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Angle Coterminal Angles Reference Angle Standard Position for Angles Trigonometry Series

Coterminal and Reference Angles Exercise 4

Author: Hannah Solomon

The following problem is solved in this video. It is recommended that you try to solve the problem before watching the video. You can click "Reveal Answer" to see the answer to the problem.

Problem: Find an angle between 0 and \(2\pi\) that is coterminal to \(\dfrac{19\pi}{6}.\) Then draw the angle in standard position and find its reference angle.

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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