x
− cos x1 + cos x
= 2cot2 x
128.
sinx1 − sinx
− sinx1 + sinx
= 2tan2 x
129. sinθ(secθ − cscθ) = tanθ − 1
130. sec2θ − sec2θ sin2θ = 1
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Prove the identities. (The answer is the process. It needs to neat and easy to follow.) 131. θθ 44 cossin − = 2sin2θ − 1
132. tan2θ 1 + cot2θ( ) = sec2θ
133. cotθ = 1 + cotθ
1 + tanθ
134.
1 − cosθsinθ
= tanθ − sinθtanθ sinθ
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Prove. (The answer is the process. It needs to neat and easy to follow.)
135.
1 + secθsecθ − 1
= 1 + cosθ1 − cosθ
136.
1 + sinθ1 − sinθ
= cscθ + 1cscθ − 1
137.
11 − sinθ
= 1 + sinθcos2θ
138.
21 − sinθ
+ 21 + sinθ = 4sec2θ
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Prove. (The answer is the process. It needs to neat and easy to follow.)
139.
tanθsecθ + 1
= secθ − 1tanθ
140.
sinθ + cosθ tanθcotθ = 2tanθ sinθ
141.
sinθ1 + cosθ
+ sinθ1 − cosθ = 2cscθ
142.
sinθ1 + cosθ
= 1 − cosθsinθ
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ADDITION IDENTIES
Addition Identities
yxyxyx
yxyxyx
sinsincoscos)cos(
sinsincoscos)cos(
+=−
−=+
yxyxyx
yxyxyx
sincoscossin)sin(
sincoscossin)sin(
−=−
+=+
Challenge #16: Simplify sin9x cos5x + cos9x sin5x using the addition identities.
Challenge #17: Evaluate cos π
6cos π
2+ sinπ
6sinπ
2.
Challenge #18: Evaluate:
cos x − π( ) .
Use the Addition Identities to simplify the following: 143. =+ xxxx 5sin9cos5cos9sin Solution.
( )x
xx14sin
59sin=
+=
144. =− xxxx 5sin8cos5cos8sin
145. cos 7x cos 5x + sin7x sin5x =
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Evaluate.
146. =+2
sin6
sin2
cos6
cos ππππ
Solution.
⎟⎠⎞⎜
⎝⎛ −=
26cos ππ
⎟⎠⎞⎜
⎝⎛−=
3cos π
= 1
2
147. =−4
sin4
sin4
cos4
cos ππππ
148. sinπ
6cos π
2+ cos π
6sinπ
2=
Use the Addition Identities to evaluate the following:
150. Evaluate: =⎟⎠⎞⎜
⎝⎛ +
2cos πx
149. Evaluate: ( ) =− πxcos Solution.
ππ sinsincoscos xx +=
= cos x −1( ) + sinx 0( )
xcos−=
− sinx
151. Evaluate: =⎟⎠⎞⎜
⎝⎛ −
23sin πx
152. Evaluate: ( ) =− xπsin Solution.
xx sincoscossin ππ −=
= 0( )cos x − −1( )sinx
xsin=
153. Evaluate: =⎟⎠⎞⎜
⎝⎛ +
2sin πx
154. Evaluate: =⎟⎠⎞⎜
⎝⎛ − x23cos π
Challenge #19: Prove that xcos− ⎟
⎠⎞⎜
⎝⎛ −= x23sin π .
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Prove. 155. xcos−
⎟⎠⎞⎜
⎝⎛ −= x23sin π 156. ( ) =− πxcos −cos x
Expand using the Addition identity xx sin
23coscos
23sin ππ −=
Simplify ( ) ( ) xx sin0cos1 −−=
Multiply and reduce xcos−=
Left = Right
157. =xsin sin π − x( ) 158. =⎟
⎠⎞⎜
⎝⎛ −
23cos πx − sinx
Challenge 20: Simplify and or combine where possible. 159. =xx sinsin
160. =+ xx sinsin
161. sinx2 sinx 2( ) =
162. sin x + x( ) =
163. =xx sinsin5
164. =+ xx sin2sin5
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28
Double-Angle Identities
Double Angle Identities xxx 2sin2cos2cos −= cos2x = 2cos2 x − 1 xx 2sin212cos −= xxx cossin22sin =
Use the double angle identities to simplify. Challenge #21: Simplify. 2sin7x cos7x.
