## Precalculus (6th Edition) Blitzer

$(-\sqrt 2,\sqrt 2)$
Step 1. See plot; move to $\theta=-\frac{\pi}{4}$ then use $r=-2$ to get the point. Step 2. Using the formulas $x=r\ cos\theta$ and $y=r\ sin\theta$, we have $x=-2cos(-\frac{\pi}{4})=-\sqrt 2$ and $y=-2sin(-\frac{\pi}{4})=\sqrt 2$, giving a point $(-\sqrt 2,\sqrt 2)$