Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 5 - Section 5.2 - Trigonometric Functions of Real Numbers - 5.2 Exercises - Page 417: 17

Answer

$a.\qquad-2 \\$ $b. \displaystyle \qquad \frac{2\sqrt{3}}{3} \\$ $c. \qquad \sqrt{3}$

Work Step by Step

See p.412: Evaluating trig. functions for any number: 1. find the reference number, 2. Find the sign , use table on p.412, 3. Find the value (using the reference number and sign) ---------------------------------- $a.\displaystyle \qquad \frac{7\pi}{6}=\pi+\frac{\pi}{6}$ lies in quadrant III, reference number : $\displaystyle \frac{\pi}{6}$, with $\displaystyle \csc\frac{\pi}{6}=2$ (table, p.410) In quadrant III, csc is negative (see table on p.412), so $\displaystyle \csc\frac{7\pi}{6}=-2.$ $b. \displaystyle \qquad -\frac{\pi}{6}=0-\frac{\pi}{6}$ lies in quadrant IV, reference number : $\displaystyle \frac{\pi}{6}$, with $\displaystyle \sec\frac{\pi}{6}=\frac{2\sqrt{3}}{3}.$ (table, p.410) In quadrant IV, sec is positive (see table on p.412), so $\displaystyle \sec(-\frac{\pi}{6})=\frac{2\sqrt{3}}{3}$ $c. \displaystyle \qquad -\frac{\pi}{6}=-\pi+\frac{\pi}{6}$ lies in quadrant III, reference number : $\displaystyle \frac{\pi}{6}$, with $\displaystyle \cot\frac{\pi}{6}=\sqrt{3}.$ (table, p.410). In quadrant III, $\cot t$ is positive (see table on p.412), so $\displaystyle \cot(-\frac{\pi}{6})=\sqrt{3}$
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.