By expressing \(\sin \theta\) and \(\cos \theta\) in terms of \(\mathrm { e } ^ { \mathrm { i } \theta }\) and \(\mathrm { e } ^ { - \mathrm { i } \theta }\), prove that
$$\sin ^ { 3 } \theta \cos ^ { 2 } \theta \equiv - \frac { 1 } { 16 } ( \sin 5 \theta - \sin 3 \theta - 2 \sin \theta )$$
Hence show that all the roots of the equation
$$\sin 5 \theta = \sin 3 \theta + 2 \sin \theta$$
are of the form \(\theta = \frac { n \pi } { k }\), where \(n\) is any integer and \(k\) is to be determined.