Answer
(a) See the graph.
(b) Quadrant II
(c) $\frac{2\pi}{3}$ and $-\frac{10\pi}{3}$
![](https://gradesaver.s3.amazonaws.com/uploads/solution/52a8468c-0051-4c42-bb30-90184bd4479f/result_image/1585083240.jpg?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAJVAXHCSURVZEX5QQ%2F20240727%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240727T015719Z&X-Amz-Expires=900&X-Amz-SignedHeaders=host&X-Amz-Signature=dc583776b793fe009a29547116a1d9393e6e2af47e41a2da5e21aa5457ec8c9d)
Work Step by Step
(b) A positive coterminal angle of $-\frac{4\pi}{3}$ is $\frac{2\pi}{3}$
$\frac{\pi}{2}\lt\frac{2\pi}{3}\lt\pi$, therefore, it lies in quadrant II.
(c)
$-\frac{4\pi}{3}+2\pi=\frac{2\pi}{3}$
$-\frac{4\pi}{3}-2\pi=-\frac{10\pi}{3}$