1. The angle \(\alpha=\frac{\pi}{6}\) is positive, so we draw an angle with its initial side on the positive \(x\)-axis and rotate counter-clockwise \(\frac{(\pi / 6)}{2 \pi}=\frac{1}{12}\) of a revolution. Thus \(\alpha\) is a Quadrant I angle.

Coterminal angles \(\theta\) are of the form \(\theta=\alpha+2 \pi k\), for some integer \(k\). To make the arithmetic a bit easier, we note that \(2 \pi=\frac{12 \pi}{6}\)

- Thus, when \(k=1\), we get

\[

\begin{aligned}

\theta &=\frac{\pi}{6}+\frac{12 \pi}{6} \\

&=\frac{13 \pi}{6}

\end{aligned}

\]

- Substituting \(k=-1\) gives

\[

\begin{aligned}

\theta &=\frac{\pi}{6}-\frac{12 \pi}{6} \\

&=-\frac{11 \pi}{6}

\end{aligned}

\]

- When we let \(k=2\), we get

\[

\begin{aligned}

\theta &=\frac{\pi}{6}+\frac{24 \pi}{6} \\

&=\frac{25 \pi}{6}

\end{aligned}

\]

2. Since \(\beta=-\frac{4 \pi}{3}\) is negative, we start at the positive \(\mathrm{x}\)-axis and rotate clockwise \(\frac{(4 \pi / 3)}{2 \pi}=\frac{2}{3}\) of a revolution. We find \(\beta\) to be a Quadrant II angle.

To find coterminal angles, we proceed as before using \(2 \pi=\frac{6 \pi}{3}\), and compute \(\theta=-\frac{4 \pi}{3}+\frac{6 \pi}{3} k\) for integer values of \(k .\) We obtain \(\frac{2 \pi}{3},-\frac{10 \pi}{3}\) and \(\frac{8 \pi}{3}\) as coterminal angles.

3. Since \(\gamma=\frac{9 \pi}{4}\) is positive, we rotate counter-clockwise from the positive \(x\)-axis. One full revolution accounts for \(2 \pi=\frac{8 \pi}{4}\) of the radian measure with \(\frac{\pi}{4}\) or \(\frac{1}{8}\) of a revolution remaining. We have \(\gamma\) as a Quadrant I angle.

All angles coterminal with \(\gamma\) are of the form \(\theta=\frac{9 \pi}{4}+\frac{8 \pi}{4} k\), where \(k\) is an integer. Working through the arithmetic, we find coterminal angles of \(\frac{\pi}{4},-\frac{7 \pi}{4}\) and \(\frac{17 \pi}{4}\).4. To graph \(\varphi=-\frac{5 \pi}{2}\), we begin our rotation clockwise from the positive \(\mathrm{x}\)-axis. As \(2 \pi=\frac{4 \pi}{2}\), after one full revolution clockwise we have \(\frac{\pi}{2}\) or \(\frac{1}{4}\) of a revolution remaining. Since the terminal side of \(\varphi\) lies on the negative \(y\)-axis, \(\varphi\) is a quadrantal angle.

To find coterminal angles, we compute \(\theta=-\frac{5 \pi}{2}+\frac{4 \pi}{2} k\) for a few integers \(k\) and obtain \(-\frac{\pi}{2}\), \(\frac{3 \pi}{2}\) and \(\frac{7 \pi}{2}\)