Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 6 - 6.6 - Inverse Trigonometric Functions - 6.6 Exercises - Page 485: 59


$tan[arccos(-\frac{2}{3})]=tan~u=-\frac{\sqrt 5}{2}$

Work Step by Step

Let $u=arccos(-\frac{2}{3})$. Then: $cos~u=-\frac{2}{3}$ The range of $arccos~x$ is $0\leq x\leq \pi$. So, since $arccos(-\frac{2}{3})\gt\frac{\pi}{2}$, then $\frac{\pi}{2}\lt u\lt \pi$ (Second Quadrant) $cos^2u+sin^2u=1$ $sin^2u=1-cos^2u$ $sin^2u=1-cos^2u$ $sin^2u=1-(-\frac{2}{3})^2=1-\frac{4}{9}=\frac{5}{9}$ $sin~u=\frac{\sqrt 5}{3}~~$ (Second Quadrant) $tan[arccos(-\frac{2}{3})]=tan~u=\frac{sin~u}{cos~u}=\frac{\frac{\sqrt 5}{3}}{-\frac{2}{3}}=-\frac{\sqrt 5}{2}$
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