Prove that sin 4A

∵ sin 4A = sin (2A + 2A)

As we know that-

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

∴ sin 4A = sin 2A cos 2A + cos 2A sin 2A

⇒ sin 4A = 2 sin 2A cos 2A

From T-ratios of multiple angle, we know that

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