Precalculus (6th Edition) Blitzer

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

Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set - Page 576: 117

Answer

$\frac{1}{2}$ $0$ $-\frac{1}{2}$ $0$ $\frac{1}{2}$
1583786091

Work Step by Step

Step 1. Given the function $y=f(x)=\frac{1}{2}cos(4x+\pi)$, we have $f(-\frac{\pi}{4})=\frac{1}{2}cos(4(-\frac{\pi}{4})+\pi)=\frac{1}{2}cos(0)=\frac{1}{2}$ Step 2. Similarly, we have $f(-\frac{\pi}{8})=\frac{1}{2}cos(4(-\frac{\pi}{8})+\pi)=\frac{1}{2}cos(\frac{\pi}{2})=0$ Step 3. We have $f(0)=\frac{1}{2}cos(4(0)+\pi)=\frac{1}{2}cos(\pi)=-\frac{1}{2}$ Step 4. We have $f(\frac{\pi}{8})=\frac{1}{2}cos(4(\frac{\pi}{8})+\pi)=\frac{1}{2}cos(\frac{3\pi}{2})=0$ Step 5. Finally, we have $f(\frac{\pi}{4})=\frac{1}{2}cos(4(\frac{\pi}{4})+\pi)=\frac{1}{2}cos(2\pi)=\frac{1}{2}$
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