Precalculus (6th Edition) Blitzer

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

Chapter 4 - Review Exercises - Page 645: 62


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

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: $(2\pi - \theta)$ First, we make the angle positive: $-\dfrac{2 \pi}{3}+2\pi=\dfrac{4\pi}{3}$ We are in the 3rd quadrant, so subtract $\pi$: $\dfrac{4\pi}{3}-\pi=\dfrac{\pi}{3}$ Thus, we have $\csc \dfrac{\pi}{3}=- \dfrac{2 \sqrt 3}{3}$ The value is negative because initially we are in the 3rd qauadrant, where $\csc$ is negative.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.