Calculus: Early Transcendentals 8th Edition

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

APPENDIX D - Trigonometry - D Exercises: 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: $sin \phi=\frac{\sqrt 5}{3}$ $cos \phi=-\frac{2}{3}$ $tan \phi=-\frac{\sqrt 5}{2}$ $csc \phi=\frac{3\sqrt 5}{5}$ $cot \phi=-\frac{2\sqrt 5}{5}$
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.