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tags:
Cosecant Cotangent Secant Trigonometry Trigonometry Series Unit Circle

Trig Functions with the Unit Circle Exercise 2

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 values of \(\csc\left(\dfrac{11\pi}{6}\right),\) \(\sec\left(45^\circ\right),\) and \(\cot\left(\dfrac{5\pi}{3}\right).\)

Answer: \(\csc\left(\dfrac{11\pi}{6}\right)=-2,\quad\) \(\sec\left(45^\circ\right)=\sqrt{2},\quad\) \(\cot\left(\dfrac{5\pi}{3}\right)=-\dfrac{\sqrt{3}}{3}\)

Solution Method: If we know the values of sine, cosine, and tangent for special angles on the unit circle, we can easily find cosecant, secant, and cotangent by taking the reciprocal of the first three functions. 
\begin{align*}
\sin{\theta} &= y \\
\cos{\theta} &= x \\
\tan{\theta} &= \frac{\sin{\theta}}{\cos{\theta}} = \frac{y}{x} \\
\csc{\theta} &= \frac{1}{y} \\
\sec{\theta} &= \frac{1}{x} \\
\cot{\theta} &= \frac{\cos{\theta}}{\sin{\theta}} = \frac{x}{y}
\end{align*}
So for \(\csc\left(\frac{11\pi}{6}\right),\) we need to find the y-coordinate at \(\frac{11\pi}{6}\) and invert it. The coordinate at \(\frac{11\pi}{6}\) is \(\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right).\) So,
    \begin{equation*}
        \csc\left(\frac{11\pi}{6}\right)=\frac{1}{y}=\frac{1}{-\frac{1}{2}} = -2
    \end{equation*}
For \(\sec(45^\circ),\) the x-coordinate at \(45^\circ\) is \(\frac{\sqrt{2}}{2}\) so
    \begin{equation*}
        \sec(45^\circ) = \frac{1}{x} = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \frac{2\sqrt{2}}{2} = \sqrt{2}
    \end{equation*}
We had to rationalize to get the final answer.

For \(\cot\left(\frac{5\pi}{3}\right),\) the coordinate at \(\frac{5\pi}{3}\) is \(\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right).\) So,
    \begin{align*}
        \cot\left(\frac{5\pi}{3}\right) &= \frac{x}{y} \\ &= \frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}} \\&= \left( -\frac{1}{2} \right) \left( \frac{2}{\sqrt{3}} \right) \\ &= -\frac{1}{\sqrt{3}}\\ &= -\frac{\sqrt{3}}{3}
    \end{align*}
Beware of Quadrantals: Notice that because \(\cot(\theta) = \frac{x}{y},\) if the y-coordinate is 0 (like at 0 and \(\pi\)) cotangent will be undefined. 

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