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}$
