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Graph each of the (oriented) angles below in standard position and classify them according to where their terminal side lies. Find three coterminal angles, at least one of which is positive and one of which is negative.

1. \(\alpha=\frac{\pi}{6}\)
2. \(\beta=-\frac{4 \pi}{3}\)
3. \(\gamma=\frac{9 \pi}{4}\)
4. \(\varphi=-\frac{5 \pi}{2}\)
in Mathematics by Platinum (132,156 points) | 170 views

1 Answer

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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 \(\frac{7 \pi}{2}\)

 

 

 

by Platinum (132,156 points)

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