Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 5 - Trigonometric Functions - Chapter Review - Review Exercises - Page 458: 34

Answer

$\dfrac{2}{3}$

Work Step by Step

The general form for the function can be expressed as: $y=A\sin \ (\omega x)$ where $A$ represents the amplitude. The $\omega$ can be found from the period by the formula: $\omega=\dfrac{2\pi}{T}$ and the phase shift is $\dfrac{\phi}{\omega}$. This means that $\phi=\omega \times \ Phase \ Shift$ We have: $A=|-2|=2$, $\omega=3 \pi$, Therefore, the period of the function can be computed as: $T=\dfrac{ 2 \pi}{\omega}=\dfrac{2 \pi}{3 \pi}=\dfrac{2}{3}$
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