## Precalculus (6th Edition) Blitzer

Let us consider the left-hand side of the given expression: ${{\sin }^{2}}\frac{\theta }{2}$ By using the trigonometric identity $si{{n}^{2}}x=\frac{1-\cos 2x}{2}$ and $\frac{1}{\sin x}=\csc x$, the above expression can be further simplified as: \begin{align} & {{\sin }^{2}}\frac{\theta }{2}=\frac{1-\cos 2\left( \frac{\theta }{2} \right)}{2} \\ & =\frac{1-\cos \theta }{2} \end{align} By multiplying the numerator and denominator by $\frac{1}{\sin \theta }$ \begin{align} & \frac{1-\cos \theta }{2}=\frac{\left( 1-\cos \theta \right)\times \frac{1}{\sin \theta }}{2\times \frac{1}{\sin \theta }} \\ & =\frac{\frac{1}{\sin \theta }-\frac{\cos \theta }{\sin \theta }}{\frac{2}{\sin \theta }} \\ & =\frac{\csc \theta -\cot \theta }{2\csc \theta } \end{align} Hence, left-hand side of the given expression is equal to the right-hand side, which is ${{\sin }^{2}}\frac{\theta }{2}=\frac{\csc \theta -\cot \theta }{2\csc \theta }$.