Answer
$\sin \theta=\frac{\sqrt 2}{2}$
$\cos \theta=\frac{-\sqrt 2}{2}$
Work Step by Step
We can consider an arbitrary point that satisfies y=-x and lies in quadrant II. Let us consider the point (-1,1).
Then, $r=\sqrt {x^{2}+y^{2}}=\sqrt {(-1)^{2}+1^{2}}=\sqrt 2$
$\sin \theta=\frac{y}{r}=\frac{1}{\sqrt 2}=\frac{\sqrt 2}{2}$
$\cos \theta=\frac{x}{r}=\frac{-1}{\sqrt 2}=\frac{-\sqrt 2}{2}$
