trigonometric equations and identities notesmathbeacon.ca/supportfiles/m12/5m12trigidents notes...

July 2011

Copyright Mathbeacon2008-2011. License Agreement Per student/Per Year: This content may be used before but not after July 2012.

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

Students visit www.mathbeacon.com for detailed solutions and support with your homework.


2

2

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


π20 ≤≤ x ?

-1

1

π 2π

7. Graph 5.04sin =x • Y=sin4x • Y=0.5


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



3

3

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 ?



4

4

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.



5

5

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


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.


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.



6

6

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 .



7

7

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°


cos x = 3

2 for 0 ≤ x ≤ 2π .


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


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 .



8

8

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 .



9

9

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 .


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


2sin2 x − 11sinx + 5 = 0 for 0 ≤ x ≤ 2π .


4cos3 x + 8cos2 x − 5cos x = 0 for 0 ≤ x ≤ 2π .



10

10

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


219sin =x


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 .



11

11

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).



12

12

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.



13

13

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

=



14

14

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.



15

15

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


96. Combining Fractions is sometimes helpful. 97. Multiplying by the conjugate is sometimes helpful.

1sin2 x

− 1 Try the following: • Common

denominators

• Combine


• Simplify

sin2 x1 − cos x

Try the following: • Multiply N & D by

the conjugate (1+cosx)

• Simplify the bottom


• Simplify



16

16

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 = ________+ ________ = ____



17

17

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( ) =



18

18

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



19

19

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



20

20


123. xx

xx

sincos1

cos1sin −=+

124. xxx

x tansecsin1

cos −=+

125. xxx

2sec2sin11

sin11 =

−+

+

126.

11 + cos x

+ 11 − cos x

= 2csc2 x



21

21


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



22

22

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



23

23

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



24

24

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



25

25

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 =



26

26

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


⎝⎛ −

23sin πx

152. Evaluate: ( ) =− xπsin Solution.

xx sincoscossin ππ −=

= 0( )cos x − −1( )sinx

xsin=


⎝⎛ +

2sin πx


⎝⎛ − x23cos π

Challenge #19: Prove that xcos− ⎟

⎠⎞⎜

⎝⎛ −= x23sin π .



27

27

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



28

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



29

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=



30

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



31

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 .



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


evaluate cos2x =

191. If cos x = 2


evaluate sin2x =

192. If cos x = 2


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 =



33

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 .



34

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).



35

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( ) .

`



36

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



37

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



38

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



39

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



40

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

+−


)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



41

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( ) =


sec2θ cos2θ( ) =

Reciprocal Identity

1cos2θ

cos2θ( ) =

Multiply and reduce 1

Left = Right

131. 44 cossin −θ 1sin2 2 −= θ 132. tan2θ 1 + cot2θ( ) θ2sec=


tan2θ csc2θ( ) =

Quotient and Reciprocal Identities

sin2θcos2θ

1sin2θ

⎛

⎝⎜

⎞

⎠⎟ =

Simplify

θ2cos1

=

Reciprocal Identity

θ2sec = Left = Right

Factor

sin2θ − cos2( ) sin2θ + cos2( ) =


sin2θ − cos2( ) 1( ) =


sin2θ − (1 − sin2θ)( ) =

Remove brackets

sin2θ − 1 + sin2θ( ) =

Simplify

1sin2 2 −θ = Left = Right



42

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


1cos1

cos

1cos1cos11

θ

θ

θ

θ ×−

+=


1sin1

sin

1sin1

1sin1

θ

θ

θ

θ ×⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−

+=

Multiply and reduce

θθcos1

1cos−

+= =

Left = Right

Multiply and reduce

θθ

sin1sin1

−+=

Left = Right



43

43


137. θsin1

1−

θθ

2cossin1 += 138.

θθ sin12

sin12

++

−

θ2sec4=


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


1sinsin

sincossin2

θθ

θθθ ×

Multiply and reduce

θθ

cossin2 2

Left = Right



44

44


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 −=


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 −=


=+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



45

45

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



46

46

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−=


xxx

cossincos2=

Reduce

xx

sincos=

Left=Right



47

47

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=


( )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



48

48

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


sinx = 0 , sinx = ±1

cos x = 0 , cos x = ±1 tanx = 0 , tanx = ∅


cos x = 0 cos x = ±1


sinx = 1

2or 3

2

& Cosine & Tangent ratios


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.



49

49

trigonometric equations and identities notesmathbeacon.ca/supportfiles/m12/5m12trigidents notes...

Documents