## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$(4,\frac{4\pi}{3})$
Solve for $r$ using the formula $r=\sqrt {x^{2}+y^{2}}$ to obtain: $$r=\sqrt {(-2)^{2}+(-2\sqrt {3})^{2}}=4$$ Solve for the angle using the formula $\alpha= \tan^{-1}(\frac{y}{x})$ to obtain: $$\alpha=\tan^{-1}\left(\frac{-2\sqrt {3}}{-2}\right)=\frac{\pi}{3}$$ Since $(-2, -2\sqrt {3})$ is in the third quadrant, then the angle measurement is: $$\theta=\alpha+\pi=\frac{\pi}{3}+\pi=\frac{4\pi}{3}$$ Thus, the polar coordinates for the point $(-2, -2\sqrt {3})$ can be given by $(4,\frac{4\pi}{3})$.