Challenge #22: Simplify. 5cos2 x + 5sin2 x
Write each expression as a single trigonometric ratio. 165. =xx 7cos7sin2 Solution:
= sin2 7x( )= sin14x
166. xx 22 sincos − 167. =+ xx 22 sin5cos5 Solution:
= 5 cos2 x + sin2 x( )= 5 1( )= 5
168. 15cos2 2 −x
169. =xx cossin4
170. =+− x2sin42 171. =− xx 22 sin2cos2 172. =xx
cos22sin
173. 14sin21 2−
174. =− xx cossin 175. =−− xx 22 sincos 176. =− 5cos5sin 22
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29
Key Points to Solving Double Angle Identities Always change xtan x
xcossin
, xcot xx
sincos
, xsec xcos1
, xcsc xsin1
Always change xxx cossin22sin = & xORxORxxx 2sin21,,12cos2,,2sin2cos2cos −−=−=
Use formula sheet Have your formula sheet in front of you
Memorize Pythagoras, Conjugate, Factoring, Distribution
Prove.
177. xtan = 1 − cos2x
sin2x
178. 12csc +x =
sinx + cos x( )2sin2x
179. xxx 2sin2
2sincot =
180. 22sin x xx 2sincot=
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30
Prove each identity.
181. x2cosx
x2
2
csc2csc −=
182. x2cos2
2cos1 x+=
183. x
x2cos1
2sin−
xx tan2csc2 −=
184. 2 1 − sin2 x( ) 12cos += x
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31
Prove.
185. xx
cos1cos1
−+
1secsec1
−+=
xx
186. xx
sincos
xx2sin
12cos +=
187. x
xsin1
cos−
xx tansec +=
188. xx cottan + x2csc2=
Challenge #23: If sinx = − 2
5and x is in the quadrant 3, evaluate sin2x .
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32
32
Find the ratio.
189. If sinx = − 2
5and x is in the quadrant 3,
evaluate sin2x = Possible solution strategy: • sin2x = 2sinx cos x
• sin2x = 2 − 2
5
⎛
⎝⎜
⎞
⎠⎟ cos x since
sinx = − 2
5=
OppHyp
• cos x = adjhyp
=a5
by pythagorus a = 21
•
sin2x = 2 − 25
⎛
⎝⎜
⎞
⎠⎟ − 21
5
⎛
⎝⎜⎜
⎞
⎠⎟⎟
(Sine and cosine are
negative in the 3rd quadrant)
ANSWER sin2x = 4 21
25
190. If sinx = − 2
5and x is in the quadrant 3,
evaluate cos2x =
191. If cos x = 2
5and x is in the quadrant 4,
evaluate sin2x =
192. If cos x = 2
5and x is in the quadrant 4,
evaluate cos2x =
Simplify using your identities sheet. 193. sin2 x + cos2 x =
194. 15sin2 x − 15cos2 x = 195. 4sin7x cos7x = 196. 4cos2 2x − 4sin2 2x =
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33
Simplify and write the following as a single trigonometric term. • Determine the best answer.
197.
cos10x + 12
cos2 5x + 1,cos2 5x,cos2 10x,cos2 20x
198.
sinx + tanxcos x + 1
sinx,cos x, tanx,secx
199.
cos12x + 12
cos2 6x,2cos6x,cos2 12x
200.
cos xcot x
+ 1cscx
2cos x, tanx,2sinx,2cot x
201.
sec2 x − 1sec2 x
sin2 x,cos2 x,tan2 x, sec2 x
202.
11 − cos x
+ 11 + cos
2cos2 x,2csc2 x,2tan2 x,2cot2 x
203.
tanx csc2 xsec2 x
cos x, tanx,sinx,cot x
204.
11 − sinx
+ 11 + sin
2sin2 x,2cos2 x,tan2 x,2sec2 x
Challenge #24: Let X be an angle in standard position such that 31tan =x & 0sin <x . Determine
the exact value of .secx
Challenge #25: Evaluate: ∑=
3
1 2cos
k
kπ .
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34
Find the exact value. 205. Let X be an angle in standard
position such that 31tan =x
& 0sin <x . Determine the exact value of .secx
Remember: adjopp
==31tan
Hyp = opp2 + adj2
Hyp = 12 + 32
Hyp = 10
Remember:
secx =
hypadj
= 103
Remember 31tan =x is (+) in 1&3
and 0sin <x in 3&4. Therefore x is in the 3rd Q.
