Using de Moivre's Theorem to derive the relation... I must use de Moivre's Theorem to derive the following relation although I'm not exactly sure where to start:
$$\sin(3 \theta) = -4 \sin^3(\theta) + 3\sin(\theta)$$
Thanks in advance.
 A: Well, DeMoivre's theorem tells us that $$\cos(n\theta)+i\sin(n\theta)=(\cos\theta+i\sin\theta)^n$$ for all integers $n.$ Consider in particular when $n=3,$ and expand the perfect cube to see what happens. Don't forget your trig identities....
A: This answer goes through most of the steps, but I've gone through pretty fast, so it will take some going over. It's correct, but do the working out :-)
De Moivre's theorem:
$$
(\cos{(x)} + i\sin{(x)})^{n} = \cos{(nx)} + i\sin{(nx) }
$$
Let $n = 3$, and use a bit of the binomial theorem:
$$\cos^{3}{(x)} + 3i\cos^{2}{(x)}\sin{(x)} - 3\cos{(x)}\sin^2{(x)} - i\sin^{3}{(x)} = \cos{(3x)} + i\sin{(3x)}$$
Now we have an identity of some description, let's try expanding that $\cos{(3x)}$, and see what happens:
$$
\cos{(2x + x)} = \cos{(2x)}\cos{(x)} - \sin{(2x)}\sin{(x)} \\
  = \cos^3{(x)} - \sin^2{(x)}\cos{(x)} - 2\sin^2{(x)}\cos{(x)}
$$
That seems pretty promising. So let's put that in and leave the identity we want on its own on the other side.
$$
i\sin{(3x)} = 3i\cos^2{(x)}\sin{(x)} - i\sin^3{(x)}
$$
Now note that $\cos^2{(x)} = 1 - \sin^2{(x)}$, and the answer is obtained.
