Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 5 - Trigonometric Functions - 5.3 Trigonometric Functions Values and Angle Measures - 5.3 Exercises - Page 533: 99

Answer

$120^{\circ}$ and $300^{\circ}$

Work Step by Step

The value of $\tan \theta$ is negative, so $\theta$ may lie in either quadrant II or IV. We know that $\tan 60^{\circ}=\sqrt 3$. Recall: $-\tan x= \tan(180^{\circ}-x)$ and $-\tan x= \tan(360^{\circ}-x)$ $\implies -\tan 60^{\circ}=-\sqrt {3}=\tan(180^{\circ}-60^{\circ})$ $=\tan 120^{\circ}$ $120^{\circ}$ which is in the second quadrant is one value of $\theta$. Now $-\tan 60^{\circ}=-\sqrt 3= \tan (360^{\circ}-60^{\circ})$ $=\tan 300^{\circ}$ $300^{\circ}$ is in the fourth quadrant. The values of $\theta$ are $ 120^{\circ}$ and $300^{\circ}$.
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.