Q9 of 76 Page 52

Prove that sin 4A = 4sinA cos3A – 4cosA sin3A (corrected question)

∵ 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


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