<s>[INST] Prove $\cos 3x =4\cos^3x-3\cos x$ How would I solve the following double angle identity. 
$$\cos 3x =4\cos^3x-3\cos x $$
I know $\,\cos 3x  = \cos(2x+x)$
So know I have $\,\cos 2x +\cos x \,$ , Which is $\,(2\cos^2x-1)\cos x$ 
But I am not sure what to do next.
 [/INST] I will attempt to answer my own question now. 
$$\begin{eqnarray}
\cos(2x)(\cos x)-\sin(2x)(\sin x)& =& (2\cos^2 x-1)(\cos x)-2\sin x\cdot\cos x\cdot\sin x\\
&=&2\cos^3x-\cos x-2\sin^2 x\cos x\\
&=&2\cos^3x-\cos x-2(1-\cos^2 x)(\cos x)\\
&=&2\cos^3x-\cos x-2\cos x+2\cos^3 x\\
&=&4\cos^3 x-3\cos x
\end{eqnarray}$$
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