310sec −=x
206. Let X be an angle in standard
position such that cot x = − 4
3
& cos x > 0 . Determine the exact value of .cscx
207. Let X be an angle in standard position such that
3csc = & 0tan <x . Determine the exact value of .cosx
208. Evaluate: ∑=
3
1 2cos
k
kπ
Solution:
2
3cos
2
2cos
2
1cos
πππ++=
= 0 + (−1) + 0= −1
209. Evaluate: ∑=
7
4 2cos
k
kπ
210. Evaluate:
∑=
4
1 6sin
k
kπ
Challenge #26: For the function, f (x) = 5sinHx + G , where H and G are positive constants, determine an expression for the smallest positive value of x that produces the maximum value of f(x).
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35
Find the appropriate value for x. 211. For the function,
f (x) = 5sinHx + G , where H and G are positive constants, determine an expression for the smallest positive value of x that produces the maximum value of f(x).
Remember
123sin,0sin
12
sin,00sin
−==
==
ππ
π
Maximum happens at 2
sin π .
Therefore, Hx = π
2. &
x = π
2H
212. For the function,
f (x) = 2sinHx + G , where H and G are positive constants, determine an expression for the smallest positive value of x that produces the smallest value of f(x).
213. For the function,
f (x) = − sinHx + G , where H and G are positive constants, determine an expression for the smallest positive value of x that produces the maximum value of f(x).
Challenge #27: The terminal arm of angle x in standard position passes through point (a,b) where a>0,b>0. Determine the value of
sin π + x( ) .
`
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36
Determine an expression for the following trig ratios. 214. The terminal arm of angle x
in standard position passes through point (a,b) where a>0,b>0. Determine the value of
sin π + x( ) .
A possible solution.
sin π + x( ) = sinπ cos x + cosπ sinx
sin π + x( ) = 0( )cos x + −1( )sinx
sin π + x( ) = − sinx
The value of − sinx = o
h. Remember
the hypotenuse is equal to a2+b2.
Therefore
− sinx = − b
a2 + b2
⎛
⎝⎜
⎞
⎠⎟
22 ba
b
+
−
215. The terminal arm of angle x in standard position passes through point (a,b) where a>0,b>0. Determine the value of
cos π + x( ) .
216. The terminal arm of angle x in standard position passes through point (a,b) where a>0,b>0. Determine the
value of csc π
2+ x
⎛
⎝⎜
⎞
⎠⎟ .
217. Determine the number of solutions for 0)cos)(sin( =−+ cxbaxa for π20 ≤≤ x , i. cba <<<1 Use any real numbers 4321 <<<
0)4cos3)(2sin2( =−+ xx Either
02sin2 =+x
1sin −=x 1 solution at
x=23π
Either 04cos3 =−x
34=cox
No solution because
?hypadj >
ii. abc <<<1 iii. cba ==<1
1
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37
Trig Identities Answers 1. 2. 2π 3. 2 4. 5. π 6. 4 7.
8.
π2
9. 8 10. 10 11. 2b 12. There are 2
solutions because the solutions occur in the first half of the cycle.
13. 4 14. 6 15. 2B 16. 2 17. 4 18. 19. 2 20. Infinite 21. The gap between
the 1st,3rd,,5th,.. is 2π. The gap between the 2nd,4th,,6th,…is 2π.
22.
π6
, 5π6
23. 2π 24.
π6
+ 2π, 5π6
+ 2π
25.
y = cos2xy = 0.5
26.
y = sinxy = x2 27.
nxnx
ππ
2189.42094.2
+=+=
28. 3.665+2πn & 5.760+ 2πn
29. 2.224 & 5.847 30.
x = 0.8959 + 2πnx = 2.2457 + 2πn
31. 0.675 & 5.608 32. 0.934 & 4.076
33.
x = −1.8955x = 0x = 1.8955
34. -1.022 & 1.022 35. x = 1.5 + 4n, x = 3.5 + 4n
cos π
2x + 1( )
cos π
2x + 1( )⎛
⎝⎜
⎞
⎠⎟
36. 65&
6ππ
37. 611&
6ππ
38. 34&
3ππ
39. 4
7,
4
ππ=x 40.
5π6
& 11π6
41. 611&
67 ππ
42. 6
11,
6
7,
6
5,
6
ππππ=x 43.
0,π,
7π
6,11π
6 44.
6
11,
6
7,
6
5,
6
ππππ 45.
x = 2π
3, 4π
3
46. x = π
6, 5π
6 47.
x = π
2, 3π
2, π3
, 5π3
48.
π12
, 5π12
49.
π18
, 5π18
50.
π36
, 3π36
51.
π20
, 7π20
52. 45
4ππ or
53. ππ 2,,0 54.
