Prove that sin 4A = 4sinA cos3A – 4 cosA sin3A.

sin 4A = sin (2A + 2A)

We know that,

sin(A + B) = sin A cos B + cos A sin B

Therefore, sin 4A = sin 2A cos 2A + cos 2A sin 2A

⇒ sin 4A = 2 sin 2A cos 2A

From T-ratios of multiple angle,

We get,

sin 2A = 2 sin A cos A and cos 2A = cos2A – sin2A

⇒ sin 4A = 2(2 sin A cos A)(cos2A – sin2A)

⇒ sin 4A = 4 sin A cos3A – 4 cos A sin3A

Hence, sin 4A = 4 sin A cos3A – 4 cos A sin3A