Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 4 - Graphs of the Circular Functions - Section 4.2 Translations of the Graphs of the Sine and Cosine Functions - 4.2 Exercises - Page 161: 19



Work Step by Step

We first write the equation in the form $y=a \cos [b(x-d)]$. Therefore, $y=3\sin (2x-4)$ becomes $y=3\sin [2(x-2)]$. Comparing the two equations, $a=3,b=2$ and $d=2$. The amplitude is $|a|=|3|=3.$ The period is $\frac{2\pi}{b}=\frac{2\pi}{2}=\pi$. The phase shift is $|d|=|2|=2.$ Therefore, the description in option B is the correct answer.
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