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: 113

Answer

makes sense

Work Step by Step

This statement makes sense as it describes the correct procedure in finding the exact value of the function. The angle $\theta=\frac{14\pi}{3}=4\pi+\frac{2\pi}{3}$ is in quadrant II with a reference angle of $\pi-\frac{2\pi}{3}=\frac{\pi}{3}$ and since $cos\frac{\pi}{3}=\frac{1}{2}$ and $cos\theta\lt0$, we have $cos\theta=-\frac{1}{2}$.
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