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Angle Coterminal Angles Standard Position for Angles Trigonometry Series

Coterminal and Reference Angles Exercise 1

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 one negative and one positive coterminal angle for \(60^\circ\) and \(\dfrac{\pi}{3}.\)

 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*}

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