Answer
(a) Reference Number: $\pi/6$
(b) Terminal Point: $(-\sqrt 3/2, -1/2)$
Work Step by Step
1. $31\pi/6 \gt 2\pi$. Subtract $2\pi$ from it until the result is less than $2\pi$.
$$\frac{31\pi}{6} - 2\pi = -\frac{31\pi}{6} - \frac{12\pi}{6} = \frac{19\pi}{6}$$ $$\frac{19\pi}{6} - 2\pi = \frac{7\pi}{6}$$
2. $7\pi/6$ is in Quadrant III, thus, the Reference Number is given by the equation:
$$t^- = \frac{7\pi}{6} - \pi = \frac{7\pi}{6} - \frac{6\pi}{6} = \frac{\pi} 6$$
3. According to the Table 1, the terminal point for $\pi/6$ is $(\sqrt 3/2, 1/2)$. In Quadrant III: x-coordinate is negative and y-coordinate is also negative. Therefore, the terminal point is: $(-\sqrt 3/2, -1/2)$