Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.2 - Sum and Difference Formulas - Exercise Set - Page 668: 3


The exact value of $\cos \left( \frac{3\pi }{4}-\frac{\pi }{6} \right)$ is $\frac{1-\sqrt{3}}{2\sqrt{2}}$.

Work Step by Step

Use the difference formula of cosines and evaluate the term as, $\cos \left( \frac{3\pi }{4}-\frac{\pi }{6} \right)=\cos \frac{3\pi }{4}\cos \frac{\pi }{6}+\sin \frac{3\pi }{4}\sin \frac{\pi }{6}$ Substitute the values $\cos \frac{3\pi }{4}=-\frac{1}{\sqrt{2}},\text{ }\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2},\text{ }\sin \frac{3\pi }{4}=\frac{1}{\sqrt{2}},\text{ and }\sin \frac{\pi }{6}=\frac{1}{2}$. $\begin{align} & \cos \left( \frac{3\pi }{4}-\frac{\pi }{6} \right)=\left( -\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2} \right)+\left( \frac{1}{\sqrt{2}}\times \frac{1}{2} \right) \\ & =\frac{1-\sqrt{3}}{2\sqrt{2}} \end{align}$ Hence, the exact value of $\cos \left( \frac{3\pi }{4}-\frac{\pi }{6} \right)$ is equivalent to $\frac{1-\sqrt{3}}{2\sqrt{2}}$.
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