Precalculus (6th Edition)

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

Chapter 6 - The Circular Functions and Their Graphs - 6.2 The Unit Circle and Circular Functions - 6.2 Exercises - Page 587: 5

Answer

$1$

Work Step by Step

$\frac{\pi}{4}=60^o$ is a special angle that intersects the unit circle at the point $(\frac{\sqrt2}{2}, \frac{\sqrt3}{2})$. Since $\tan{\theta}=\frac{y}{x}$, then $\tan{\frac{\pi}{4}} =\dfrac{\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}=1$
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.