## Precalculus (6th Edition) Blitzer

The result of $\cos \left( \frac{\pi }{6}+\alpha \right)\cos \left( \frac{\pi }{6}-\alpha \right)-\sin \left( \frac{\pi }{6}+\alpha \right)\sin \left( \frac{\pi }{6}-\alpha \right)$ is $\frac{1}{2}$.
Let us consider the given expression, $\cos \left( \frac{\pi }{6}+\alpha \right)\cos \left( \frac{\pi }{6}-\alpha \right)-\sin \left( \frac{\pi }{6}+\alpha \right)\sin \left( \frac{\pi }{6}-\alpha \right)$ By using the trigonometric identity, $\cos \left( \alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta$ Now, the above expression can be further simplified as, \begin{align} & \cos \left( \frac{\pi }{6}+\alpha \right)\cos \left( \frac{\pi }{6}-\alpha \right)-\sin \left( \frac{\pi }{6}+\alpha \right)\sin \left( \frac{\pi }{6}-\alpha \right)=\cos \left[ \left( \frac{\pi }{6}+\alpha \right)+\left( \frac{\pi }{6}-\alpha \right) \right] \\ & =\cos \left( \frac{\pi }{6}+\alpha +\frac{\pi }{6}-\alpha \right) \\ & =\cos \left( \frac{\pi }{6}+\frac{\pi }{6} \right) \\ & =\cos \left( \frac{2\pi }{6} \right) \end{align} Use, $\cos \frac{\pi }{3}=\frac{1}{2}$. \begin{align} & \cos \left( \frac{2\pi }{6} \right)=\cos \left( \frac{\pi }{3} \right) \\ & =\frac{1}{2} \end{align} Thus, $\cos \left( \frac{\pi }{6}+\alpha \right)\cos \left( \frac{\pi }{6}-\alpha \right)-\sin \left( \frac{\pi }{6}+\alpha \right)\sin \left( \frac{\pi }{6}-\alpha \right)$ can be simplified as $\frac{1}{2}$. Hence, the result of $\cos \left( \frac{\pi }{6}+\alpha \right)\cos \left( \frac{\pi }{6}-\alpha \right)-\sin \left( \frac{\pi }{6}+\alpha \right)\sin \left( \frac{\pi }{6}-\alpha \right)$ is $\frac{1}{2}$.