Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

APPENDIX D - Trigonometry - D Exercises - Page A 32: 31

Answer

$sin \phi=\frac{\sqrt 5}{3}$ $cos \phi=-\frac{2}{3}$ $tan \phi=-\frac{\sqrt 5}{2}$ $csc \phi=\frac{3\sqrt 5}{5}$ $cot \phi=-\frac{2\sqrt 5}{5}$

Work Step by Step

Since $sec \phi=-1.5=-\frac{3}{2}$, we can label the hypotenuse side as having length 3 and the adjacent side as having length 2. The Pythagorean Theorem gives opposite side $=\sqrt {3^{2}-2^{2}}=\sqrt 5$ The other five trigonometric functions are given as follows: $sin \phi=\frac{\sqrt 5}{3}$ $cos \phi=-\frac{2}{3}$ $tan \phi=-\frac{\sqrt 5}{2}$ $csc \phi=\frac{3\sqrt 5}{5}$ $cot \phi=-\frac{2\sqrt 5}{5}$
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