Trigonometry (10th Edition)

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

Chapter 4 - Graphs of the Circular Functions - Section 4.2 Translations of the Graphs of the Sine and Cosine Functions - 4.2 Exercises - Page 156: 16

Answer

D

Work Step by Step

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