Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 3 - Radian Measure and the Unit Circle - Section 3.3 The Unit Circle and Circular Functions - 3.3 Exercises - Page 117: 16

Answer

$-\sqrt2$

Work Step by Step

RECALL: $\sin{s} = y \\\cos{s} = x \\\tan{s} = \frac{y}{x} \\\cot{s} = \frac{x}{y} \\\sec{s} = \frac{1}{x} \\\csc{s}=\frac{1}{y}$ (refer to Figure 11 , page 111 of the textbook) The angle $\frac{5\pi}{4}$ intersects the unit circle at the point $(-\frac{\sqrt2}{2}, -\frac{\sqrt2}{2})$. This point has: $x= -\frac{\sqrt2}{2}$ $y=-\frac{\sqrt2}{2}$ Thus, $\sec{\frac{5\pi}{4}} \\= \frac{1}{x} \\=\frac{1}{-\frac{\sqrt2}{2}} \\=1 \cdot -\frac{2}{\sqrt2} \\=-\frac{2}{\sqrt2}$ Rationalize the denominator by multiplying $\sqrt2$ to both the numerator and the denominator to obtain: $=-\frac{2}{\sqrt2} \cdot \frac{\sqrt2}{\sqrt2} \\=-\frac{2\sqrt2}{2} \\=-\sqrt2 $
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.