Problem

By considering the binomial expansion of $\big(z - \frac{1}{z}\big)^5$, where $z = \cos\theta + \mathrm{i}\sin\theta$, use de Moivre’s theorem to show that

$$\begin{align*} \cosec^5\theta =&\, \frac{a}{\sin 5\theta + b\sin 3\theta + c\sin\theta}, \end{align*}$$

where $a$, $b$ and $c$ are integers to be determined.