Answer
(a) Reference Number: $\pi/6$
(b) Terminal Point: $(-\sqrt 3/2, 1/2)$
Work Step by Step
1. $41\pi/6 \gt 2\pi$. Subtract $2\pi$ from it until the result is less than $2\pi$.
$$\frac{41\pi}{6} - 2\pi = \frac{41\pi}{6} -\frac{12\pi}{6} = \frac{29\pi}{6}$$ $$\frac{29\pi}{6} - 2\pi = \frac{29\pi}{6} -\frac{12\pi}{6} = \frac{17\pi}{6}$$ $$\frac{17\pi}{6} - 2\pi = \frac{17\pi}{6} -\frac{12\pi}{6} = \frac{5\pi}{6}$$
2. $5\pi/6$ is in Quadrant II, thus, the Reference Number is given by the equation:
$$t^- = \pi - \frac{5\pi}{6} = \frac{6\pi}{6} - \frac{5\pi}{6} = \frac {\pi} 6$$
3. According to the Table 1, the terminal point for $\pi/6$ is $(\sqrt 3/2, 1/2)$. Since the terminal point is in Quadrant II, the x-coordinate is negative, and the y-coordinate is positive. Therefore, the terminal point for $t = 41\pi/6$ is $(-\sqrt 3/2, 1/2)$
