Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 6 - Trigonometric Functions - 6.4 Graphs of the Sine and Cosine Functions - 6.4 Assess Your Understanding - Page 408: 19

Answer

amplitude: $\frac{1}{2}$ period: $\frac{4\pi}{3}$

Work Step by Step

If a function has form $y=A\cos(\omega x)$, then there $|A|$ stands for the amplitude. First we identify $\omega$ and $A$ from the form of $y=A\cos{(\omega x)}$. Hence, $A=\left|-\frac{1}{2}\right|=\frac{1}{2}$, $\omega=1.5.$ To get the period $T$ we can use the formula: $T=\frac{2\pi}{\omega}$ to obtain $T=\dfrac{2\pi}{\frac{3}{2}}=\dfrac{4\pi}{3}$
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