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Trigonometric Identities. r. y. x. SOH. θ. CAH. TOA. Pythagoras Theorem:. SOH. Definitions:. CAH. Reciprocal. TOA. Trigonometric identities. (1) Reciprocals. (2) Shaded Triangles. (3) Clockwise direction. (4) Anti-clockwise direction. - PowerPoint PPT Presentation

### Transcript of Trigonometric Identities

x

yrθ

sin

cos

tan

x

ry

rx

y

sin

cos

tan (cos 0)

x y

r rx r

r y

x

y

2 2 2 2

2

2

2

1

x y x y

r r r

r

r

TOA

CAH

SOH

Pythagoras Theorem:2 2 2r x y

2 2sin cos 1x x

Definitions:

sin

cos

tan

opp

hyp

hyp

opp

TOA

CAH

SOH1

cosecsin

1sec

cos

1cot

tan

Reciprocal

Trigonometric identities2 2sin cos 1x x

xx

x

x

xx

22

2

2

22

cos

1

cos

cos

cos

sin:cos

2 2tan 1 secx x

xx

x

x

xx

22

2

2

22

sin

1

sin

cos

sin

sin:sin

2 21 cot cosecx x

1 1 1cosec , sec , cot

sin cos tan

(1) Reciprocals2 2

2 2

2 2

sin cos 1

tan 1 sec

1 cot cosec

sin cos cottan , sin , cos ,...

cos cot cosecetc

(3) Clockwise direction

cos sin tancot , cos , sin ,...

sin tan secetc

(4) Anti-clockwise direction

Find the angles between 0 and 360 which satisfy:

sec x = 2

sec x = 2

2

1cos

2cos

1

x

x

x = 60 or 360 − 60 = 300

Find the angles between 0 and 360 which satisfy:

010sec23sec5 2 xx

x = 78.5 or 281.5

05sec2sec5 xx 5secor5

2sec xx

5

1cosor

2

5cos xx

54.28146.78360or 46.78 xx

(reject)

Given that A is obtuse and cot A 3 , find the exact value of

secA cosecA

1cot 3 tan

3A A 1. Change to basic trigo function

2. Draw ∆, use pythagoras1

3

10

1sec

cos

10

3

AA

3. Find basic trigo function

1cosec

sin

10

AA

AS

T C

cot A cosecA

1cot

tan1

2

AA

2

1

5

1cos

5

5

5

A

1cosec

sin12

5

5

2

AA

2tan A Atan cos AGiven that and that and have opposite signs, find the value of

cos A

AS

T C

Solve the following equations for 0 360o o x

2. Find the new range

sec( ) x 2

o o360 0x

3. Find B.A

1cos( )

2x 1. Change to basic trigo function

1 1cos 60

2BA

60 , 300x

4. Find the quadrants using ASTC

60 ,300x

5. Solve unknown in ( )

6. Solve x

Solve the following equations for 0 360o o x

cot 2sin 25x

2. Find B.A

1tan

2sin 25x 1. Change to basic trigo function

1 1. tan 49.79

2sin 25B A

49.8 ,229.8x

3. Find the quadrants using ASTCAS

T C

Solve the following equations for 0 360o o x

sec 2cosx x

2

2

sec 2cos

12cos

cos

2cos 1

1cos

2

x x

xx

x

x

45 ,135 ,225 ,315x

1 1. cos 45

2B A

2sin 3x 2cos 1y

4)1()3( 22 yx

Given that and , show that

For “showing” questions, always start from one side and end with the other

2 2

2 2

2 2

2 2

2 2

( 3) ( 1)

( 3) ( 1)

4sin 4cos

4(sin cos )

4 ( sin cos 1

2si 2c

)

osn 3 1

LHS x y

0sin2cos5 xx

0 360x

2sin 5cosx xsin 5

cos 2

x

x

5tan

2x

1 5. tan 68.198...

2B A

111.8 ,291.8x

AS

T C

22 tan 4 secx x

0 360x

22 tan 4 ( 1)tanx x

tan 3 tan 1x or x

. 71.565 45B A or

45 ,225 ,108.4 ,288.4x

AS

T C

2tan 2 tan 3 0x x (tan 3)(tan 1) 0x x

2nd/4th 1st /3rd

3sin 2 2 0x

0 360x

2sin 2

3x

. 41.81B A

221.8 ,318.2 ,581.8 ,678.22x

AS

T C

3rd /4th

0 2 720x

110.9 ,159.1 ,290.9 ,339.1x

2sin tany y0 360x

180 ,60 ,300y

sin2sin

cos

yy

y

2sin cos sin 0

sin (2cos 1) 0

1sin 0 cos

2

y y y

y y

y or y

1st /4thAll

. 0 60B A or

2cos 2 1 2x

0 360x

2 78.05 ,281.95 ,438.05 ,641.95

39.0 ,141.0 ,219.0 ,321.0

x

x

2 1cos 2

2x

1st /4th

. 78.05B A

0 2 720x

2(cos 2)(cos 1) sinx x x

0 360x

180x

2 2cos cos 2 1 cosx x x

2nd /3rd

. 90B A

22cos cos 3 0x x (2cos 3)(cos 1) 0x x

3cos ( ) cos 1

2x rej or x

2 22sin 5sin cos 3cosx x x x

0 360x

71.6 ,251.6 ,153.4 ,333.4x

2 22sin 5sin cos 3cos 0x x x x

. 26.565 71.565B A or

(2sin cos )(sin 3cos ) 0x x x x

2sin cos 0 sin 3cos 0x x or x x

2sin cos sin 3cos

1tan tan 3

2

x x or x x

x or x

1st /3rd 2nd / 4th

10sec5tan2 2

tanIf is obtuse and

, find the value of without using a calculator.

22(sec 1) 5sec 10 22sec 5sec 12 0

(2sec 3)(sec 4) 0 3

sec sec 42or

1cos ( is ob

2c tuo

4s se)

3reor j

3

2

55

tan2