Problem: Prove that the average of the numbers $n \sin n^{\circ}$ for
$n = 2,4,6,\dots,180$ is $\cot 1^{\circ}$.

Solution: Because
\[ n \sin n\dg + (180-n) \sin(180\dg-n\dg) = 180 \sin n\dg \]
So enough to show that
\[ \sum_{n=0}^{89} \sin (2n)\dg = \cot1\dg \]
Let $\zeta = \cos2\dg + i \sin 2 \dg$ be a primitive root.
Then
\begin{align*}
  \sum_{n=0}^{89} \frac{\zeta^n - \zeta^{-n}}{2i}
  &= \frac{1}{2i}
  \left[ \frac{\zeta^{90}-1}{\zeta-1} - \frac{\zeta^{-90} - 1}{\zeta^{-1}-1} \right] \\
  &= \frac{1}{2i}
  \left[ \frac{-2}{\zeta-1} - \frac{-2}{\zeta^{-1}-1} \right] \\
  &= \frac{1}{-i} \frac{\zeta\inv-\zeta}{(\zeta-1)(\zeta\inv-1)}
  = i \cdot \frac{\zeta+1}{\zeta-1}.
\end{align*}
Also,
\begin{align*}
  \cot 1\dg &= \frac{\cos 1\dg}{\sin 1\dg}
  = \frac{(\cos1\dg)^2}{\cos1\dg\sin1\dg} \\
  &= \frac{\frac{\cos2\dg+1}{2}}{\frac{\sin2\dg}{2}}
  = \frac{\half(\zeta+\zeta\inv)+1}{\frac{1}{2i}(\zeta-\zeta\inv)} \\
  &= i \cdot \frac{(\zeta+1)^2}{\zeta^2-1} = i \cdot \frac{\zeta+1}{\zeta-1}.
\end{align*}
So we're done.