I am stuck on trying to prove a trig identity using De Moivre's theorem.
I have to prove, $$\cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta)$$
I am not sure where to even start, I broke the LHS down to $$\cos(3\theta) + i\sin(3\theta)$$
but I have no idea where to go from here, or if this is fully correct.
If I could get some pointers or a simple worked example that I could follow it would be great.
Thanks
De Moivre's formula reads $$(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta)$$ Of course this identity implies the real part should be also equality. That is $$\cos(n\theta)=\Re\{(\cos\theta+i\sin\theta)^n\}$$ Hence we have $$\cos(3\theta)=\Re\{\cos^3\theta+3i\cos^2\theta\sin\theta-3\cos\theta\sin^2\theta-i\sin^3\theta\}=\cos^3\theta-3\cos\theta\sin^2\theta$$
Hint: $$\cos(n \phi) = \Re((\cos(\phi) + i\sin (\phi))^n)$$ Then use the Binomial Theorem to expand the expression.
(Where $\Re$ is the real part of the expression.)
Hint: Using de Moivre's identity: $$ \cos 3\theta = \mathrm{Re}\left(e^{i3\theta}\right) $$ Now, $e^{i3\theta} = \left(e^{i\theta}\right)^3$ and (by definition of $e^{i\theta}$) $$\left(e^{i\theta}\right)^3 = \left(\cos\theta+i\sin \theta\right)^3 $$ Expand the RHS (it's only a cube, so it's straightforward), and simplify the $i^3$, $i^2$; then take the real part to get $\mathrm{Re}\left(e^{i3\theta}\right)$.
