Answer 1

\[LHS = sin 4x\]

\[ = 2\sin2x \cos2x \left( \because \sin2\theta = 2sin\theta cos\theta \right)\]

Now, using the identities

\[\sin2\alpha = 2\sin\alpha\cos\alpha \text{ and }  \cos2\alpha = \cos^2 \alpha - \sin^2 \alpha\], we get

\[LHS = 2(2\text{ sin } x \text{ cos } x) . ( \cos^2 x - \sin^2 x)\]

\[ = 4\text{ sin } x \cos^3 x - 4 \sin^3 x \text{ cos } x = RHS\]

\[\text{ Hence proved } .\]