Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.5 - Graphs of Sine and Cosine Functions - Exercise Set - Page 595: 8

Answer

The amplitude and period of the function are $1$ and $\frac{\pi }{2}$, respectively.

Work Step by Step

We have the trigonometric function $y=\sin 4x$ The amplitude is the maximum value of $y$. The maximum value of the given trigonometric function is $1$. So, the amplitude of the trigonometric function is $1$. The function after a certain interval starts repeating itself, and this interval is known as the period of the function. The period $\left( B \right)$ of the trigonometric function is $\begin{align} & B=\frac{2\pi }{4} \\ & =\frac{\pi }{2} \end{align}$ And the quarter period is $\frac{\left( \frac{\pi }{2} \right)}{\left( 4 \right)}$ or $\frac{\pi }{8}$. 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 $\begin{align} & {{x}_{2}}=0+\frac{\pi }{8} \\ & =\frac{\pi }{8} \end{align}$ Third key point is $\begin{align} & {{x}_{3}}=\frac{\pi }{8}+\frac{\pi }{8} \\ & =\frac{\pi }{4} \end{align}$ Fourth key point is $\begin{align} & {{x}_{4}}=\frac{\pi }{4}+\frac{\pi }{8} \\ & =\frac{3\pi }{8} \end{align}$ Fifth key point is $\begin{align} & {{x}_{5}}=\frac{3\pi }{8}+\frac{\pi }{8} \\ & =\frac{\pi }{2} \end{align}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.