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

Can someone teach me how to do this using method FROM LHS/RHS-example-1
User Dick Eshelman
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1 Answer

9 votes
9 votes

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}

User Tushar Nallan
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