Answer
$\left(\sqrt2,-\sqrt2\right)$
Work Step by Step
By definition: $x=r\cdot \cos(\theta)$ and $y=r\cdot \sin(\theta)$.
Here $r=-2, \theta=\frac{3\pi}{4}$.
Hence, after plugging these in:
$x=-2\cdot \cos(\frac{3\pi}{4})=-2\cdot\left(-\frac{\sqrt2}{2}\right)=\sqrt2$
$y=-2\cdot \sin(\frac{3\pi}{4})=-2\cdot\frac{\sqrt2}{2}=-\sqrt2.$
The point has coordinates $(\sqrt2,-\sqrt2)$ in the rectangular coordinate system.