Precalculus (6th Edition) Blitzer

Published by Pearson

Chapter 4 - Section 4.5 - Graphs of Sine and Cosine Functions - Exercise Set - Page 595: 3

Answer

The amplitude of the function is $\frac{1}{3}$.

Work Step by Step

We have the trigonometric function $y=\frac{1}{3}\sin x$ The amplitude is the maximum value of function. The maximum value of the given trigonometric function is $\frac{1}{3}$. So, the amplitude of the trigonometric function is $\frac{1}{3}$. The period of the trigonometric function is $2\pi$. The quarter period is $\frac{2\pi }{4}$ or $\frac{\pi }{2}$. The cycle begins at $x=0$. Add quarter periods to find out the key points. First key point is ${{x}_{1}}=0$ Second key point is \begin{align} & {{x}_{2}}=0+\frac{\pi }{2} \\ & =\frac{\pi }{2} \end{align} Third key point is \begin{align} & {{x}_{3}}=\frac{\pi }{2}+\frac{\pi }{2} \\ & =\pi \end{align} Fourth key point is \begin{align} & {{x}_{4}}=\pi +\frac{\pi }{2} \\ & =\frac{3\pi }{2} \end{align} Fifth key point is \begin{align} & {{x}_{5}}=\frac{3\pi }{2}+\frac{\pi }{2} \\ & =2\pi \end{align}

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