Calculus: Early Transcendentals 8th Edition

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

APPENDIX D - Trigonometry - D Exercises: 32

Answer

$sinx=-\frac{2\sqrt 2}{3}$ $tan x=2\sqrt 2$ $cscx=-\frac{3}{2\sqrt 2}=-\frac{3\sqrt 2}{4}$ $sec x=-3$ $cot x=\frac{1}{2\sqrt 2}=\frac{\sqrt 2}{4}$

Work Step by Step

Since $cosx=-\frac{1}{3}$, we can label the adjacent side as having length 1 and the hypotenuse as having length 3. Then Pythagorean Theorem gives opposite side $=\sqrt {3^{2}-1^{2}}=2\sqrt 2$ The other five trigonometric functions are given as follows: $sinx=-\frac{2\sqrt 2}{3}$ $tan x=2\sqrt 2$ $cscx=-\frac{3}{2\sqrt 2}=-\frac{3\sqrt 2}{4}$ $sec x=-3$ $cot x=\frac{1}{2\sqrt 2}=\frac{\sqrt 2}{4}$
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.