Trigonometric Identities2 2 1sin x cos x

sin xtan x

cos x

cos xcot x

sin x

2 21 tan x sec x

2 21 cot x cosec x

ExamplesProve that (1 – cos A)(1 + sec A) sin A tan A

L.H.S. (1 – cos A)(1 + sec A) = 1 + sec A – cos A –Cos A sec A = 1 + sec A – cos A - 1

= sec A – cos A

1cos A

cos A

21 cos A

cos A

2sin A

cos A

sin Asin A

cos A

= sin A tan A= R.H.S.

ExamplesProve that cot A + tan A sec A cosec A

L.H.S.

2 2cos A sin A

sin Acos A

1

sin Acos A

cos ecAsec A = R.H.S.

cos A sin Acot A tan A

sin A cos A

Examples

R.H.S.

2

1

1 sin

2

1

cos

2sec = R.H.S.

1

1cos ec sin

cos ec sin sinsin

2 cos ecPr ovethat sec

cos ec sin

Solving equationsSolve 2 tan2 x – 7 sec x + 8 = 0 for 0 x 360

2 (sec2x – 1) – 7 sec x + 8 = 0 2 sec2x – 2 – 7 sec x + 8 = 0

2 sec2x – 7 sec x + 6 = 0

(2 sec x – 3)(sec x – 2)= 0

sec x = 3/2 or sec x = 2

cos x = 2/3 or cos x = ½

x = 48.2 or x = 60

or: x = 360 – 48.2 or x = 360 - 60

complete solution: x = 48.2 or 60 or 300 or 311.8

Solving equationsSolve 2 cos x = cot x for 0 x 360

2 cos x = cos x/ sin x

2 cos x sin x = cos x

2 cos x sin x – cos x= 0

cos x(2 sin x – 1)= 0

cos x = 0 or sin x = ½

cos x = 0 x = 90 or 270

sin x = ½ x = 30 or 330

complete solution: x = 30 or 90 or 270 or 30

Solving equationsSolve 3 cot2 x – 10 cot x + 3 = 0 for 0 x 2

(3 cot x - 1)(cot x – 3) = 0

cot x = 1/3 or cot x = 3

tan x = 3 or tan x = 1/3

tan x = 3 x = 1.24c or 4.39c

tan x = 1/3 x = 0.32c or 3.46c

complete solution: x = 0.32c or 1.24c or 3.46c or 4.39c

Solving equationsSolve 5 cot2 x – 2 cosec x + 2 = 0 for 0 x 2

5(cosec2 x – 1) – 2 cosec x + 2 = 0

5cosec2 x – 5 – 2 cosec x + 2 = 0

5cosec2 x – 2 cosec x - 3 = 0

22

5 23 5 2 3 0sin x sin x

sin x sin x

23 2 5 0sin x sin x

3 5 1 0( sin x )(sin x )

sin x = -5/3 not possible or sin x = 1 x = /2

Additional formulaesin (A + B) = sin A cos B + sin B cos Asin (A - B) = sin A cos B - sin B cos Acos (A + B) = cos A cos B - sin A sin Bcos (A - B) = cos A cos B + sin A sin B

1

tan A tan Btan( A B )

tan Atan B

1

tan A tan Btan( A B )

tan Atan B

ExamplesFind the exact value of sin 75

sin (A + B) = sin A cos B + sin B cos Asin (30 + 45) = sin 30 cos 45 + sin 45 cos 30

1 2 2 3

2 2 2 2

2 6 2 6

4 4 4

ExamplesExpress cos (x + /3) in terms of cos x and sin x

1 3

2 2cos x sin x

cos (A + B) = cos A cos B - sin A sin B

cos (x + /3) = cos x cos /3 - sin /3 sin x

Examplessin( A B )

Pr ovethat tan A tan Bcos Acos B

L.H.S.sin A sin B

tan A tan Bcos A cos B

sin Acos B sin B cos A

cos Acos B

sin( A B )

cos Acos B

=

R.H.S.

