Solution: Coterminal angles for \(60^\circ:\) \(420^\circ\) and \(-300^\circ\)

 Coterminal angles for\(\dfrac{\pi}{3}.\) and \(\frac{7\pi}{3}\) 

 Note: \(60^\circ\) and \(\dfrac{\pi}{3}\) are the same angle. Also, these are two coterminal angles, but there are many other coterminal angles.

Solution method: 
Coterminal Angle: Angles in standard position (with the initial side on the positive x-axis) that have a common terminal side. Coterminal angles differ by multiples of \(360^\circ\) or 2\(\pi\) in radians

First, we draw \(60^\circ\) in standard position. Standard position means the angle's initial side is the positive x-axis, so we go up \(60^\circ\) from the positive x-axis, and draw the terminal side. 

Now, coterminal angles have the same terminal side. So if I go around another \(360^\circ\) counterclockwise, I'll end up back here at the same terminal side as \(60^\circ.\) So that means that the initial \(60^\circ\) plus the \(360^\circ\) rotation, \(60^\circ+360^\circ= 420^\circ\) is a positive coterminal angle to \(60^\circ.\) 

To get a negative coterminal angle to \(60^\circ\) we go clockwise (negative rotation) \(360^\circ.\) In other words, we subtract \(360^\circ.\) So \(60^\circ-360^\circ=-300^\circ\) is a negative coterminal angle to \(60^\circ.\)

To find a positive and negative coterminal angle in radians, we add and subtract \(2\pi\). 

To convert \(60^\circ\) to radians:\begin{align*}
        60^\circ &= 60 \times \frac{\pi}{180} \text{ radians}\\ 
        &= \frac{60\pi}{180} \text{ radians}\\
        &= \frac{\pi}{3} \text{ radians}
    \end{align*}So the coterminal angles to \(\frac{\pi}{3}\) are\begin{align*}
        \frac{\pi}{3} + 2\pi &= \frac{\pi}{3} + \frac{6\pi}{3} = \frac{7\pi}{3} \\
        \frac{\pi}{3} - 2\pi &= \frac{\pi}{3} - \frac{6\pi}{3} = -\frac{5\pi}{3}
    \end{align*}

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Coterminal and Reference Angles Exercise 1

Answer: Reference Angle: \(60^\circ\)

Solution Method:  First, to draw the angle in standard position we need to place its vertex on the origin, and its initial side on the positive x-axis. Then we just determine where the terminal side needs to be. The \(120^\circ\) angle is positive so it's measured counterclockwise from the initial side. Each quadrant is \(90^\circ\) so \(120^\circ\) will lie 30 degrees inside quadrant 2. 

Now to find the reference angle. Recall, the reference angle is the smallest positive acute angle formed by the x-axis and the terminal side of the angle in standard position. The \(120^\circ\) is in Q2 so the reference angle is drawn between the terminal side and the negative x-axis here. We see then that the reference angle is the difference of \(180^\circ\) and \(120^\circ.\)
\begin{equation*}
    180^\circ - 120^\circ = 60^\circ
\end{equation*}

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Coterminal and Reference Angles Exercise 2

Answer: \(\dfrac{\pi}{4}\)

Solution Method: First, to draw the angle in standard position we need to place its vertex on the origin, and its initial side on the positive x-axis. Then we just determine where the terminal side needs to be. The \(\frac{7\pi}{4}\) angle is positive so it's measured counterclockwise from the initial side. Each quadrant is \(\frac{\pi}{2}\) so \(\frac{7\pi}{4}\) lies between \(\frac{3\pi}{2}\) and \(2\pi\) in Q4.  

