Compound Angle Formulae

1. Addition Formulae

cos( ) cos cos sin sin

cos( ) cos cos sin sin

A B A B A B

A B A B A B

sin( ) sin cos cos sin

cos( ) sin cos cos sin

A B A B A B

A B A B A B

Example:

3sin cossin( 30) sin cos30 cos sin30

2o x x

x x x

2. Formulae Involving Double Angle (2A)

sin 2 2sin cosA A A2 2

2

2

cos2 cos sin

2cos 1

1 2sin

A A A

A

A

cos2Two further formulae derived from the formulae.A

2 12

2 12

cos (1 cos2 )

sin (1 cos2 )

A A

A A

Mixed Examples:4

tan sin 2 cos23

Given that is an acute angle and , calculate and .A A A A

sin 4

cos 3

A

A 2 2sin cos 1A A Substitute form the tan

(sin/cos) equation2 23

4sin ( sin ) 1A A

16sin

25

4

5A +ve because A is acute

Similarly: cos3

5A

3-4-5 triangle !!!

2sin24

sin 25

cos2

AA A

2 2 9 16cos sin

7cos

252

25A A A

A is greater than 45 degrees – hence 2A is greater than 90 degrees.

sin75 .Find the exact value of o

sin(75 ) sin(45 30)o 30o1

1

2

45o

1

23

sin(75 ) sin 45cos30 cos45sin30o

1 3 1 1 1 3

2 22 2 2 2

sin( )tan tan

cos cosProve that

sin( ) sin cos cos sin

cos cos cos cos

sin sin

cos cos

tan tan Q.E.D.

1

L

M

N

3

3

5cos .

5For the diagram opposite show that LMN

cos cos( )LMN

18 3 2Length of LM

3 2

10Length of MN

10

cos( ) cos cos sin sin

1 3 1 1

2 10 2 102 2 1

20 4 5

5

5

5

sin.

cos( )Show that, for the triangle ABC in the diagram,

ba

(A Higher Question)

2

A

BC a

b c

The sine rulesin sin sin

a b c

a b c

From the diagram:

2sin sin( [ ])

a b

The sum of the angles of a triangle=180

2sin( [ ])

b

cos( )

b

2sin( ) cos

sin

cos( )

ba

As required

4 4cos sin cos2 .Prove that,

2 2 ( )( )x y x y x y

4 4 2 2 2 2cos sin (cos ) (sin )

2 2 2 2(cos sin )(cos sin )

2 2cos sin 1 2 2cos sin

cos( ) cos cos sin sin

cos2

TRIGONOMETRIC EQUATIONS

Double angle formulae (like cos2A or sin2A) often occur in trig equations. We can solve these equations by substituting the expressions derived in the previous sections.

sin2A = 2sinAcosA when replacing sin2A

cos2A = 2cos2A – 1 if cosA is also in the equationcos2A = 1 – 2sin2A if sinA is also in the equation

when replacing cos2A

Use

cos2 4sin 5 0 0 360 .Solve: for o o ox x x

cos2x and sin x, so substitute 1-2sin2

2(1 2sin ) 4sin 5 0x x

26 4sin 2sin 0x x 26 4 2 0cp. w. z z

(6 2sin )(1 sin ) 0x x

sin 1 sin 3 or x x 0 sin 1 for all real anglesx

90ox

5cos2 cos 2 0 360 .Solve: for o o ox x x

cos 2x and cos x, so substitute 2cos2 -125(2cos 1) cos 2x x

210cos cos 3 0x x

(5cos 3)(2cos 1) 0x x

3 1cos cos

5 2 or x x

c

as

t

All S_ Talk C*&p ??

0y

0.6y

0.5y

2 3

2

360

51.3

308.7

51.3

oro

o

x

x

150

210

90 60

270 60

oro

o

x

x

( ) sin ( ) sin

0 360 .

The diagram shows the graphs of and

for

o o

o

f x a bx g x c x

x

y

x

( )y f x

( )y g x

360o

-2

0

2

4

-4

x

y

Three problems concerning this graph follow.

, .i) State the values of and a b cy

x

( )y f x

( )y g x

360o

-2

0

2

4

-4

x

y

( ) sin of x a bx

The max & min values of asinbx are 3 and -3 resp.

The max & min values of sinbx are 1 and -1 resp.

3a

f(x) goes through 2 complete cycles from 0 – 360o

2b

( ) sin og x c x

The max & min values of csinx are 2 and -2 resp.

2c

( ) ( )ii) Solve the equation algebraically.f x g x

From the previous problem we now have:

( ) 3sin 2 ( ) 2sin and f x x g x x

Hence, the equation to solve is: 3sin 2 2sinx x

Expand sin 2x3(2sin cos ) 2sinx x x

6sin cos 2sin 0x x x Divide both sides by 2

3sin cos sin 0x x x Spot the common factor in the terms?

sin (3cos 1) 0x x

Is satisfied by all values of x for which:

1sin 0 cos

3 or x x

0 360iii) find the coordinates of the points of intersection of the graphs for .ox

From the previous problem we have:1

sin 0 cos3

or x x

sin 0x

0

180

360

or

or

o

o

o

x

x

x

Hence:

1cos

3x

(36

70.5

289.

0 70.

5

5)

or

o

o

ox

x

x