Question

Prove that sin 4A = 4 sin A × cos³A - 4 cos A × sin³A

Answer 1

$$\begin{gathered} \mathbf{LHS}\mathbf{}=\sin 4\mathrm{A} \\ =\sin \left(2\mathrm{A}+2\mathrm{A}\right) \\ =\sin 2\mathrm{A}.\cos 2\mathrm{A}+\cos 2\mathrm{A}.\sin 2\mathrm{A}\left[\mathbf{As}\mathbf{,}\mathbf{}\mathbf{sin}\left(\mathbf{x}\mathbf{+}\mathbf{y}\right)\mathbf{}\mathbf{=}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{x}\mathbf{}\mathbf{.}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{y}\mathbf{}\mathbf{+}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{x}\mathbf{}\mathbf{.}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{y}\right] \\ =2\sin 2\mathrm{A}.\cos 2\mathrm{A} \\ =2\left(2\sin \mathrm{A}.\cos \mathrm{A}\right)\left({\cos }^2\mathrm{A}-{\sin }^2\mathrm{A}\right) \\ =4\sin \mathrm{A}.\cos \mathrm{A}\left({\cos }^2\mathrm{A}-{\sin }^2\mathrm{A}\right) \\ =4\sin \mathrm{A}{\cos }^3\mathrm{A}-4{\sin }^3\mathrm{A}.\cos \mathrm{A} \\ =\mathbf{RHS} \end{gathered}$$