Prove $\sin^6x = \frac1{32}(10 − 15\cos2x + 6\cos4x − \cos6x)$ Im trying to prove the following equation:
$$\sin^6x = \frac1{32}(10 − 15\cos2x + 6\cos4x − \cos6x)$$
I have tried to do it several times now but get stuck at different places. I don't know if it's a calculation error or the question is defective. It would be really helpful if someone could point out the method.
we are to use $$z - \frac1{z} = 2i\sin x$$
$$z^n + (\frac1{z})^n = 2\cos nx$$

The problem is simple but you may have made a calculation mistake. Let's start with the identity $$\sin^{2}x = \frac{1 - \cos 2x}{2}$$ and then cube it to get $$\sin^{6}x = \frac{1}{8}(1 - 3\cos 2x + 3\cos^{2}2x - \cos^{3}2x)$$ and then we need to note that $$\cos^{2}2x = \frac{1 + \cos 4x}{2},\,\cos 6x = 4\cos^{3}2x - 3\cos 2x$$ and using these we get $$\sin^{6}x = \frac{1}{8}\left(1 - 3\cos 2x + \frac{3 + 3\cos 4x}{2} - \frac{\cos 6x + 3\cos 2x}{4}\right)$$ and on simplifying we get $$\sin^{6}x = \frac{10 - 15\cos 2x + 6\cos 4x - \cos 6x}{32} $$
