Answer
(a) Reference number: $\pi/4$
(b) Terminal point: $(-\sqrt 2/2, -\sqrt 2/2)$
Work Step by Step
1. $13\pi/4 \gt 2\pi$. Subtract $2\pi$ from it until the result is less than $2\pi$.
$$\frac{13\pi}{4} - 2\pi = \frac{13\pi}{4} -\frac{8\pi}{4} = \frac{5\pi}{4}$$
2. $5\pi/4$ is in Quadrant III, thus, the Reference Number is given by the equation:
$$t^- = \frac{5\pi}{4} - \pi = \frac{\pi}{4}$$
3. According to the Table 1, the terminal point for $\pi/4$ is $(\sqrt 2/2, \sqrt 2/2)$. Since the terminal point must be in Quadrant III, the x-coordinate is negative, and the y-coordinate is also negative. Therefore, the terminal point for $t = 13\pi/4$ is $(-\sqrt 2/2, -\sqrt 2/2)$
