## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Section 5.2 - Sum and Difference Formulas - Exercise Set - Page 669: 30

#### Answer

The expression $\sin \frac{7\pi }{12}\cos \frac{\pi }{12}-\cos \frac{7\pi }{12}\sin \frac{\pi }{12}$ is written as $\sin \frac{\pi }{2}$ and the exact value of $\sin \frac{\pi }{2}$ is $1$.

#### Work Step by Step

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{7\pi }{12}-\frac{\pi }{12} \right)=\sin \frac{7\pi }{12}\cos \frac{\pi }{12}-\cos \frac{7\pi }{12}\sin \frac{\pi }{12} \\ & \sin \left( \frac{\pi }{2} \right)=\sin \frac{7\pi }{12}\cos \frac{\pi }{12}-\cos \frac{7\pi }{12}\sin \frac{\pi }{12} \end{align} Therefore, the expression $\sin \frac{7\pi }{12}\cos \frac{\pi }{12}-\cos \frac{7\pi }{12}\sin \frac{\pi }{12}$ is equivalent to $\sin \frac{\pi }{2}$. From the knowledge of trigonometric ratios defined for sine of an angle, the exact value of $\sin \frac{\pi }{2}$ is $1$.

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