## Precalculus (6th Edition) Blitzer

Let us consider the left side of the given expression: $\sin {{60}^{\circ }}\sin {{30}^{\circ }}$ Now, put the values of the trigonometric functions to find the exact value as given below: \begin{align} & \sin {{60}^{\circ }}\sin {{30}^{\circ }}=\frac{\sqrt{3}}{2}\cdot \frac{1}{2} \\ & =\frac{\sqrt{3}}{4} \end{align} Also, consider the right side of the given expression: $\frac{1}{2}\left[ \cos \left( {{60}^{\circ }}-{{30}^{\circ }} \right)-\cos \left( {{60}^{\circ }}+{{30}^{\circ }} \right) \right]$ Then, put the values of the trigonometric functions to find the exact value as shown below: \begin{align} & \frac{1}{2}\left[ \cos \left( {{60}^{\circ }}-{{30}^{\circ }} \right)-\cos \left( {{60}^{\circ }}+{{30}^{\circ }} \right) \right]=\frac{1}{2}\left[ \cos {{30}^{\circ }}-\cos {{90}^{\circ }} \right] \\ & =\frac{1}{2}\left[ \frac{\sqrt{3}}{2}-0 \right] \\ & =\frac{1}{2}\cdot \frac{\sqrt{3}}{2} \\ & =\frac{\sqrt{3}}{4} \end{align} Thus, the left side of the expression is equal to the right side, which is $\sin {{60}^{\circ }}\sin {{30}^{\circ }}=\frac{1}{2}\left[ \cos \left( {{60}^{\circ }}-{{30}^{\circ }} \right)-\cos \left( {{60}^{\circ }}+{{30}^{\circ }} \right) \right]$.