Precalculus (6th Edition) Blitzer

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

Chapter 4 - Mid-Chapter Check Point - Page 578: 23


$-\dfrac{\sqrt 2}{2}$

Work Step by Step

The reference angle of an angle $0 \leq \theta \lt 2\pi $ based on its position can be computed by using the following steps: a) Quadrant- I: $\theta $ b) Quadrant -II: $(\pi-\theta)$ c) Quadrant- III: $(\theta - \pi)$ d) Quadrant - IV: $(2\pi - \theta)$ Thus, we have the reference angle $540^{\circ}-495^{\circ}=45^{\circ}$ So, $\cos 45^{\circ}=\dfrac{\sqrt 2}{2}$ Thus, we have $\cos 495^{\circ}=-\dfrac{\sqrt 2}{2}$ because $\theta $ lies in Quadrant-II.
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.