## Trigonometry 7th Edition

Published by Cengage Learning

# Chapter 1 - Section 1.3 - Definition I: Trigonometric Functions - 1.3 Problem Set - Page 32: 19

#### Answer

$\sin\theta$ = $\frac{6}{\sqrt 85}$ $\cos\theta$ =$\frac{-7}{\sqrt 85}$ $\tan\theta$ = - $\frac{6}{7}$

#### 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 (-7, 6) Now, we may apply Definition I to find required trigonometric functions- We got $x = -7, y = 6$ Therefore r= $\sqrt (x^{2} + y^{2})$ = $\sqrt ((-7)^{2} + 6^{2})$ = $\sqrt (49 + 36)$ = $\sqrt 85$ i.e. $x = -7, y = 6,$ and $r= \sqrt 85$ Applying Definition I- $\sin\theta$ =$\frac{y}{r}$ = $\frac{6}{\sqrt 85}$ $\cos\theta$ =$\frac{x}{r}$ =$\frac{-7}{\sqrt 85}$ $\tan\theta$ =$\frac{y}{x}$ =$\frac{6}{-7}$ = - $\frac{6}{7}$

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