Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 1 - Section 1.3 - Definition I: Trigonometric Functions - 1.3 Problem Set - Page 31: 6

Answer

$\sin\theta$ = - $ \frac{4}{5}$ $\cos\theta$ = -$ \frac{3}{5}$ $\tan\theta$ = $ \frac{4}{3}$ $\cot\theta$ = $ \frac{3}{4}$ $\sec\theta$ = -$ \frac{5}{3}$ $\csc\theta$ = -$ \frac{5}{4}$

Work Step by Step

Given, point (-3, -4) is on the terminal side of $\theta$, we may apply Definition I to find all six trigonometric functions- We got $ x = -3, y = -4$ Therefore r= $\sqrt (x^{2} + y^{2})$ = $\sqrt ((-3)^{2} + (-4)^{2})$ = $\sqrt (9 + 16)$ = $\sqrt (25)$ = 5 i.e. $ x = -3, y = -4,$ and $ r= 5$ Applying Definition I- $\sin\theta$ =$ \frac{y}{r}$ = $ \frac{-4}{5}$ = - $ \frac{4}{5}$ $\cos\theta$ =$ \frac{x}{r}$ =$ \frac{-3}{5}$ = -$ \frac{3}{5}$ $\tan\theta$ =$ \frac{y}{x}$ =$ \frac{-4}{-3}$ = $ \frac{4}{3}$ $\cot\theta$ =$ \frac{x}{y}$ =$ \frac{-3}{-4}$ = $ \frac{3}{4}$ $\sec\theta$ =$ \frac{r}{x}$ =$ \frac{5}{-3}$ = -$ \frac{5}{3}$ $\csc\theta$ =$ \frac{r}{y}$ =$ \frac{5}{-4}$= -$ \frac{5}{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.