Trigonometric Sum

I'll post a long, detailed solution. Perhaps there's an easier one, I don't know.

Let us consider the sum:

$\large \displaystyle \sum_{n=1}^{90} \cos(2nx)$

This is the same as ( $j$ is the imaginary unit):

$\large \displaystyle \sum_{n=1}^{90} \cos(2nx) = \sum_{n=1}^{90} \Re \left [\ e^{j2nx} \right ]$

$\large \displaystyle \sum_{n=1}^{90} \cos(2nx) = \Re \left [\sum_{n=1}^{90} e^{j2nx} \right ]$

By geometric progression sum formula:

$\large \displaystyle \sum_{n=1}^{90} \cos(2nx) = \Re \left [ \frac{e^{j182x} - e^{j2x}}{e^{j2x}-1} \right ]$

$\large \displaystyle \sum_{n=1}^{90} \cos(2nx) = \Re \left [ \frac{(e^{j182x} - e^{j2x}) \cdot ( e^{-j2x}-1)}{(e^{j2x}-1) \cdot (e^{-j2x}-1)} \right ]$

$\large \displaystyle \sum_{n=1}^{90} \cos(2nx) = \Re \left [ \frac{e^{j180x} - e^{j182x} - 1 + e^{j2x}}{2 - 2 \cos(2x)} \right ]$

$\large \displaystyle \sum_{n=1}^{90} \cos(2nx) = \frac{\cos(180x) - \cos(182x) - 1 + \cos(2x)}{2 - 2 \cos(2x)}$

Since $\cos(2x) = 1 - 2\sin^2(x)$ :

$\large \displaystyle \sum_{n=1}^{90} \cos(2nx) = \frac{\cos(180x) - \cos(182x) - 1 + \cos(2x)}{4\sin^2(x)}$

Differentiating on both sides with respect to $x$ :

$\large \displaystyle \sum_{n=1}^{90} -2n \sin(2nx) = \frac{ [-180 \sin(180x) + 182 \sin(182x) - 2\sin(2x)] \cdot [4 \sin^2(x)] - [\cos(180x) - \cos(182x) - 1 + \cos(2x)]\cdot[4\sin(2x)] }{ 16 \sin^4(x) }$

Make $x = 1^o$ and considering that $\cos(180^o+x^o) = -\cos(x^o)$ and $\sin(180^o+x^o) = -\sin(x^o)$ :

$\large \displaystyle \sum_{n=1}^{90} -2n \sin(2n^o) = \frac{ (-184 \sin(2^o))(4 \sin^2(1^o)) - ( 2 \cos(2^o) - 2)(4\sin(2^o))}{16 \sin^4(1^o) }$

Recalling again that $\cos(2^o) = 1 - 2\sin^2(1^o)$ :

$\large \displaystyle \sum_{n=1}^{90} 2n \sin(2n^o) = \frac{ (184 \sin(2^o))(4 \sin^2(1^o)) + ( -4 \sin^2(1^o))(4\sin(2^o))}{16 \sin^4(1^o) }$

Dividing above and below by $4 \sin^2(1^o)$ :

$\large \displaystyle \sum_{n=1}^{90} 2n \sin(2n^o) = \frac{ 184 \sin(2^o) - 4\sin(2^o)}{4 \sin^2(1^o) }$

$\large \displaystyle \sum_{n=1}^{90} 2n \sin(2n^o) = \frac{ 180 \sin(2^o) }{4 \sin^2(1^o) }$

Since $\sin(2^o) = 2\sin(1^o) \cos(1^o)$ :

$\large \displaystyle \sum_{n=1}^{90} 2n \sin(2n^o) = \frac{ 360 \sin(1^o) \cos(1^o) }{4 \sin^2(1^o) }$

$\large \displaystyle \sum_{n=1}^{90} 2n \sin(2n^o) = \frac{ 90 \cos(1^o) }{\sin(1^o) }$

$\large \displaystyle \sum_{n=1}^{90} 2n \sin(2n^o) = 90 \cot(1^o)$

$\color{#20A900} \boxed{ \large \displaystyle \frac{1}{90} \sum_{n=1}^{90} 2n \sin(2n^o) = \cot(1^o) }$

Which is the average we are looking for. Thus:

$\color{#3D99F6} \boxed{ \large \displaystyle x = 1}$

We have $\begin{aligned} S & = \; \tfrac{1}{90}\sum_{n=1}^{90}2n \sin(2n)^\circ \; = \; \tfrac{1}{90}\sum_{n=1}^{89}(2n)\sin(2n)^\circ \\ & = \; \frac{1}{90}\left[ \sum_{n=1}^{44}(2n)\sin(2n)^\circ + 90 + \sum_{n=1}^{44}(180-2n)\sin(180-2n)^\circ\right] \; = \; 1 + 2\sum_{n=1}^{44}\sin(2n)^\circ \\ & = \; 1 + 2\sum_{n=1}^{44}\sin \tfrac{n\pi}{90} \; = \; 1 + 2\mathfrak{Im}\left[\sum_{n=1}^{44} e^{\frac{n\pi i}{90}}\right] \; = \; 1 + 2\mathfrak{Im}\left(\frac{e^{\frac{\pi i}{2}} - 1}{e^{\frac{\pi i}{90}} - 1}\right) \; = \; 1 + 2\frac{\sin\frac{\pi}{4}\sin\big(\frac{\pi}{4} - \frac{\pi}{180}\big)}{\sin\frac{\pi}{180}} \\ & = \; 1 + \frac{\cos\frac{\pi}{180} - \cos\big(\frac{\pi}{2} - \frac{\pi}{180}\big)}{\sin\frac{\pi}{180}} \; = \; \cot\tfrac{\pi}{180} \; = \; \cot 1^\circ \end{aligned}$ making the answer $\boxed{1}$ .