Chapter 8 - Triangle Trigonometry and Polar Coordinates - 8.3 Polar Coordinates - Exercises and Problems for Section 8.3 - Exercises and Problems - Page 347: 20

$$\left(-\sqrt{\frac{3}{2}},\:-\sqrt{\frac{3}{2}}\right)$$

Converting the polar coordinates to Cartesian coordinates, we find: $$x=\sqrt{3}\cos \left(\left(-\frac{3\pi }{4}\right)\right)= -\sqrt{\frac{3}{2}} \\ y=\sqrt{3}\sin \left(\left(-\frac{3\pi }{4}\right)\right) = -\sqrt{\frac{3}{2}} \\ \left(-\sqrt{\frac{3}{2}},\:-\sqrt{\frac{3}{2}}\right)$$