## 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: 31

#### Answer

The amplitude is $\frac{1}{4}$, the period is $4\pi$, there is no vertical translation and the phase shift is $\pi$ to the left since $d$ is less than zero.

#### Work Step by Step

We first write the equation in the form $y=c+a \sin [b(x-d)]$. Therefore, $y=-\frac{1}{4}\cos (\frac{1}{2}x+\frac{\pi}{2})$ becomes $y=0-\frac{1}{4}\cos [\frac{1}{2}(x+\pi)]$. Comparing the two equations, $a=-\frac{1}{4},b=\frac{1}{2}$,c=0 and $d=-\pi$. The amplitude is $|a|=|-\frac{1}{4}|=\frac{1}{4}.$ The period is $\frac{2\pi}{b}=\frac{2\pi}{0.5}=4\pi$. The vertical translation is $c=0$. The phase shift is $|d|=|-\pi|=\pi$ Therefore, the amplitude is $\frac{1}{4}$, the period is $4\pi$, there is no vertical translation and the phase shift is $\pi$ to the left since $d$ is less than zero.

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