## Precalculus (6th Edition) Blitzer

The right side is equal to $\sin \frac{x}{2}$.
By using the identity $\sin \left( \alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta$ , the above expression can be simplified as: \begin{align} & \sin \ \frac{5x}{2}\ \cos \ 2x-\ \cos \ \frac{5x}{2}\ \sin \ 2x=\sin \left( \frac{5x}{2}-2x \right) \\ & =\sin \left( \frac{5x-4x}{2} \right) \\ & =\sin \frac{x}{2} \end{align} Thus, the right side of the equation is equal to $\sin \frac{x}{2}$. Thus, it is proved that left side is equal to the right side.