Answer
$\sin\theta$ = - $ \frac{1}{\sqrt 10}$
$\cos\theta$ = - $ \frac{3}{\sqrt 10}$
$\tan\theta$ = $ \frac{1}{3}$
Work Step by Step
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}$
