Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Trigonometric Identities - Section 5.5 Double-Angle Identities - 5.5 Exercises - Page 232: 70c

Answer

$W = -100.245~cos~240\pi t+100.245$ $a = -100.245$ $c = 100.245$ $\omega = 240\pi$

Work Step by Step

We can use the identity: $cos~2x = 1-2~sin^2~x$ Note that: $~~sin^2~x = \frac{1-cos~2x}{2}$ $W = VI$ $W = (163~sin~120\pi t)(1.23~sin~120\pi t)$ $W = 200.49~sin^2~120\pi t$ $W = 200.49~[\frac{1-cos~(2)(120\pi t)}{2}]$ $W = 100.245-100.245~cos~240\pi t$ $W = -100.245~cos~240\pi t+100.245$ $a = -100.245$ $c = 100.245$ $\omega = 240\pi$
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