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.
 A: I might as well: there is in fact a useful recurrence (Chebyshev) that you can use for expressing $\cos\,nx$ entirely in terms of powers of $\cos\,x$:
$$\cos((n+1)x)=2\cos(x)\cos(nx)-\cos((n-1)x)$$
In this case, you can start with $\cos\,x$ and $\cos\,2x=2\cos^2 x-1$:
$$\begin{align*}
\cos\,3x&=(2\cos\,x)(2\cos^2 x-1)-\cos\,x\\
&=4\cos^3 x-2\cos\,x-\cos\,x=4\cos^3 x-3\cos\,x
\end{align*}$$
A: I will attempt to answer my own question now. 
$$\begin{eqnarray}
\cos(2x)(\cos x)-\sin(2x)(\sin x)& =& (2\cos^2 x-1)(\cos x)-2\sin x\cdot\cos x\cdot\sin x\\
&=&2\cos^3x-\cos x-2\sin^2 x\cos x\\
&=&2\cos^3x-\cos x-2(1-\cos^2 x)(\cos x)\\
&=&2\cos^3x-\cos x-2\cos x+2\cos^3 x\\
&=&4\cos^3 x-3\cos x
\end{eqnarray}$$
A: $$e^{ix}=\cos x+i\sin x\Rightarrow e^{3ix}=(\cos x+i\sin x)^3\Rightarrow\cos 3x+i\sin 3x=(\cos x+i\sin x)^3$$
Now expand the cube and equate the real and imaginary parts of both sides to get the answer.
A: \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}
A: In general, a product of sin and cos functions can be easily solved via Fourier series expansion and convolution.
