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: 18

Answer

$a.\qquad-\sqrt{2} \\$ $b. \displaystyle \qquad -\frac{1}{2} \\$ $c. \displaystyle \qquad -\frac{\sqrt{3}}{3}$

Work Step by Step

See p.412: Evaluating trig. functions for any number: 1. find the reference number, 2. Find the sign 3. Find the value (using the reference number and sign) ---------------------------------- $a.\displaystyle \qquad \frac{3\pi}{4}=\pi-\frac{\pi}{4}$ lies in quadrant II, reference number : $\displaystyle \frac{\pi}{4}$, with $\displaystyle \sec\frac{\pi}{4}=\sqrt{2} $ (table, p.410) In quadrant II, $\sec t$ is negative , (table, p.412) so $\displaystyle \sec\frac{3\pi}{4}=-\sqrt{2}.$ b$.\displaystyle \qquad -\frac{2\pi}{3}=-\pi+\frac{\pi}{3}$ lies in quadrant III, reference number : $\displaystyle \frac{\pi}{3}$, with $\displaystyle \cos\frac{\pi}{3}=\frac{1}{2} $ (table, p.410) In quadrant II, $\cos t$ is negative , (table, p.412) so $\displaystyle \cos(-\frac{2\pi}{3})=-\frac{1}{2}.$ $ c.\qquad -\displaystyle \frac{7\pi}{6}=-\pi-\frac{\pi}{6}$ lies in quadrant II, reference number : $\displaystyle \frac{\pi}{6}$, with $\displaystyle \tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}.$ (table, p.410) In quadrant II, $\tan t$ is negative (table, p.412), so $\displaystyle \tan(-\frac{7\pi}{6})=-\frac{\sqrt{3}}{3}.$
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