Using Euler's Formula to prove $\sin^32x\cos^23x = -\frac1{16}(\sin12x-3\sin8x+2\sin6x+3\sin 4x-6\sin 2x) $

Question

I am asked to show the following using Euler's formula, but I cannot see, how this relates to Euler's formula.

$$ \sin^32x\cos^23x = -\frac{1}{16}(\sin 12x - 3\sin 8x + 2\sin 6x + 3\sin 4x - 6\sin 2x) $$

Answer 1

By Euler the left hand side is $$ \left(\frac{\zeta^2-\zeta^{-2}}{2i}\right)^3 \left(\frac{\zeta^3+\zeta^{-3}}{2}\right)^2 $$ where $\zeta:=e^{ix}$. It is easy to multiply this out and identify every term on the right hand side.