Interesting Average

First of all, sorry for using $sin1$ in stead of $sin1^{\circ}$ i.e. in m solution, i won't be giving that degree symbol everywhere ...

By considering the terms of $sin 90$ and $sin 180$ separately, $90 sin 90 +180 sin 180 = 90$

About the remaining, the first fact we use here is $sin \theta = sin(180^{\circ}-\theta)$ ... and here , they will sum to give co-efficient 180 for all the sine terms.

Thus given expression transforms to

$=180 sin 2 + 180 sin 4 + ... +180 sin 86 + 180 sin 88 + 90$

$=180 \displaystyle \sum_{k=1}^ {44} sin (2k^{\circ}) + 90$

The average of this is expression divided by 90 (there are 90 terms).

$= 2 \displaystyle \sum_{k=1}^ {44} sin (2k^{\circ}) + 1$

Here as our angles will be in an arithmetic progression, we will use this identity

Thus $average = 2 \dfrac{sin (\frac{44\times 2}{2})}{sin1} \times sin (2 + \frac{43\times 2}{2}) +1$

$= 2 \dfrac{sin44}{sin1} sin 45 + 1$

$= \dfrac{\sqrt{2}sin (45-1) +sin1} {sin1}$

$= \dfrac{\sqrt{2}( \frac{cos1}{\sqrt{2}} - \frac{sin1}{\sqrt{2}} ) +sin 1}{sin1} = \dfrac{cos1 -sin1 +sin 1}{sin1}$

$= \dfrac{cos1}{sin1} = \boxed{\Huge{cot 1^{\circ}}}$