$\frac{\sin 4\theta }{\sin \theta }=$

We know that,

$\sin n\theta ={}^n{C}_1\cdot {\cos }^{n-1}\theta \cdot \sin \theta -{}^n{C}_3\cdot {\cos }^{n-3}\theta \cdot {\sin }^3\theta +\cdots$

$\Rightarrow \sin 4\theta =4{\cos }^3\theta \cdot \sin \theta -4\cos \theta \cdot {\sin }^3\theta$

$\Rightarrow \frac{\sin 4\theta }{\sin \theta }=4{\cos }^3\theta -4\cos \theta {\sin }^2\theta$

$=4{\cos }^3\theta -4\cos \theta (1-{\cos }^2\theta )$ $=8{\cos }^3\theta -4\cos \theta$.