Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.4 Double-Angle and Half-Angle Identities - 7.4 Exercises - Page 694: 65


$-\displaystyle \frac{\sqrt{42}}{12}$

Work Step by Step

Letting $A=2x, \displaystyle \quad\cos A=-\frac{5}{12}$ we express $x=\displaystyle \frac{A}{2}, \displaystyle \quad\cos x=\cos\frac{A}{2}$ ($\cos x$ is negative as x terminates in Q.II.) Half-Angle identity: $\displaystyle \cos x=\cos\frac{A}{2}=-\sqrt{\frac{1+\cos A}{2}}$ $=-\sqrt{\dfrac{1+(-\dfrac{5}{12})}{2}}$ $=-\sqrt{\dfrac{12-5}{24}}=-\sqrt{\dfrac{7}{24}}$ $=-\displaystyle \dfrac{\sqrt{7}}{2\sqrt{6}}\cdot\dfrac{\sqrt{6}}{\sqrt{6}}=-\dfrac{\sqrt{42}}{12}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.