Trigonometry (10th Edition)

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

Chapter 4 - Graphs of the Circular Functions - Section 4.1 Graphs of the Sine and Cosine Functions - 4.1 Exercises - Page 146: 55c

Answer

For $t=0$, $E=5$ For $t = 0.03$, $E=1.545$ For $t = 0.06$, $E=-4.054$ For $t = 0.09$, $E=-4.054$ For $t = 0.12$, $E=1.545$

Work Step by Step

We substitute each of the five values of $t$ in the formula to find the value of E for each value of $t$: For $t=0$, $E = 5 \cos 120\pi(0)=5 \cos 0\pi=5\cos 0=5(1)=5$ For $t = 0.03$, $E=5 \cos 120\pi(0.03)=5\cos 3.6\pi=5(0.3090)=1.545$ For $t = 0.06$, $E=5 \cos 120\pi(0.06)=5\cos 7.2\pi=5(-0.8090)=-4.054$ For $t = 0.09$, $E=5 \cos 120\pi(0.09)=5\cos 10.8\pi=5(-0.8090)=-4.054$ For $t = 0.12$, $E=5 \cos 120\pi(0.12)=5\cos 14.4\pi=5(0.3090)=1.545$
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