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: 2


The exact value of $\cos \left( 120{}^\circ -45{}^\circ \right)$ 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( 120{}^\circ -45{}^\circ \right)=\cos 120{}^\circ \cos 45{}^\circ +\sin 120{}^\circ \sin 45{}^\circ $ Substitute the values $\cos 120{}^\circ =-\frac{1}{2},\text{ }\cos 45{}^\circ =\frac{1}{\sqrt{2}},\text{ }\sin 120{}^\circ =\frac{\sqrt{3}}{2},\text{ and }\sin 45{}^\circ =\frac{1}{\sqrt{2}}$. $\begin{align} & \cos \left( 120{}^\circ -45{}^\circ \right)=\left( -\frac{1}{2}\times \frac{1}{\sqrt{2}} \right)+\left( \frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}} \right) \\ & =\frac{-1+\sqrt{3}}{2\sqrt{2}} \end{align}$ Hence, the exact value of $\cos \left( 120{}^\circ -45{}^\circ \right)$ is equivalent to $\frac{-1+\sqrt{3}}{2\sqrt{2}}$.
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