## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 5 - Trigonometric Identities - Section 5.5 Double-Angle Identities - 5.5 Exercises - Page 238: 69

#### Answer

$W = 1771.27~sin^2~120\pi t$ $W = -885.63~cos~240\pi t+885.63$ When we graph these two functions on the same coordinate axes, we can see that they are equal.

#### Work Step by Step

We can use the identity: $cos~2x = 1-2~sin^2~x$ Note that: $~~sin^2~x = \frac{1-cos~2x}{2}$ $V = 163~sin(120 \pi t)$ $R = 15$ We can find an expression for $W$: $W = \frac{V^2}{R}$ $W = \frac{(163~sin~120\pi t)^2}{15}$ $W = \frac{(163)^2~sin^2~120\pi t}{15}$ $W = 1771.27~sin^2~120\pi t$ We can make another expression for $W$: $W = 1771.27~sin^2~120\pi t$ $W = 1771.27\cdot \frac{1-cos~240\pi t}{2}$ $W = -885.63~cos~240\pi t+885.63$ Therefore: $a = -885.63$ $c = 885.63$ $\omega = 240\pi$ $W = a~cos~\omega t+c$ $W = -885.63~cos~240\pi t+885.63$

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