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