## Trigonometry 7th Edition

$\sin\theta$ = - $\frac{1}{\sqrt 10}$ $\cos\theta$ = - $\frac{3}{\sqrt 10}$ $\tan\theta$ = $\frac{1}{3}$
Given $\theta$ is in standard position. Spotting a Point P on terminal side of $\theta$, in the given diagram- We find point P (-6, -2) Now, we may apply Definition I to find required trigonometric functions- We got $x = -6, y = -2$ Therefore r= $\sqrt (x^{2} + y^{2})$ = $\sqrt ((-6)^{2} + (-2)^{2})$ = $\sqrt (36+4)$ = $\sqrt 40$ = $2\sqrt 10$ i.e. $x = -6, y = -2,$ and $r= 2\sqrt 10$ Applying Definition I- $\sin\theta$ =$\frac{y}{r}$ = $\frac{-2}{2\sqrt 10}$ = - $\frac{1}{\sqrt 10}$ $\cos\theta$ =$\frac{x}{r}$ =$\frac{-6}{2\sqrt 10}$ = - $\frac{3}{\sqrt 10}$ $\tan\theta$ =$\frac{y}{x}$ =$\frac{-2}{-6}$ = $\frac{1}{3}$