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 637: 26

Answer

Amplitude $=\dfrac{1}{3}$ ; $ f=\dfrac{1}{\pi}$ and Period $=\pi $

Work Step by Step

Amplitude, $|A|=|\dfrac{1}{3}|=\dfrac{1}{3}$ Given: $\omega =2$ So, $ f=\dfrac{\omega}{2 \pi}=\dfrac{2}{2 \pi}=\dfrac{1}{\pi}$ Now, period, $ P=\dfrac{2 \pi}{\omega}=\dfrac{2 \pi}{2}=\pi $ Hence, Amplitude $=\dfrac{1}{3}$ ; $ f=\dfrac{1}{\pi}$ and Period $=\pi $
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