Precalculus (6th Edition) Blitzer

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

Chapter 4 - Test - Page 647: 3

Answer

a) $\dfrac{4 \pi}{3}$ b ) $\dfrac{\pi}{3}$

Work Step by Step

The co-terminal and reference angle of an angle $0 \leq \theta \lt 2\pi $ based on its position can be computed by using the following steps: a) Quadrant- I: $\theta $ b) Quadrant -II: $(\pi-\theta)$ c) Quadrant- III: $(\theta - \pi)$ d) Quadrant - IV: $(2\pi - \theta)$ a) The positive angle is less than $2 \pi $; this means the co-terminal with $\dfrac{16 \pi}{3}$ is $\dfrac{16\pi}{6} -4 \pi =\dfrac{4 \pi}{3}$ b ) Thus, the reference angle of $\dfrac{16 \pi}{3}$ or, $960^{\circ}$ is $\dfrac{\pi}{3}$ or, $60^{\circ}$
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