## Precalculus (6th Edition) Blitzer

The required value of the $\sin \frac{x}{2}$ is $\pm \sqrt{\frac{1-\cos x}{2}}$ .
In order to find the value of $\sin \frac{x}{2}$, the half angle formula is used that is as given below: ${{\sin }^{2}}\alpha =\frac{1-\cos 2\alpha }{2}$ Now, replace $\alpha$ with $\frac{x}{2}$ \begin{align} & {{\sin }^{2}}\frac{x}{2}=\frac{1-\cos 2\left( \frac{x}{2} \right)}{2} \\ & =\frac{1-\cos x}{2} \\ & =\pm \sqrt{\frac{1-\cos x}{2}} \end{align}