trigonometry - Using De Moivre's Theorem to prove $cos(3theta) = 4cos^3(theta) - 3cos(theta)$ trig identity

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

Answer

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 $$\cos(3\theta)=\Re\{\cos^3\theta+3i\cos^2\theta\sin\theta-3\cos\theta\sin^2\theta-i\sin^3\theta\}=\cos^3\theta-3\cos\theta\sin^2\theta$$