Question

Can someone teach me how to do this using method FROM LHS/RHS

Answer 1

We will prove:

`\sin (3x)=3\sin x-4\sin ^3x`

First, review these equations:

`\begin{gathered} \sin (a+b)=\sin a\cos b+\cos a\sin b \\ \cos (2a)=1-2\sin ^2a \\ \sin ^2a+\cos ^2a=1 \\ \sin 2a=2\sin a\cos a \end{gathered}`

So, for the given formula:

`\begin{gathered} \text{LHS}=\sin (3x)=\sin (2x+x) \\ \end{gathered}`

Using the first equation:

`\begin{gathered} \sin (2x+x)=\sin 2x\cos x+\cos 2x\sin x \\ =(2\sin x\cos x)\cdot\cos x+(1-2\sin ^2x)\sin x \\ =2\sin x\cos ^2x+\sin x-2\sin ^3x \\ =2\sin x(1-\sin ^2x)+\sin x-2\sin ^3x \\ =2\sin x-2\sin ^3x+\sin x-2\sin ^3x \\ =3\sin x-4\sin ^3x=\text{RHS} \end{gathered}`