23
2ππ or
55. Ø since 11 ≤≤− Cosx
56. x = 0, π
4, 5π
4,π 57.
45,
4,
23,
2ππππ
58. 47,
45,
43,
4ππππ
59. − 29
2 60.
5
41 61.
− 5
2
62. 22 nm
m
+
63.
6114
64.
3714
65. 3
66.
4912
67. 2 68. =+
acabbc
69. acb
70. cb
71. 11
−+
aa
72. abab
−+
73.
sinxcos x
74.
1cos x
75.
cos xsinx
76.
1sinx
77. 2sinx 78. xcos
79. cscx 80.
2sinx
81. 1sin1sin
−+
xx
82. xsin 83. 1 84. xsec
85. xcsc 86. xcos 87. cot x 88. xcos2 89. xtan 90. xcot
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38
There are many correct ways of proving the identities. 91. 1tancsccos =xxx 92. xxx seccotcsc =
93. xxx sin
sectan =
xx
xx
cossin
sin1cos
Reduce 1
1=
Left =Right
xsin1
xx
xcos1
sincos=
Cancel cosx
xsin1=
=
x
xx
cos1
cossin
Multiply N&D by cosx =xsin
94. 1 95. csc2x 96. cot2 x 97. 1 + cos x
98. xxx csc1)sin1(csc +=+
99. xx
xx tan1cos
cossin +=+ 100.
xxxxxx
cossincossin)cot1(sec +=+
=+ )sin1(sin1 xx
Expand
=+xx
x sinsin
sin1
Simplify =+ 1cscx
xx
cossin1 +=
Get Common D
xx
xx
cossin
coscos +=
xxx
cossincos +=
=⎟⎠⎞⎜
⎝⎛ +
xx
x sincos1
cos1
Expand
=+xx
xx sincos
coscos1
Simplify
=+xx sin
1cos1
xxx
cossincossin +
101. xtan 102. x2cos 103. x2cos− 104. x2cos2 105. x2cot− 106. 4
107. 8 108. x2sec 109. 5sin2 x 110. 14csc2 x 111. cos2 x 112. sin2 x There are many correct ways of proving the identities.
113. xxx 222 cotsinsin1 =− 114. xxxx seccossinsec 2 =+
x2cos
xxx 2
22
sincossin
Reduce
x2cos left=right
xxx
cossincos1 2 +
Add fractions by creating common D
xx
xx
coscos
cossin 22
+
Add Numerators
xxx
coscossin 22 +
Simplify by Pythagorean Identity
xcos1
left=right
xcos1
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39
There are many correct ways of proving the identities. 115. xxxx seccsccottan =+ 116. xxx 2cot)1)(csc1(csc =−+
xx
xx
sincos
cossin +
Determine common D and add
xxx
xxx
cossincos
cossinsin 22
+
xxxx
cossincossin 22 +
Simplify using Pythagorean Identity
xx cossin1
Left=right
xx cos1
sin1 ×=
xx cossin1=
1csc2 −x Simplify using Pythagorean Identity
x2cot left=right
117. x
xx 2
22
tan1tansin+
= 118. xxxx sincoscot
sincos3 −=
xx
2
2
sectan
Write in terms of sinx & cosx
x
xx
2
2
2
cos1
cossin
Multiply N&D by cos2x
1sin2 x
left=right
xx
xx sincos
sincos −
Determine common D
xxx
xx
sinsincos
sincos 2
−
xxxx
sinsincoscos 2−
Factor out cos x
xxx
sin)sin1(cos 2−
Use Pythagorean Identity
xxx
sin)(coscos 2
xx
sincos3
left=right
119. xxx sin
csc1sin1 =
++
120. xxxx 2244 cossincossin −=−
x
x
sin11
sin1
+
+
Multiply N&D by sinx
1sinsin)sin1(+
+x
xx
xsin Left=right
Factor
)cos)(sincos(sin 2222 xxxx −+
Pythagorean identity
)cos)(sin1( 22 xx − Left=right
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40
There are many correct ways of proving the identities. 121. xx
xxx csccos
sin1cotcos =
++
122. 1cot
tan1cot1tan1
−−=
