Question

Solve:

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

Answer 1

We have,

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

$=\frac{\sin \theta (1-2{\sin }^2\theta )}{\cos \theta (2{\cos }^2\theta -1)}$

We know that

$\cos 2x=1-2{\sin }^{2}x=2{\cos }^{2}x-1$

Therefore,

$=\frac{\sin \theta (\cos 2\theta )}{\cos \theta \left(\cos 2\theta \right)}$

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

$=\tan \theta$

Hence, this is the answer.