Trigonometry (10th Edition)

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

Chapter 4 - Test - Page 188: 13b

Answer

The amplitude is $16.5$, the period is $12$, the vertical translation is $67.5$ units up and the phase shift is $4$ units to the right.

Work Step by Step

We first write the equation in the form $y=c+a \cos [b(x-d)]$. Therefore, $y=16.5\sin [\frac{\pi}{6}(x-4)]+67.5$ becomes $y=67.5+16.5\sin [\frac{\pi}{6}(x-4)]$. Comparing the two equations, $a=16.5,b=\frac{\pi}{6},c=67.5$ and $d=4$. The amplitude is $|a|=|16.5|=16.5$ The period is $\frac{2\pi}{b}=\frac{2\pi}{\pi/6}=12$ The vertical translation is $c=67.5$ The phase shift is $|d|=|4|=4$ Therefore, the amplitude is $16.5$, the period is $12$, the vertical translation is $67.5$ units up as $c$ is more than zero and the phase shift is $4$ units to the right since $d$ is more than zero.
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