++
xx
xx
xxxx
sin1sincoscos
+
+
Multiply N&D by sinx
xxx
xxxx
sin)sin1(sin
sincossincos
+
+
Factor out cosx
xxxxsin)sin1(
)1(sincos+
+
Reduce fraction
xx
sincos
Left=right
xx
sin1cos
xx
sincos
xxxx
sincos1cossin1
+
+
Multiply N&D by sinxcosx
xxxxxx
2
2
coscossinsincossin
++
Factor out GCF
)sin(coscos)cos(sinsin
xxxxxx
++
Reduce
xx
cossin
Left=right
1sincos
cossin1
−
−=
xx
xx
Multiply N&D by sinxcosx
xxxxxx
cossincossincossin
2
2
−−
Factor out GCF
)sin(coscos)sin(cossin
xxxxxx
−−
Reduce
xx
cossin
123. xx
xx
sincos1
cos1sin −=+
124. xxx
x tansecsin1
cos −=+
Multiply by the conjugate
xx
xx
cos1cos1
cos1sin
−−×
+
xxx
2cos1)cos1(sin
−−
Pythagorean Identity
xxx
2sin)cos1(sin −
Reduce
xx
sincos1 −
Left=right
xx
x cossin
cos1 −
xx
cossin1 −
Multiply by the conjugate
xx
xx
sin1sin1
cossin1
++×−
)sin1(cossin1 2
xxx
+−
Pythagorean Identity
)sin1(coscos2
xxx
+
Reduce
xx
sin1cos+
Left=right
125. xxx
2sec2sin11
sin11 =
−+
+ 126. x
xx2csc2
cos11
cos11 =
−+
+
Find a common D
1(1 − sinx) + 1(1 + sinx)1 − sin2 x
Simplify & Pythagorean identity
x2cos2
Left=right
x2cos12×
x2cos2
Find a common D
1(1 − cos x) + 1(1 + cos x)1 − cos2 x
Simplify & Pythagorean identity
x2sin2
Left=right
x2csc12×
x2sin2
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41
There are many correct ways of proving the identities.
127.
cos x1 − cos x
− cos x1 + cos x
= 2cot2 x 128.
sinx1 − sinx
+ sinx1 + sinx
= 2tan2 x
Create common denominator
cos x(1 + cos x)1 − cos x
− cos x(1 − cos x)1 + cos x
Simplify(watch the negative)
cos x + cos2 x − cos x + cos2 x1 − cos2 x
Simplify
xx
2
2
sincos2
xx
2
2
sincos2
Left=right
Create common denominator
sinx(1 + sinx)1 − sinx
− sinx(1 − sinx)1 + sinx
Simplify(watch the negative)
sinx + sin2 x − sinx + sin2 x1 − sin2 x
Simplify
xx
2
2
cossin2
xx
2
2
cossin2
Left=right
129. )csc(secsin θθθ − 1tan −= θ 130. θθ 222 sinsecsec − 1=
Reciprocal Identity
sinθ 1
cosθ− 1
sinθ
⎛
⎝⎜
⎞
⎠⎟ =
DistributeMultiply by sinθ
sinθcosθ
− sinθsinθ
⎛
⎝⎜
⎞
⎠⎟ =
Reciprocal Identity and reduce
tanθ − 1 = Left = Right
Factor
sec2θ 1 − sin2θ( ) =
Pythagorean identity
sec2θ cos2θ( ) =
Reciprocal Identity
1cos2θ
cos2θ( ) =
Multiply and reduce 1
Left = Right
131. 44 cossin −θ 1sin2 2 −= θ 132. tan2θ 1 + cot2θ( ) θ2sec=
Pythagorean identity
tan2θ csc2θ( ) =
Quotient and Reciprocal Identities
sin2θcos2θ
1sin2θ
⎛
⎝⎜
⎞
⎠⎟ =
Simplify
θ2cos1
=
Reciprocal Identity
θ2sec = Left = Right
Factor
sin2θ − cos2( ) sin2θ + cos2( ) =
Pythagorean identity
sin2θ − cos2( ) 1( ) =
Pythagorean identity
sin2θ − (1 − sin2θ)( ) =
Remove brackets
sin2θ − 1 + sin2θ( ) =
Simplify
1sin2 2 −θ = Left = Right
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42
There are many correct ways of proving the identities. 133. θcot
= 1 + cotθ
1 + tanθ 134.
