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Question
Prove that: \[\sin 4x = 4 \sin x \cos^3 x - 4 \cos x \sin^3 x\]
Solution
\[LHS = sin 4x\]
\[ = 2\sin2x \cos2x \left( \because \sin2\theta = 2sin\theta cos\theta \right)\]
Now, using the identities
\[\sin2\alpha = 2\sin\alpha\cos\alpha \text{ and } \cos2\alpha = \cos^2 \alpha - \sin^2 \alpha\], we get
\[LHS = 2(2\text{ sin } x \text{ cos } x) . ( \cos^2 x - \sin^2 x)\]
\[ = 4\text{ sin } x \cos^3 x - 4 \sin^3 x \text{ cos } x = RHS\]
\[\text{ Hence proved } .\]
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