Precalculus (6th Edition) Blitzer

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

Chapter 4 - Mid-Chapter Check Point - Page 578: 22

Answer

$\dfrac{-2\sqrt 3}{3}$

Work Step by Step

The 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: $(2\pi - \theta)$ We can see that the angle $\dfrac{22\pi}{3}$ is close to $\dfrac{21\pi}{3}=7\pi $. Thus, the reference angle of $\dfrac{22\pi}{3}$ is: $ \dfrac{22\pi}{3}-7 \pi =\dfrac{22\pi}{3} - \dfrac{21\pi}{3}=\dfrac{\pi}{3}$ $\implies \csc \dfrac{\pi}{3} =\dfrac{2\sqrt 3}{3}$ So, $\csc (\dfrac{22\pi}{3})=\dfrac{-2\sqrt 3}{3}$ because $\theta $ lies in Quadrant-III.
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