Now to find the reference angle. Recall, the reference angle is the smallest positive acute angle formed by the x-axis and the terminal side of the angle in standard position. The \(\frac{7\pi}{4}\) is in Q4 so the reference angle is drawn between the terminal side and the positive x-axis here. We see then that the reference angle is the difference of \(2\pi\) and \(\frac{7\pi}{4}\) 
\begin{align*}
    2\pi - \frac{7\pi}{4} &= \frac{8\pi}{4}-\frac{7\pi}{4} \\
    &= \frac{\pi}{4}
\end{align*}

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Coterminal and Reference Angles Exercise 3

Answer: Coterminal Angle: \(\dfrac{7\pi}{6}, \quad \) Reference Angle: \(\dfrac{\pi}{6}\)

Solution Method: First, we need to find a coterminal angle between 0 and \(2\pi\). We see that \(\frac{19\pi}{6}\) is greater than \(2\pi\) because \(2\pi=\frac{12\pi}{6}\) and \(\frac{19\pi}{6}>\frac{12\pi}{6}\). So we need to subtract a rotation, \(2\pi\) from \(\frac{19\pi}{6}\). 
\begin{align*}
    \frac{19\pi}{6} - 2\pi &= \frac{19\pi}{6} - \frac{12\pi}{6} \\ 
    &= \frac{7\pi}{6}
\end{align*}
And, indeed, \(\frac{7\pi}{6}\) is between 0 and \(2\pi\). 

Now to draw the angle in standard position. As a reminder, an angle is in standard position when the vertex is at the origin and the initial side is the positive x-axis. So the angle in standard position is then determined by its terminal side. \(\frac{7\pi}{6}\) is positive so the angle will be measured counterclockwise from the initial side. To determine which quadrant \(\frac{7\pi}{6}\) is in we need to think about which quadrantal angles it lies between, \(0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\). 
\begin{equation*}
\pi = \frac{6\pi}{6}<\frac{7\pi}{6}<\frac{9\pi}{6}=\frac{3\pi}{2}
\end{equation*}
So the terminal side of \(\frac{7\pi}{6}\) lies in Q3 by just \(\frac{\pi}{6}\). 

Now to find the reference angle. Recall, the reference angle is the smallest positive acute angle formed by the x-axis and the terminal side of the angle in standard position. \(\frac{7\pi}{6}\) is in Q3, so the reference angle will be drawn between the negative x-axis and the terminal side. Then to find the reference angle's measure we can subtract off the part of \(\frac{7\pi}{6}\) that lies in Q1 and Q2, which is \(\pi\). 
\begin{align*}
    \frac{7\pi}{6} - \pi &= \frac{7\pi}{6} - \frac{6\pi}{6} \\
    &= \frac{\pi}{6}
\end{align*}

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Coterminal and Reference Angles Exercise 4

Answer: The Coterminal and Reference Angle are both \(60^\circ.\)

Solution Method: First, we need to find a coterminal angle between 0 and \(360^\circ.\) We see that \(-660^\circ\) is less than 0 so we need to add a \(360^\circ\) rotation. 
\begin{align*}
    -660^\circ + 360^\circ = -300^\circ
\end{align*}
But this measure is still less than 0 so while \(-300^\circ\)is coterminal to \(-660^\circ,\) it's not in the range we want. So we'll add \(360^\circ\) again.
\begin{align*}
    -300^\circ + 360^\circ = 60^\circ
\end{align*}
And, indeed, \(60^\circ\) is between \(0^\circ\) and \(360^\circ.\) 

Now to draw the angle in standard position. As a reminder, an angle is in standard position when the vertex is at the origin and the initial side is the positive x-axis. So the angle in standard position is then determined by its terminal side. \(60^\circ\) is positive so the angle will be measured counterclockwise from the initial side. To determine which quadrant \(60^\circ\) is in we need to think about which quadrantal angles it lies between, \(0^\circ,\) \(90^\circ,\) \(180^\circ,\) \(270^\circ,\) or \(360^\circ.\) So \(60^\circ\) is in Q1. 

We also could have found this terminal side by going clockwise \(300^\circ\) because they are coterminal. 

Now to find the reference angle. Recall, the reference angle is the smallest positive acute angle formed by the x-axis and the terminal side of the angle in standard position. But our terminal side for \(60^\circ\) lies in Q1 and is acute so it is the reference angle. 

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Coterminal and Reference Angles Exercise 5