1 − cosθsinθ
= tanθ − sinθ
tanθ sinθ
= cosθ
sinθ
Reciprocal Identity
=1 + cosθ
sinθ
1 + sinθcosθ
Quotient Identity
=
sinθcosθ
− sinθ
sinθcosθ
⎛
⎝⎜
⎞
⎠⎟ sinθ
Eliminate complex fraction by multiplying by Denominators
=
1 + cosθsinθ
⎛
⎝⎜
⎞
⎠⎟
1 + sinθcosθ
⎛
⎝⎜
⎞
⎠⎟
×
sinθ cosθ1
sinθ cosθ1
Eliminate Complex Fraction by Multiplying by Denominator
=
sinθcosθ
− sinθ
sinθcosθ
⎛
⎝⎜
⎞
⎠⎟ sinθ
×
cosθ1
cosθ1
Distribution
= sinθ cosθ + cos2θ
sinθ cosθ + sin2θ
Multiply
= sinθ − sinθ cosθ
sin2θ
Factor
=cosθ sinθ + cosθ( )sinθ cosθ + sinθ( )
Factor
=
sinθ 1 − cosθ( )sin2θ
Reduce
θθ
sincos=
Left = Right
Reduce
θθ
sincos1 −=
Left =Right
135. 1sec
sec1−
+θ
θ
θθ
cos1cos1
−+= 136.
θθ
sin1sin1
−+
1csc1csc
−+=
θθ
Reciprocal Identity
1cos1cos11
−
+
θ
θ =
Reciprocal Identity
1sin1
1sin1
−
+=
θ
θ
Eliminate complex fraction by multiplying by Denominators
1cos1
cos
1cos1cos11
θ
θ
θ
θ ×−
+=
Eliminate complex fraction by multiplying by Denominators
1sin1
sin
1sin1
1sin1
θ
θ
θ
θ ×⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
−
+=
Multiply and reduce
θθcos1
1cos−
+= =
Left = Right
Multiply and reduce
θθ
sin1sin1
−+=
Left = Right
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43
There are many correct ways of proving the identities.
137. θsin1
1−
θθ
2cossin1 += 138.
θθ sin12
sin12
++
−
θ2sec4=
Pythagorean Identity
θθ2sin1
sin1−+=
Common D and add Fractions
θ
θ
θθ
θ
θ sin1
sin1
sin1
2
sin1
sin1
sin1
2
−
−×
++
+
+×
−
θ2cos
4
Factor the Denominator
( )( )θθ
θ
sin1sin1
sin1
−+
+=
Multiply
θθθ
2sin1sin22sin22
−−++
Reduce
( )θsin11
−=
Left =Right
Simplify on top and Pythagorean Identity on the bottom
θ2cos4
Left =Right
139. 1sec
tan+θθ
θ
θtan
1sec −= 140. θ
θθθcot
tancossin +
θθ sintan2=
Multiply by conjugate
1sec1sec
1sectan
−−×
+ θθ
θθ
Quotient Identity
θθθθθ
cotcossincossin ⎟
⎠⎞⎜
⎝⎛+
θθθ sin
cossin2 ⎟
⎠⎞⎜
⎝⎛=
Multiply ( )
1sec1sectan
2 −−
θθθ
simplify
θθθ
cotsinsin +
⎟⎟⎠
⎞⎜⎜⎝
⎛=
θθ
cossin2 2
Pythagorean Identity and reduce
( )θθθ
2tan1sectan −
Quotient Identity
θθθ
sincossin2
Reduce
θθ
tan1sec −=
Left = Right
Eliminate complex fraction by multiplying by Denominators
1sinsin
sincossin2
θθ
θθθ ×
Multiply and reduce
θθ
cossin2 2
Left = Right
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There are many correct ways of proving the identities.
141.
sinθ1 + cosθ
+ sinθ1 − cosθ
= 2cscθ
142. θ
θcos1
sin+
= 1 − cosθ
sinθ
Common D and add Fractions
sinθ1 + cosθ
× 1 − cosθ1 − cosθ
+ sinθ1 − cosθ
× 1 + cosθ1 + cosθ
2sinθ
Multiply by conjugate
sinθ1 + cosθ
× 1 − cosθ1 − cosθ
Multiply
sinθ − sinθ cosθ + sinθ + sinθ cosθ1 − cos2θ
Multiply
sinθ 1 − cosθ( )1 − cos2θ
Simplify on top and Pythagorean Identity on the bottom
2sinθsin2θ
Pythagorean Identity and
reduce
sinθ 1 − cosθ( )sin2θ
Reduce
2sinθ
Left =Right
Reduce
1 − cosθ( )sinθ
Left = Right
143. = sin14x 144. x3sin 145. x2cos
146. 21−=
147. 0 148.
