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: 21



Work Step by Step

Figure 13 on page 579 of this book shows that the unit circle point $(-\frac{\sqrt2}{2}, \frac{\sqrt2}{2})$ corresponds to the real number $\dfrac{3\pi}{4}$. RECALL: (i) $\cos{s} = x$ (ii) $\sin{s} = y$ (iii) $\tan{s}=\dfrac{y}{x}$ Use the coordinates of the unit circle point above and the formula in (iii) above to obtain: $\color{blue}{\tan{(\frac{3\pi}{4})}=\frac{\frac{\sqrt2}{2}}{-\frac{\sqrt2}{2}}=-1}$
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.