## 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: 36

#### Answer

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

#### Work Step by Step

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

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.