## Precalculus (6th Edition) Blitzer

The required solution is $2\sin \frac{\pi }{4}\sin \frac{\pi }{6}$, $\frac{\sqrt{2}}{2}$.
One of the sum-to-product formula is $\cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$. So, in this question, according to the above-mentioned formula, the value of $\alpha$ is $\frac{\pi }{12}$ and the value of $\beta$ is $\frac{5\pi }{12}$. Now, the expression can be evaluated as provided below: \begin{align} & \cos \frac{\pi }{12}-\cos \frac{5\pi }{12}=-2\sin \left( \frac{\frac{\pi }{12}+\frac{5\pi }{12}}{2} \right)\sin \left( \frac{\frac{\pi }{12}-\frac{5\pi }{12}}{2} \right) \\ & =-2\sin \left( \frac{\frac{6\pi }{12}}{2} \right)\sin \left( \frac{\frac{-4\pi }{12}}{2} \right) \\ & =-2\sin \left( \frac{6\pi }{24} \right)\sin \left( -\frac{4\pi }{24} \right) \\ & =-2\sin \left( \frac{\pi }{4} \right)\sin \left( -\frac{\pi }{6} \right) \end{align} Thus, applying the even-odd identity, which is $sin(-x)=-\sin x$, the expression can be further evaluated as given below: $-2\sin \left( \frac{\pi }{4} \right)\sin \left( -\frac{\pi }{6} \right)=2\sin \frac{\pi }{4}\sin \frac{\pi }{6}$ Therefore, the respective values are put to find the product’s exact value: \begin{align} & 2\sin \frac{\pi }{4}\sin \frac{\pi }{6}=2\left( \frac{\sqrt{2}}{2} \right)\left( \frac{1}{2} \right) \\ & =\frac{\sqrt{2}}{2} \end{align} Hence, the provided expression can be written as $2\sin \frac{\pi }{4}\sin \frac{\pi }{6}$. And the product’s exact value is $\frac{\sqrt{2}}{2}$.