Compound Angle Formulae
1. Addition Formulae
cos( A  B)  cos A cos B  sin A sin B
cos( A  B)  cos A cos B  sin A sin B
sin( A  B)  sin A cos B  cos A sin B
cos( A  B)  sin A cos B  cos A sin B
Example:
sin( x  30)o  sin x cos30  cos x sin 30 
3 sin x  cos x
2
2. Formulae Involving Double Angle (2A)
sin 2 A  2sin Acos A
cos 2 A  cos 2 A  sin 2 A
 2cos 2 A  1
 1  2sin 2 A
Two further formulae derived from the cos2 A formulae.
cos 2 A  12 (1  cos 2 A)
sin 2 A  12 (1  cos 2 A)
Mixed Examples:
4
Given that A is an acute angle and tan A  , calculate sin 2 A and cos 2 A.
3
sin A 4

cos A 3
sin 2 A  cos 2 A  1
sin 2 A  ( 43 sin A)2  1
sin A  
Similarly:
sin 2 A  2sin A cos A 
cos A 
16

25
3
5
4
5
Substitute form the tan
(sin/cos) equation
+ve because A is acute
3-4-5 triangle
!!!
24
25
cos 2 A  cos2 A  sin 2 A 
9  16

25
7
25
A is greater than 45
degrees – hence 2A is
greater than 90
degrees.
Find the exact value of sin 75o.
2
sin(75o )  sin(45  30)
sin(75o )  sin 45cos30  cos 45sin30
45
1
o
1

Prove that
sin(   )
 tan   tan 
cos cos 

sin  cos   cos sin 
cos cos 
sin  sin 

cos  cos 
 tan   tan 
30 o
1
1 3 1 1
1 3


2 2
22
2 2
sin(   )

cos cos 
2
Q.E.D.
3
For the diagram opposite show that cos LMN 
5
.
5
M

cos LMN  cos(   )
3 2
Length of LM 
18  3 2
Length of MN 
10


1 3
1 1

2 10
2 10
2
2


20
4 5
5
5
10
3
3
cos(   )  cos cos   sin  sin 


1
5
L
1
N
(A Higher Question)
Show that, for the triangle ABC in the diagram, a 
a
b
c


sin a sin b sin c
b sin 
.
cos(   )

The sine
rule
b

From the diagram:
C
a
b

sin  sin(  [  2   ])


a
b
sin( 2  [   ])
The sum of the angles of
a triangle=180
sin( 2   )  cos
b
cos(   )
b sin 
cos(   )
As required
A
2
c

a
B
Prove that,
cos 4   sin 4   cos 2 .
x 2  y 2  ( x  y)( x  y )
cos4   sin 4   (cos2  )2  (sin 2  )2
 (cos2   sin 2  )(cos2   sin 2  )
cos 2   sin 2   1
 cos 2   sin 2 
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.
Use
sin2A = 2sinAcosA
when replacing sin2A
cos2A = 2cos2A – 1
cos2A = 1 – 2sin2A
if cosA is also in the equation
if sinA is also in the equation
when replacing cos2A
Solve:
cos 2 x o  4sin x o  5  0 for 0  x  360o.
cos2x and sin x, so substitute 1-2sin2
(1  2sin 2 x)  4sin x  5  0
6  4sin x  2sin 2 x  0
cp. w. 6  4 z  2 z 2  0
(6  2sin x)(1  sin x)  0
sin x  1 or sin x  3

x  90o
0  sin x  1 for all real angles
Solve:
5cos 2 x o  cos x o  2
for 0  x  360o.
cos 2x and cos x, so substitute 2cos2 -1
5(2cos2 x  1)  cos x  2
10cos 2 x  cos x  3  0
(5cos x  3)(2cos x  1)  0
3
1
cos x  or cos x  
5
2
s
a
t
c
All S_ Talk C*&p ??

x  51.3 or
x  90  60  150o or
x  360  51.3
x  270  60  210
o
 308.7 o
o
2
y  0.6
y0
y  0.5

3
2
The diagram shows the graphs of
f ( x)  a sin bx o
and g ( x)  c sin x o
for 0  x  360o.
yy
4
y  f ( x)
2
360o
0
-2
xx
y  g ( x)
-4
Three problems concerning this graph follow.
i)
State the values of a, b and c.
yy
4
f ( x)  a sin bxo
The max & min values of asinbx
are 3 and -3 resp.
The max & min values of sinbx
are 1 and -1 resp.
 a3
2
g ( x)  c sin x o
The max & min values of csinx are
2 and -2 resp.
 c2
360o
0
-2
-4
f(x) goes through 2 complete cycles from 0 – 360o
 b2
y  f ( x)
y  g ( x)
xx
ii) Solve the equation
f ( x)  g ( x) algebraically.
From the previous problem we now have:
f ( x)  3sin 2 x
and
g ( x)  2sin x
Hence, the equation to solve is:
3sin 2 x  2sin x
Expand sin 2x
3(2sin x cos x)  2sin x
6sin x cos x  2sin x  0
Divide both sides by 2
3sin x cos x  sin x  0
Spot the common factor in the terms?
sin x(3cos x  1)  0
Is satisfied by all values of x for which:
sin x  0 or
cos x 
1
3
iii) find the coordinates of the points of intersection of the graphs for 0  x  360o.
From the previous problem we have:
sin x  0 or
sin x  0
cos x 
cos x 
1
3
Hence:
x
 0o
or
x
 70.5o
x
x
 180o
 360o
or
x
x
 (360  70.5)o
 289.5o
or
1
3