Precalculus (6th Edition) Blitzer

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

Chapter 4 - Section 4.8 - Applications of Trigonometric Functions - Exercise Set - Page 639: 61

Answer

$\sin 528 \pi t $

Work Step by Step

Amplitude, $ A=|1|=1$ Now, $ f=\dfrac{\omega}{2 \pi} \implies \omega =2 \pi f $ and period, $ P=\dfrac{2 \pi}{\omega} $ So, $\omega =2 \pi (264)=528 \pi $ Therefore, $ d=A \sin \omega t= \sin 528 \pi t $
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