(a) To use the formulae
sin (A  B), cos (A
 B) and tan (A  B).
(b) To derive and use the double angle
formulae
(c) To derive and use the half angle formulae
(d) Find the value of an angle without using
table or calculator
Compound Angles Formulae
sin(A  B)  sin A cos B  cos Asin B
cos(A  B)  cos Acos B + sin Asin B
tan A tanB
tan(AB) 
1+ tan AtanB
If we substitute B with A into the compound angle
formulae, we have
sin(A + A)  sin A cos A  cos A sin A
 2 sin A cos A
cos(A + A)  cos A cos A  sin A sin A
2
2
 cos A  sin A
tan A  tan A
tan(A + A) 
1  tan A tan A
2 ta n A

2
1  ta n A
Thus,
Double Angle Formulae
sin 2A  2 sin A cos A
2
2
cos 2A  cos A  sin A
2
 2cos A 1
2
 1 2sin A
tan 2A 
2 tan A
2
1  tan A
or
or
2
2
sin x  cos x  1
By substituting A with A in the double
2
angle formulae.
Half- Angle Formulae
A
A
sin A  2 sin
cos
2
2
2A
cos A  cos
2
 sin
2A
A
2 tan
2
tan A 
2 A
1  tan
2
2
or
or
2A
 2cos
2
1
2A
 1 2sin
2
2A
from cos A  1 2sin
2
A
1

cos
A
sin  
2
2
2A
from cos A  2cos
1
2
A
1

cos
A
cos  
2
2
Example 1
3
5
If sin A and cos B  , A and B are acute
5
13
angles, find without using calculators, the
value of
a) sin (A+B)
b) cos (A-B)
c) tan (A+B)
Solution
5
3
A
13
12
B
4
Given
3
sin A
5
4
cos A
5
3
tan A
4
5
5
cos B
13
12
sin B
13
12
tan B
5
a) sin (A+B)  sin A cos B  cos Asin B
3 5 4 12
   
5 13 5 13
15  48

65
63

65
b) cos (A-B)  cos AcosB  sin AsinB
4 5 3 12
   
5 13 5 13
20  36

65
56

65
c) tan (A+B)  tan A  tan B
1 tan A tan B
3 12

4
5

3 12
1 
4 5
15  48
20

9
1
5
63 5


20 4
63

16
sin( A  B )  sin A cos B  co s A sin B
Example 2
Simplify:
a) sin 4x cos x – cos 4x sin x
Solution
a) sin 4x cos x – cos 4x sin x
= sin (4x – x)
= sin 3x
ta n 2 x  ta n x
b)
1  ta n 2 x ta n x
= tan (2x –x)
= tan x
ta n A  ta n B
ta n ( A  B ) 
1  ta n A ta n B
c o s ( A  B )  c o s A c o s B  s in A s in B
Example 3
Find the values of the followings
without using calculator:
a) cos 170o cos 70o – sin 170o sin 70o
= cos ( 170o + 70o)
= cos 240o
= - cos 60o
1
 
2
240o
-ve
60o
b) 
o
o
tan 107  tan 47

 1  tan107 o tan 47 o


2
o
  tan 60 
o
o
 tan(107  47 )


=3
3

2

2




2
Example 4
Find the exact value:
a) sin 15o = sin (600 - 450 )
= sin 600 cos 450- cos 60o sin 45o


3
2
1
2

 
2
2
2
2
6 
4
2

 
b) tan
 tan   
3 4
12


ta n
 ta n
3
4



1  ta n
ta n
3
4
3 1

1 3
3 1 1


1 3 1
 2 
3
3
3
Example 5
8
Given that sin A   ,
17
where A in the fourth quadrant, find
without using calculator:
a) tan 2A
b) cos 2A
A
c) cos
2
A
d ) s in
2
Solution
Given A in the fourth quadrant.
8
sin A  
,
17
15
A
17
8
15
cos A 
,
17
8
tan A  
15
a) tan 2A 
2 ta n A
2
1  ta n A
8 

2

15 


2
8 

1 

 15 
16

15

64
1
225
16 225
 
15 161
240

161
2
2
b) cos 2A  cos A  sin A
2
 15 
 8 
   

 17 
 17 
225 64

289
161

289
2
A
c ) cos A  2 cos
1
2
15
2A
 2cos
1
17
2
Since A is in the 4th
32
2 A
quadrant
2 cos

2 17
o
o
A
32
16
270

A

360
cos  

2
34
17
o A
o
135   180
2
4
 
A
4
17
 cos  
2
17
2
A
A
d) sin A  2 sin
cos
2
2
8
A 4 
  2sin  

17
2  17 
A
8
17
sin  

2
17
8

17
17
Example 6
Find the exact value of
a)
45o
1
2 tan 67
2
o
o
1
1  tan 2 67
2
135o
o

1

 tan 2  67


2


= tan 135o
= - tan 45o
= -1
 45 
b ) s i n 2 2 .5  s i n 

 2 
o
cos A  1 2sin
o
2A
2
A
1  cos A
sin

2
2
Since 22.5o is in the 1st quadrant
A
sin

2
1  cos A
2
REMEMBER?????!!!!!
2A
from cos A  1 2sin
2
A
1

cos
A
sin  
2
2
2A
from cos A  2cos
1
2
A
1

cos
A
cos  
2
2
Let A = 45o
sin
22.5o

1  cos 45
2

2
1
2
2

2 2
4

2
2
2
o
Example 7
Prove the identity
sin(x  y)
 tan x  tan y
cos x cos y
Solution
sin(x  y) sin xcos y  cos xsin y

cos xcos y
cos x cos y
sin x cos y cos xsin y


cos x cos y cos xcos y
sin x sin y


 tan x  tan y
cos x cos y
Example 8
Prove the identity
tan   co t   2 cosec 2 
Solution
s in 
cos 

tan   cot  
cos 
s in 
sin   cos 

cos  sin 
2
1


sin  cos  2
2

2sin  cos 
2
2

sin 2
2
 2cos ec2
Example 9
Show that






tan   A   tan   A   2sec 2A
4

4

Solution






tan   A   tan   A 
4

4



tan  tan A tan  tan A
4
4




1  tan tan A 1  tan tan A
4
4
1  tan A 1  tan A


1  tan A 1  tan A
1  tan A   1  tan A 


2
1  tan A
2
2
1  2 tan A  tan A  1  2 tan A  tan A

2
1  tan A
2
2  2 tan A
2

2
2
1  tan A
cos
A

2
2
cos A  sin A
2
2(1  tan A)

2

cos A
2
1  tan A
2
2
cos
A
2


2sec A
2
2
2
cos
A
cos
A

sin
A

2
sin A
2
1
2

 2sec 2A
cos A
2
2
cos 2A
Exercise:
a) sin 4A  4  sin A cos A  cos A sin A 
3
3
b) sin 3A  3sin A  4sin A
3
sin(A  B)  sin A cos B  cos Asin B
cos(A  B)  cos Acos B + sin Asin B
tan A tanB
tan(AB) 
1+ tan AtanB