By considering the binomial expansion of \(\left(z-\frac{1}{z}\right)^{5}\), where \(z=\cos \theta+\mathrm{i} \sin \theta\), use de Moivre's theorem to show that
\(\operatorname{cosec}^{5} \theta=\frac{a}{\sin 5 \theta+b \sin 3 \theta+c \sin \theta},\)
where \(a, b\) and \(c\) are integers to be determined.