Question

How do you prove the following?

$\frac{\sin \left(3\theta \right)}{\sin \theta }-\frac{\cos \left(3\theta \right)}{\cos \theta }=2$

Answer 1

Prove the given identity:

Proving the above by solving Left Hand Side and showing it equals to Right Hand Side:

$\mathrm{Left}\mathrm{Hand}\mathrm{Side}=\frac{\sin \left(3\theta \right)}{\sin \theta }-\frac{\cos \left(3\theta \right)}{\cos \theta }$

Getting denominators alike by cross multiplying numerators and denominators of each term as follows:

$\begin{array}{rcl}\mathrm{LHS}& =& \frac{\sin \left(3\theta \right)\cos \theta -\cos \left(3\theta \right)\sin \theta }{\sin \theta \cos \theta }\\ & =& \frac{\sin (3\theta -\theta )}{\sin \theta \cos \theta }\left[\mathrm{By}\mathrm{using}\mathrm{the}\mathrm{compound}\mathrm{angle}\mathrm{formula}\sin (a-b)=\sin \left(a\right)\cos \left(b\right)-\cos \left(a\right)\sin \left(b\right)\right]\\ & =& \frac{\sin (3\theta -\theta )}{{\displaystyle \frac{1}{2}\times 2\sin \theta \cos \theta }}\left[\mathrm{Adjustment}\mathrm{for}\mathrm{using}\mathrm{doubled}\mathrm{angle}\mathrm{trigonometric}\mathrm{identities}\mathrm{of}\sin 2\theta =2\sin \theta \cos \theta \right]\\ & =& 2\frac{\sin 2\theta }{\sin 2\theta }\\ & =& 2\end{array}$

Hence, $\frac{\sin \left(3\theta \right)}{\sin \theta }-\frac{\cos \left(3\theta \right)}{\cos \theta }=2$ is proved.