Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Section 6.2 - The Law of Cosines - Exercise Set - Page 732: 70


The value of $\cos \frac{5\pi }{12}$ is $\frac{\sqrt{6}-\sqrt{2}}{4}$

Work Step by Step

By using the trigonometric identity, we get $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ So, $\begin{align} & \cos \frac{5\pi }{12}=\cos \left( \frac{\pi }{6}+\frac{\pi }{4} \right) \\ & =\cos \frac{\pi }{6}\cdot \cos \frac{\pi }{4}-\sin \frac{\pi }{6}\cdot \sin \frac{\pi }{4} \\ & =\frac{\sqrt{3}}{2}\cdot \frac{1}{\sqrt{2}}-\frac{1}{2}\cdot \frac{1}{\sqrt{2}} \\ & =\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4} \end{align}$ Therefore, $\cos \frac{5\pi }{12}=\frac{\sqrt{6}-\sqrt{2}}{4}$
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