Answer
$\sin\theta$ = - $ \frac{2}{\sqrt 5}$
$\cos\theta$ =$ \frac{1}{\sqrt 5}$
Work Step by Step
Given, terminal side of $\theta$ lies along the line-
y = - 2x in QIV
To find a point on this terminal side, let's substitute an arbitrary value of 'x' , i.e. x = 1 in above equation, we get-
y = -2
Therefore point (1, -2) in Q IV lies on the terminal side of $\theta$ so we may apply Definition I to find required trigonometric functions-
We got $ x = 1, y = -2$
Therefore r= $\sqrt (x^{2} + y^{2})$
= $\sqrt ((1)^{2} + (-2)^{2})$
= $\sqrt (1 + 4)$
= $\sqrt (5)$
i.e. $ x = 1, y = -2, $ and $ r= \sqrt 5$
Applying Definition I-
$\sin\theta$ =$ \frac{y}{r}$ = $ \frac{-2}{\sqrt 5}$ = - $ \frac{2}{\sqrt 5}$
$\cos\theta$ =$ \frac{x}{r}$ =$ \frac{1}{\sqrt 5}$
