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 32: 20

Answer

$\sin\theta$ = $ \frac{-8}{\sqrt 73}$ $\cos\theta$ =$ \frac{-3}{\sqrt 73}$ $\tan\theta$ = $ \frac{8}{3}$

Work Step by Step

Given $\theta$ is in standard position. Finding a Point P on terminal side of $\theta$, in the given diagram- We find point P (-3, -8) Now, we may apply Definition I to find required trigonometric functions- We got $ x = -3, y = -8$ Therefore r= $\sqrt (x^{2} + y^{2})$ = $\sqrt ((-3)^{2} + (-8)^{2})$ = $\sqrt (9 + 64)$ = $\sqrt 73$ i.e. $ x = -3, y = -8,$ and $ r= \sqrt 73$ Applying Definition I- $\sin\theta$ =$ \frac{y}{r}$ = $ \frac{-8}{\sqrt 73}$ $\cos\theta$ =$ \frac{x}{r}$ =$ \frac{-3}{\sqrt 73}$ $\tan\theta$ =$ \frac{y}{x}$ =$ \frac{-8}{-3}$ = $ \frac{8}{3}$
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