DOUBLE ANGLES

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

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.)

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

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°.

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

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

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