Trig Functions with the Unit Circle Exercise 2

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.