Proving trig identity using De Moivre's Theorem 

Question: Prove $$\cos(3x) = \cos^3(x) - 3\cos(x)\sin^2(x) $$ by using De'Moivres Theorem


So far (learning complex numbers at the moment) that De Moivre's theorem states that
if $z$ $=$ $r\text{cis}(\theta)$ then $z^n = r^n\text{cis}(n\theta)$
so with this question I was thinking if
$$ z = \cos(3\theta) + i\sin(3\theta) $$
then 
$$ z = (\cos(\theta) + i\sin(\theta))^3 $$
and then expanding and comparing the real part? Is that the right way to go for this question?

The solution can be completed in this manner:

We know, by De-Moivre's Theorem, 
$$(\cos x + i \sin x)^3=\cos 3x + i \sin 3x$$
Therefore, we can write
$$\cos^3 x + 3i\cos^2x\sin x + 3i^2\cos x\sin^2 x + i^3 \sin^3 x=\cos 3x + i \sin 3x$$
or, $$\cos^3 x + 3i\cos^2x\sin x - 3\cos x\sin^2 x - i \sin^3 x=\cos 3x + i \sin 3x$$
or, $$(\cos^3 x- 3\cos x\sin^2 x)  + i(3\cos^2x\sin x -  \sin^3 x)=\cos 3x + i \sin 3x$$
As far as your problem is concerned, just compare the real parts of this equation.
Hope this helps you.