23
149. xcos−= 150. - xsin 151. xcos 152. xsin= 153. xcos 154. xsin− 155. xcos−
= sin 3π
2− x
⎛
⎝⎜
⎞
⎠⎟
156. ( ) =− πxcos xcos−=
Expand using the Addition identity
xx sin23coscos
23sin ππ −=
Expand using the Addition identity
cos x cosπ + sinx sinπ =
Simplify ( ) ( ) xx sin0cos1 −−=
Simplify =+− )0sin(sin)1(cos xx
Multiply and reduce xcos−=
Left = Right
Multiply and reduce =− xcos
Left = Right
157. =xsin ( ) =− xπsin 158. =⎟
⎠⎞⎜
⎝⎛ −
23cos πx
xsin−=
Expand using the Addition identity xx sincoscossin ππ −=
Expand using the Addition identity
=+23sinsin
23coscos ππ xx
Simplify xx sin)1(cos)0( −−=
Expand using the Addition identity =−+ )1sin(sin)0cos(cos xx
Multiply and reduce xsin=
Left = Right
Multiply and reduce =− xsin
Left = Right
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159. x2sin 160. xsin2 161. sin2 x2( ) 162. x2sin 163. x2sin5 164. xsin7
165.
= sin 14x( )= sin2x
166. x2cos=
167.
= 5 cos2 x + sin2 x( )= 5 1( )= 5
168.
= cos 2 × 5x( )= cos10x
169.
= 2 2sinx cos x( )= 2sin2x
170.
= −2 1 − 2sin2 x( )= −2cos2x
171.
= 2 cos2 x − sin2 x( )= 2cos2x
172.
xx
xx
sincos2
cossin2
=
= 173.
= cos 2 × 14( )= cos28
174.
=− 2sin x cos x( )
2
=− sin2x
2
175.
= − cos2 x + sin2 x( )= − 1( )= −1
176.
= sin2 5 − cos2 5= − − sin2 5 + cos2 5( )= − cos10( )
There are many correct ways of proving the identities. 177. xtan
= 1 − cos2x
sin2x
178. 12csc +x
=
sinx + cos x( )2sin2x
xx
cossin Double angle identity
=
1 − 1 − 2sin2 x( )2sinx cos x
Reciprocal Identity Left
12sin1 +=x
Expand Right
= sin2 x + 2sinx cos x + cos2 x
sin2x
Remove brackets
xxx
cossin2sin211 2+−=
Common Denominator Left
xx
x 2sin2sin
2sin1 +=
Group Right
= sin2 x + cos2 x + 2sinx cos x
sin2x
1-1=0
xxx
cossin2sin2 2
=
Combine Fraction Left
xx
2sin2sin1 +=
Pythagorean Identity Right
=
1( ) + 2sinx cos x
sin2x
reduce
xx
cossin=
Left=Right
Double angle Identity Right
xx
2sin2sin1 +=
Left=Right 179. xcot
xx2sin22sin= 180.
22sin x xx 2sincot=
xx
sincos= Double angle Identity left
2cossin2 xx=
Double angle Identity
xxxx
sinsin2cossin2=
Reduce
xx
sincos=
Left=Right
Reduce Left
xx sincos=
Reciprocal Identity Right
1sin
sincos 2 x
xx=
Reduce Right xx sincos=
Left=Right
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181. x2cos
xx2
2
csc2csc −= 182. x2cos
22cos1 x+=
Reciprocal Identity
x
x
2
2
sin1
2sin
1 −=
x2sin1 −= Double angle Identity
=
1 + 2cos2 x − 1( )2
Eliminate Complex fraction by multiplying by denominator
xx
x
x2
2
2
2
sinsin
sin1
2sin
1
×−
=
Remove Brackets
21cos21 2 −+= x
Multiply
1sin21 2 x−=
1+(-1)=0
2cos2 2 x=
Double angle Identity x2cos=
Left=Right
Reduce x2cos=
Left=Right 183.
xx2cos1
2sin−
xx tan2csc2 −= 184. 2 1 − sin2 x( )
12cos += x
Double Angle Identity
= 2sinx cos x1 − 1 − 2sin2 x( )
Reciprocal Identity
xx
x cossin
2sin2 −=
Pythagorean Identity Left
= 2 cos2 x( )
Double angle Identity Right
= 2cos2 x − 1( ) + 1
Remove brackets Left
xxx
2sin2cossin2=
Double Angle Identity ⎟⎠⎞⎜
⎝⎛−=
xx
xx
xx sinsin
cossin
cossin22
x2cos2= Remove brackets and collect like terms
x2cos2= Left=Right
Reduce
xx
sincos=
Reduce
xxx
xx cossinsin
cossin1 2
−=
Combine Fraction
xxx
cossinsin1 2−=
Pythagorean Identity
xxx
cossincos2=
Reduce
xx
sincos=
Left=Right
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There are many correct ways of proving the identities. 185.
