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: 9

Answer

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

Work Step by Step

Given, point (-1, -2) is on the terminal side of $\theta$, we may apply Definition I to find all six trigonometric functions- We got $ x = -1, y = -2$ Therefore r= $\sqrt (x^{2} + y^{2})$ = $\sqrt ((-1)^{2} + (-2)^{2})$ = $\sqrt (1 + 4)$ = $\sqrt 5$ i.e. $ x = -1, y = -2,$ and $ r= = \sqrt 5$ Applying Definition I- $\sin\theta$ =$ \frac{y}{r}$ = $ \frac{-2}{\sqrt 5}$ = - $ \frac{2}{\sqrt 5}$ $\cos\theta$ =$ \frac{x}{r}$ =$ \frac{-1}{\sqrt 5}$ = - $ \frac{1}{\sqrt 5}$ $\tan\theta$ =$ \frac{y}{x}$ =$ \frac{-2}{-1}$ = 2 $\cot\theta$ =$ \frac{x}{y}$ =$ \frac{-1}{-2}$ = $ \frac{1}{2}$ $\sec\theta$ =$ \frac{r}{x}$ =$ \frac{\sqrt 5}{-1}$ = -$ \sqrt 5$ $\csc\theta$ =$ \frac{r}{y}$ =$ \frac{\sqrt 5}{-2}$ = -$ \frac{\sqrt 5}{2}$
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