Solution
= Solution
Set $z=e^{i\theta}$, so $2i\sin\theta=z-z^{-1}$. The <binomial theorem> gives
$$
(2i\sin\theta)^5
=(z^5-z^{-5})-5(z^3-z^{-3})+10(z-z^{-1}).
$$
Use $z^k-z^{-k}=2i\sin(k\theta)$ and $(2i)^5=32i$ to obtain
$$
\boxed{\sin^5\theta=\frac{10\sin\theta-5\sin3\theta+\sin5\theta}{16}.}
$$
This is a finite <Fourier sine series>, with only odd harmonics.