## Algebra and Trigonometry 10th Edition

(a) See the graph. (b) Quadrant I. (c) $\frac{20\pi}{9}$ and $-\frac{16\pi}{9}$
(b) $0\lt\frac{2\pi}{9}\lt\frac{\pi}{2}$, therefore, it lies in quadrant I. (c) $\frac{2\pi}{9}+2\pi=\frac{20\pi}{9}$ $\frac{2\pi}{9}-2\pi=-\frac{16\pi}{9}$