## Precalculus (6th Edition) Blitzer

The result is $\frac{\sqrt{3}}{2}$.
Let us consider the expression: $\sin \left( \frac{\pi }{3}-\alpha \right)\cos \left( \frac{\pi }{3}+\alpha \right)+\cos \left( \frac{\pi }{3}-\alpha \right)\sin \left( \frac{\pi }{3}+\alpha \right)$ By using the trigonometric identity, $\sin \left( \alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta$ Now, the above expression can be further simplified as: \begin{align} & \sin \left( \frac{\pi }{3}-\alpha \right)\cos \left( \frac{\pi }{3}+\alpha \right)+\cos \left( \frac{\pi }{3}-\alpha \right)\sin \left( \frac{\pi }{3}+\alpha \right)=\sin \left[ \left( \frac{\pi }{3}-\alpha \right)+\left( \frac{\pi }{3}+\alpha \right) \right] \\ & =\sin \left( \frac{\pi }{3}-\alpha +\frac{\pi }{3}+\alpha \right) \\ & =\sin \left( \frac{\pi }{3}+\frac{\pi }{3} \right) \\ & =\sin \left( \frac{2\pi }{3} \right) \end{align} Use, $\sin \frac{2\pi }{3}=\frac{\sqrt{3}}{2}$. $\sin \left( \frac{2\pi }{3} \right)=\frac{\sqrt{3}}{2}$ Thus, $\sin \left( \frac{\pi }{3}-\alpha \right)\cos \left( \frac{\pi }{3}+\alpha \right)+\cos \left( \frac{\pi }{3}-\alpha \right)\sin \left( \frac{\pi }{3}+\alpha \right)$ can be simplified as $\frac{\sqrt{3}}{2}$.