xx
cos1cos1
−+
1secsec1
−+=
xx
186. xx
sincos
xx2sin
12cos +=
Reciprocal Identity
1cos
1cos
11
−
+=
x
x
Eliminate Complex fraction by multiplying by denominator
=1 + 1
cos x1
cos x− 1
cos x1
cos x1
⎛
⎝
⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟
Multiply and Reduce
= 1 + cos x
1 − cos x Left=Right
Double angle Identity
=
2cos2− 1( ) + 1
2sinx cos x
Remove brackets and collect like terms
xxx
cossin2cos2 2
= Reduce
xx
sincos=
Left=Right
187. x
xsin1
cos−
xx tansec += 188. xx cottan + x2csc2= Multiply by conjugate
⎟⎠⎞⎜
⎝⎛
++
−=
xx
xx
sin1sin1
sin1cos
Reciprocal Identity
xx
x cossin
cos1 +=
Reciprocal Identity Left
xx
xx
sincos
cossin +=
Reciprocal Identity Right
x2sin2=
Multiply Left
( )xxx
2sin1sin1cos
−+=
Combine Fraction
xx
cossin1 +=
Common Denominator
⎟⎠⎞⎜
⎝⎛+⎟
⎠⎞⎜
⎝⎛=
xx
xx
xx
xx
coscos
sincos
sinsin
cossin
Double Angle Identity Right
xx cossin22=
Pythagorean Identity
( )x
xx2cossin1cos +=
Multiply and Collect like terms
xxxx
cossincossin 22 +=
Reduce Right
xx cossin1=
Reduce
xx
cossin1 +=
Left=Right Pythagorean Identity
xx cossin1=
Left=Right
189.
4 2125
190.
1725
191.
−4 2125
192. − 17
25
193. 1 194. x2cos15−
195. x14sin2 196. x4cos4 197. x5cos2 198. tanx 199. x6cos2 200. xsin2
201. sin2 x 202. 2csc2 x 203. xcot 204. 2sec2 x 205.
− 10
3 206.
− 5
3(Q4)
207. 322− (Q2)
208. -1 209. 0 210.
3 + 2 32
211.
π2H
212.
3π2H
213.
3π2H
214.
22 ba
b
+
− 215. 22 ba
a
+
− 216.
aba 22 +
217. 1,3,3
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Trig equations and Identities Summary Page 1. Solving with a Graphing Calculator Determine the period if applicable, Set up Window, Graph both functions Press 2ndF Press CALC Choose Intsct Repeat process for the next intersection point, State solution and or General Solution 2. Exact Solutions
Sine Wave
The Unit Circle
Cosine Wave
30º,60º,90º
45º,45º,90º
Helps solve questions like
sinx = 0 sinx = ±1
Helps solve questions like
sinx = 0 , sinx = ±1
cos x = 0 , cos x = ±1 tanx = 0 , tanx = ∅
Helps solve questions like
cos x = 0 cos x = ±1
Helps solve questions like
sinx = 1
2or 3
2
& Cosine & Tangent ratios
Helps solve questions like
sinx = cos x = 1
2
tanx = 1 3. Proving Trigonometric Identities Helpful Facts*
xtan xx
cossin , xcot
xx
sincos , xsec
xcos1 , xcsc
xsin1
Helpful Facts*
xxx cossin22sin = &
cos2x = cos2 x − sin2 x,
cos2x = 2cos2 x − 1,
cos2x = 1 − 2sin2 x
Pythagoras* sin2 x + cos2 x = 1 , 1 + cot2 x = csc2 x , tan2 x + 1 = sec2 x
One of these may help
Factoring sin2 x − sinx cos x = sinx sinx − cos x( )
Distribution
1cos x
cos xsinx
+ 1⎛
⎝⎜
⎞
⎠⎟ = cos x
cos x sinx+ 1
cos x
Combine Fractions
1cos x
+ 1sinx
= sinx + cos xcos x sinx
Conjugate 1 − cos x( ) 1 + cos x( ) = 1 − cos2 x = sin2 x
4. Addition Identities
Addition Identities*
cos(x + y) = cos x cosy − sinx sinycos(x − y) = cos x cosy + sinx siny
sin(x + y) = sinx cosy + cos x sinysin(x − y) = sinx cosy − cos x siny
*Starred items will be given on all quizzes, tests and final exams.
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