## Precalculus (6th Edition) Blitzer

The given expression on the left side $\frac{\sin 2\theta -\sin \theta }{\cos 2\theta +\cos \theta }$ can be simplified by applying the double angle formula $\sin 2\theta =2\sin \theta \cos \theta$ and $\cos 2\theta =2{{\cos }^{2}}\theta -1$. \begin{align} & \frac{\sin 2\theta -\sin \theta }{\cos 2\theta +\cos \theta }=\frac{2\sin \theta \cos \theta -\sin \theta }{2{{\cos }^{2}}\theta -1+\cos \theta } \\ & =\frac{\sin \theta \left( 2\cos \theta -1 \right)}{\left( 2\cos \theta -1 \right)\left( \cos \theta +1 \right)} \\ & =\frac{\sin \theta }{\cos \theta +1} \end{align} Rationalizing the expression gives: \begin{align} & \frac{\sin \theta }{\cos \theta +1}=\frac{\sin \theta }{\cos \theta +1}.\frac{\cos \theta -1}{\cos \theta -1} \\ & =\frac{\sin \theta \left( \cos \theta -1 \right)}{{{\cos }^{2}}\theta -1} \\ & =\frac{\sin \theta \left( \cos \theta -1 \right)}{-{{\sin }^{2}}\theta } \\ & =\frac{-\left( \cos \theta -1 \right)}{\sin \theta } \end{align} Now, the expression can be written as: $\frac{-\left( \cos \theta -1 \right)}{\sin \theta }=\frac{1-\cos \theta }{\sin \theta }$ Hence, the expression on the left-side is equal to the expression on the right-side.