Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 6 - Trigonometric Functions - 6.3 Properties of the Trigonometric Functions - 6.3 Assess Your Understanding - Page 394: 25


$\dfrac {\sqrt {3}}{3}$

Work Step by Step

$\tan \alpha =\tan \left( \alpha -\pi k\right) \\\Rightarrow \tan \dfrac {19\pi }{6}=\tan \left( \dfrac {19\pi }{6}-3\pi \right) \\=\tan \dfrac {\pi }{6} \\=\dfrac {\sin \frac {\pi }{6}}{\cos \frac {\pi }{6}} \\=\dfrac {\frac {1}{2}}{\frac {\sqrt {3}}{2}} \\=\dfrac {\sqrt {3}}{3}$
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.