<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] \begin{eqnarray}
\cos(3x) &=& \cos(2x+x)\\
&=& \cos(2x)\cos x - \sin(2x)\sin x\\
&=& (\cos^2 x - \sin^2 x)\cos x - 2\sin^2 x\cos x\\
&=& \cos^3 x -(1-\cos^2 x)\cos x - 2 (1 -\cos^2 x)\cos x\\
&=& 2 \cos^3 x -\cos x + \cos^3 x -2 \cos x\\
&=& 4 \cos^3 x - 3 \cos x
\end{eqnarray}
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