Coterminal and Reference Angles
Instructions
- The first videos below explain the concepts in this section.
- This page includes exercises that you should attempt to solve yourself. You can check your answers and watch the videos explaining how to solve the exercises.
When you are done, you can use the graph to pick another section or use the buttons to go to the next section.
Learning Objectives
- Draw angles in standard position
- Defining coterminal angles
- Calculate negative and positive coterminal angles
- Locating and calculating the measure of referene angles
Concept Video(s)
Exercises
1.
Solution: \(-300^\circ\) and \(420^\circ\) in degrees
\(-\dfrac{5\pi}{3}\) and \(\dfrac{7\pi}{3}\) in radians
(This is one possible solution. There are infinitely many positive and negative 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*}
\(-\dfrac{5\pi}{3}\) and \(\dfrac{7\pi}{3}\) in radians
(This is one possible solution. There are infinitely many positive and negative 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*}
2.
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*}
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*}
3.
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*}
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*}
4.
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*}
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*}
5.
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.
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.