## Trigonometry (11th Edition) Clone

Published by Pearson

# 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: 35

#### Answer

The amplitude is $1$, the period is $\frac{2\pi}{3}$, the vertical translation is $2$ units up as $c$ is greater than zero and the phase shift is $\frac{\pi}{15}$ units to the right since $d$ is more than zero.

#### Work Step by Step

We first write the equation in the form $y=c+a \sin[b(x-d)]$. Therefore, $y=2-\sin [3x-\frac{\pi}{5})]$ becomes $y=2-\sin [3(x-\frac{\pi}{15})]$. Comparing the two equations, $a=-1,b=3,c=2$ and $d=\frac{\pi}{15}$. The amplitude is $|a|=|-1|=1.$ The period is $\frac{2\pi}{b}=\frac{2\pi}{3}$. The vertical translation is $c=2$. The phase shift is $|d|=|\frac{\pi}{15}|=\frac{\pi}{15}$ Therefore, the amplitude is $1$, the period is $\frac{2\pi}{3}$, the vertical translation is $2$ units up as $c$ is greater than zero and the phase shift is $\frac{\pi}{15}$ units to the right since $d$ is more than zero.

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