1. DOUBLE ANGLES

2. sin (A + B) = sinA cosB + cosA sinB
From the compound angle formula:
Let B =A
sin (A + A) = sinA cosA + cosA sinA
sin 2A = 2 sin A cos A
cos (A + B) = cosA cosB – sinA sinB
From the compound angle formula:
Let B =A
cos (A + A) = cosA cosA – sinA sinA
cos 2A = cos 2A – sin 2A
. . . (1)
Since, cos 2A + sin 2A = 1
From equation (1):
cos 2A = cos 2A – ( 1 – cos 2A )
cos 2A = 2cos2A – 1
Or, from equation (1):
cos 2A = ( 1 – sin 2A ) – sin 2A
cos 2A = 1 – 2sin2A

3. Example 1: Solve the equation 3cos 2x – sin x = 2 for 0 ≤ x ≤ 360°.
Using:
cos 2A = 1 – 2sin2A
3cos 2x – sin x = 2
The sine of an angle is positive
in the 1st and 2nd quadrants.
3( 1 – 2 sin 2 x ) – sin x = 2
α = sin-1 1 3
3 – 6 sin 2 x – sin x = 2 
0 = 6 sin 2 x + sin x – 1
= 19.47°
0 = 6 s 2 + s – 1
The sine of an angle is negative
in the 3rd and 4th quadrants.
0 = ( 3s – 1 )( 2s + 1 )
α1 = sin-1 1 2
sin x = 1 3 1 2 –
sin x = or 1
= 30°
x = 19.5°, 160.5°, 210°, 330°. ( 1 d.p.)

4. Example 2: Solve the equation sin 2x – sin x = 0 for 0 ≤ x ≤ 2 π.
Using:
sin 2A = 2 sin A cos A
sin 2x – sin x = 0
2 sin x cos x – sin x = 0
The cosine of an angle is positive
in the 1st and 4th quadrants.
sin x ( 2 cos x – 1 ) = 0
cos x = or 1 2
α = cos-1 1 2
sin x = 0  = π 3
A quick sketch of
y = sin x for this
special case:
x = π 3 , 5π 3
x = 0, π , 2π

5. Example 3: Solve the equation tan 2x = cot x for 0 ≤ x ≤ 180°.
tan (A + B) =
tanA + tanB
1 – tanA tanB
From the compound angle formula:
tan 2A =
2 tan A
1 – tan2A
Let B =A
Example 3: Solve the equation tan 2x = cot x for 0 ≤ x ≤ 180°.
2 tan x
1 – tan2x = 1
tan x
tan 2x = cot x
The tangent of an angle is
positive in the 1st and 3rd
quadrants, and negative
in the 2nd and 4th. 2t
1 – t2 = 1 t
Let tan x = t:
α = tan-1 1 
2t2 = 1 – t2
3t2 = 1
= 30°
t2 = 1 3 1
tan x =
x = 30°, 150°.

6. Example 4: Prove the identity: sin 2A 1 + cos 2A ≡ tan A
sin 2A = 2 sin A cos A
cos 2A = 2cos2A – 1
Using:
LHS =
sin 2A
1 + cos 2A =
2 sin A cos A
1 + ( 2cos2A – 1 )
2 sin A cos A
2cos2A = =
sin A
cos A =
tan A = RHS

7. Example 5: Prove the identity: cosec x – cot x tan x 2 ≡
Let x = 2A, and hence prove:
cosec 2A – cot 2A tan A ≡
LHS = cosec 2A – cot 2A = 1
sin 2A
cos 2A
sin 2A –
cos 2A = 1 – 2sin2A
sin 2A = 2 sin A cos A
Using: =
1 – cos 2A
sin 2A =
1 – ( 1 – 2sin2A )
2sin A cos A =
2sin2A
2sin A cos A =
sin A
cos A
= tan A = RHS

8. The double angle identities are:
Summary of key points:
The double angle identities are:
sin 2A = 2 sin A cos A
There are 3 versions for cos 2A:
cos 2A = cos 2A – sin 2A
cos 2A = 2cos2A – 1
cos 2A = 1 – 2sin2A
tan 2A =
2 tan A
1 – tan2A
Also:
This PowerPoint produced by R.Collins ; Updated Mar. 2010