## Precalculus (6th Edition) Blitzer

The expression $\sin \frac{5\pi }{12}\cos \frac{\pi }{4}-\cos \frac{5\pi }{12}\sin \frac{\pi }{4}$ is written as $\sin \frac{\pi }{6}$ and the exact value of $\sin \frac{\pi }{6}$ is $\frac{1}{2}$.
Use the difference formula of sine and rewrite the expression as the difference of angles to obtain the sine of the angle as, \begin{align} & \sin \left( \frac{5\pi }{12}-\frac{\pi }{4} \right)=\sin \frac{5\pi }{12}\cos \frac{\pi }{4}-\cos \frac{5\pi }{12}\sin \frac{\pi }{4} \\ & \sin \left( \frac{\pi }{6} \right)=\sin \frac{5\pi }{12}\cos \frac{\pi }{4}-\cos \frac{5\pi }{12}\sin \frac{\pi }{4} \end{align} Therefore, the expression $\sin \frac{5\pi }{12}\cos \frac{\pi }{4}-\cos \frac{5\pi }{12}\sin \frac{\pi }{4}$ is equivalent to $\sin \frac{\pi }{6}$. From the knowledge of trigonometric ratios defined for sine of an angle, the exact value of $\sin \frac{\pi }{6}$ is $\frac{1}{2}$.