Double angle formulaesin (A + B) = sin A cos B + sin B cos Asin (A + A) = sin A cos A + sin A cos A sin 2A = 2 sin A cos A

cos (A + B) = cos A cos B - sin A sin Bcos (A + A) = cos A cos A- sin A sin Acos (A + A) = cos2A - sin2A

cos 2A = cos2A - sin2A

cos 2A = 2cos2A - 1

cos 2A = 1 – 2sin2A

Double angle formulae

1

tan A tan Btan( A B )

tan Atan B

1

tan A tan Atan( A A )

tan Atan A

2

22

1

tan Atan A

tan A

2 2Pr ovethat tan A tan A tan Asec A 2 2

2 2 2

2 2 1 1

1 1 1

tan A tan A tan A( tan A ) tan A( tan A )tan A

tan A tan A tan A

2

22

1 2

tan Asec A tan Atan Asec A

tan A cos A

ExamplesGiven that cos A = 2/3, find the exact value of cos 2A.

cos 2A = 2cos2A - 12

22 1

3

8 11

9 9

Given that sin A = ¼ , find the exact value of sin 2A.

sin 2A = 2 sin A cos A

1 15 152

4 4 8 A

4 1

15

Solving equationsSolve cos 2A + 3 + 4 cos A = 0 for 0 x 2

=2 cos2A - 1+ 3 + 4 cos A = 0

=2 cos2A + 4 cos A + 2= 0

= cos2A + 2 cos A + 1 = 0

= cos2A + 2 cos A + 1 = 0

= (cos A + 1)2 = 0

= cos A = - 1

A =

Solving equationsSolve sin 2A = sin A for - x

=2sin A cos A = sin A

=2 sin A cos A – sin A = 0

= sin A(2 cos A – 1) = 0

sin A = 0 or cos A = ½

sin A = 0 A = - or 0 or

cos A = ½ A = - /3 or /3

Complete solution: A = - or - /3 or 0 or /3 or

Solving equationsSolve tan 2A + 5 tan A = 0 for 0 x 2

Complete solution: A= 0.97 , 2.27, 4.01, 5.41c 0, or 2

22

25 2 5 1 0

1

tan Atan A tan A tan A( tan A )

tan A

22 5 1 0tan A[ ( tan A )]

22 5 5 0tan A[ tan A )

27 5 0tan A[ tan A )

tan A = 0 A = 0 or or 2

7 – 5tan2 A = 0 tan A = 7/5 A = 0.97 , 2.27, 4.01 or 5.41c

Harmonic form

If a and b are positive a sin x + b cos x can be written in the form R sin( x + )

a cos x + b sin x can be written in the form R cos( x - )

a sin x - b cos x can be written in the form R sin( x - )

a cos x - b sin x can be written in the form R cos( x + )

2 2R a b

R cos a and R sin b

ExamplesExpress 3 cos x + 4 sin x in the form R cos( x - )

R cos( x - ) = R cos x cos + R sin x sin

3 cos x + 4 sin x = R cos x cos + R sin x sin

R cos = 3 [1] R sin = 4 [2]

[1]2 + [2]2 : R2 sin2 x + R2 cos2 x = 32 + 42

R2(sin2 x + cos2 x ) = 32 + 42

R2= 32 + 42 = 25 R = 5

[2] [1]: tan = 4/3 = 53.1

3 cos x + 4 sin x = 5 cos( x + 53.1 )

ExamplesExpress 12 cos x + 5 sin x in the form R sin( x + )

R sin( x + ) = R sin x cos + R cos x sin

12 cos x + 5 sin x = R sin x cos + R cos x sin

R cos = 12 [1] R sin = 5 [2]

[1]2 + [2]2 : R2 cos2 x + R2 sin2 x = 122 + 52

R2(cos2 x + sin2 x ) = 122 + 52

R2= 122 + 52 = 169 R = 13

[2] [1]: tan = 5/12 = 22.6

12 cos x + 5 sin x = 13 sin( x + 22.6 )

ExamplesExpress cos x - 3 sin x in the form R cos( x + )

R cos( x + ) = R cos x cos - R sin x sin

cos x - 3 sin x = R cos x cos - R sin x sin

R cos = 1 [1] R sin = 3 [2]

[1]2 + [2]2 : R2 cos2 x + R2 sin2 x = 12 + (3 ) 2

R2(cos2 x + sin2 x ) = 12 + 3

R2= 1 + 3 = 4 R = 2

[2] [1]: tan = 3 = 60

cos x + 3 sin x = 2 cos( x + 60 )

Solving equationsSolve 7 sin x + 3 cos x = 6 for 0 x 2

R sin( x + ) = R sin x cos + R cos x sin

7 sin x + 3 cos x = R sin x cos + R cos x sin

R cos = 7 [1] R sin = 3 [2]

R2 = 72 + 32 R = 7.62

[2] [1]: tan = 3/7 = 0.405c (Radians)

7 sin x + 3 cos x = 7.62 sin( x + 0.405)

7.62 sin( x + 0.405 ) = 6 x + 0.405 = sin-1(6/7.62)

x + 0.405 = 0.907 or 2.235

x = 0.502c or 1.830c