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 $$z - \frac1{z} = 2i\sin x$$
$$z^n + (\frac1{z})^n = 2\cos nx$$

Raising to the $6^{th}$ power we get
$$
\eqalign{
64i^6\sin^6 x &= \bigl(z-{1\over z}\bigr)^6 \cr
&= z^6 - 6z^5{1\over z} + 15z^4{1\over z^2} -20z^3{1\over z^3} + 15z^2{1\over z^4} - 6z{1\over z^5} + {1\over z^6} \cr
&= z^6 - 6z^4 + 15z^2 -20 + 15{1\over z^2} - 6{1\over z^4} + {1\over z^6} \cr
&= z^6 + {1\over z^6} - 6\bigl(z^4+{1\over z^4}\bigr) + 15\bigl(z^2+{1\over z^2}\bigr) -20   \cr
&= 2\cos6x -6\cdot 2\cos4x +15\cdot 2\cos2x -20 \ \ \ so \ \ \ that \cr
-32\sin^6x &= \cos6x -6\cos4x +15\cos2x -10  \cr
}
$$
