Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 1 - Trigonometric Functions - Section 1.3 Trigonometric Functions - 1.3 Exercises - Page 26: 45

Answer

$\sin\theta=\frac{y}{r}=\frac{-4}{2\sqrt 5}=\frac{-2\sqrt 5}{5}$ $\cos\theta=\frac{x}{r}=\frac{2}{2\sqrt 5}=\frac{\sqrt 5}{5}$ $\tan\theta=\frac{y}{x}=\frac{-4}{2}=-2$ $\csc\theta=\frac{1}{\sin\theta}=-\frac{\sqrt 5}{2}$ $\sec\theta=\frac{1}{\cos\theta}=\sqrt 5$ $\cot\theta=\frac{1}{\tan\theta}=-\frac{1}{2}$

Work Step by Step

$x=2$ $y=-4$ $r=\sqrt {(2)^{2}+(-4)^{2}}=\sqrt {4+16}=\sqrt {20}= 2\sqrt 5$
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.