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 575: 14



Work Step by Step

Here, $ y= -1$; $ x=0$ and $ r=\sqrt {(0)^2+(-1)^2}=1$ The trigonometric ratios are as follows: $\cos \theta =\dfrac{x}{r}$ Then, we have $\cos (3\pi/2) =\dfrac{0}{-1}=0$
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