Answer
a. $\quad \displaystyle \overline{t}= \frac{\pi}{6}$
b. $\displaystyle \quad P(- \frac{\sqrt{3}}{2}, \displaystyle \frac{1}{2})$
Work Step by Step
The reference number associated with the real number $t$ is the shortest distance along the unit circle between the terminal point determined by $t$ and the x-axis.
For t, find its terminal point on the unit circle (positive=counterclockwise) and associate it with the terminal point of some t between 0 and $ 2\pi$
If the terminal point "lands" in quadrants II, III or IV,
choose the symmetric terminal number in quadrant I$:$
t in Q.II $\Rightarrow \overline{t}=\pi-t$
t in Q.III$\Rightarrow \overline{t}=t-\pi$
t in Q.IV$\Rightarrow \overline{t}=2\pi-t$
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The terminal point of $-\displaystyle \frac{7\pi}{6}=-\pi-\frac{\pi}{6}$ is in Q.II (clockwise, $-\pi=- \displaystyle \frac{6\pi}{6}$),
the same as the terminal point of $\displaystyle \pi-\frac{\pi}{6}=\frac{5\pi}{6}$
its reference number is $\displaystyle \pi-\frac{5\pi}{6}=\frac{\pi}{6} $
The terminal point for $\displaystyle \frac{\pi}{6}$ is $( \displaystyle \frac{\sqrt{3}}{2}, \displaystyle \frac{1}{2})$ ... Table 1, fig.6.
In Q.II, the x-coordinate is negative, the y positive, so by symmetry
the terminal point for $-\displaystyle \frac{7\pi}{6}$ is $(- \displaystyle \frac{\sqrt{3}}{2}, \displaystyle \frac{1}{2})$
