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tags:
Cosine Cotangent Trigonometry Trigonometry Series Unit Circle

Trig Functions with the Unit Circle Exercise 3

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 the exact value of \(\cos(5\pi)\) and \(\cot(480^\circ).\)

Answer: \(\cos(5\pi)=-1,\quad\) \(\cot(480^\circ)=-\frac{\sqrt{3}}{3}

Solution Method: To find the trig value of an angle bigger than \(360^\circ,\) or \(2\pi\), we find its coterminal angle on the unit circle, that means between 0 and \(360^\circ\) (or 0 and \(2\pi\)). Since coterminal angles all have the same terminal side, the trig values of coterminal angles are the same. 
 
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.
(For more info see the coterminal angles section.)
To find \(\cos(5\pi),\)) we must first find the coterminal angle to \(5\pi\) in the unit circle. Since \(5\pi\) is greater than \(2\pi\), I'm going to subtract a rotation from \(5\pi\). A full rotation is 2\(\pi\), so \(5\pi-2\pi=3\pi\). So \(3\pi\) is coterminal to \(5\pi\). But \(3\pi\) is still greater than \(2\pi\). So I need to subtract another rotation until I get a value between 0 and \(2\pi\). So \(3\pi-2\pi=\pi\). So \(5\pi\) is coterminal to \(\pi\). Since \(5\pi\) and \(\pi\) are coterminal, \(\cos(5\pi)=\cos(\pi).\) The \(x\)-coordinate at \(\pi\) is -1, so \(\cos(5\pi)=-1.\)
Note: All even multiples of \(\pi\) are coterminal to \(2\pi\) and all odd multiple of \(\pi\) are coterminal to \(\pi\).

To find \(cot(480^\circ),\) we need to find the coterminal angle to \(480^\circ\) on the unit circle. In degrees, a rotation is \(360^\circ.\) So we do \(480^\circ -360^\circ=120^\circ,\) which is between 0 and 360 degrees. So \(120^\circ\) is coterminal to \(480^\circ\) and \(\cot(480^\circ=\cot (120^\circ.\) The coordinate at \(120^\circ\) is \(\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right).\) Cotangent is \(\frac{\cos{\theta}}{\sin{\theta}}=\frac{x}{y},\) so 
\begin{align*}
    \cot(480^\circ) = \frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}
\end{align*}
 

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