Find the $\frac mn$ if $T=\sin 5°+\sin10°+\sin 15°+\cdots+\sin175°=\tan \frac mn$ It's really embarrassing to be able to doesn't solve this simple-looking trigonometry question.

$$T=\sin(5^\circ) +\sin(10^\circ) + \sin(15^\circ)  + \cdots +\sin(175^\circ) =\tan \frac mn$$
Find the $\frac mn=?$, where $m$ and $n$ are positive integer numbers.

Attepmts:
$$T=2\Big(\sin(5°)+\sin(10°) + \cdots + \sin(85°)\Big) + 1 = 2\Big((\sin(5°) + \cos(5°))+(\sin(10°)+ \cos(10°))+\cdots + (\sin(40°)+\cos(40°))\Big)+1 = 2\Big(\sqrt 2((\sin50°+\sin55°)+\cdots+\sin(80°))\Big)+1.$$
and then I can not see an any way...
 A: Not exactly the most elementary approach, but with complex numbers we could write
$$
T = \operatorname{Im}\left(1 + e^{\frac{\pi}{36}i} + e^{2\frac{\pi}{36}i} + \cdots + e^{35\frac{\pi}{36}i}\right) \\
=\operatorname{Im}\left(1 + z + z^2 + \cdots + z^{35}\right) \quad \left(z = e^{\frac{\pi}{36}i}\right)\\
= \operatorname{Im}\left(\frac{1 - z^{36}}{1 - z}\right) = \operatorname{Im}\left(\frac{2}{(1 - \cos(5^\circ)) - i\sin(5^\circ)}\right)\\
= \operatorname{Im}\left(\frac{2}{[1 - \cos(5^\circ)]^2 + \sin^2(5^\circ)}[(1 - \cos(5^\circ)) + i\sin(5^\circ)]\right)
\\ = \frac{2 \sin(5^\circ)}{[1 - \cos(5^\circ)]^2 + \sin^2(5^\circ)}
\\ = \frac{2 \sin(5^\circ)}{2 - 2\cos(5^\circ)} = \frac{\sin(5^\circ)}{1 - \cos(5^\circ)}
$$
As the commenter below points out, we have $\displaystyle\frac{\sin x}{1 - \cos x} = \tan\left(90^\circ - \frac{x}{2}\right)$.  It follows that our answer is
$$
\frac{\sin(5^\circ)}{1 - \cos(5^\circ)} = \tan\left(90^\circ - \frac{5^\circ}{2}\right) = \tan(87.5^\circ) = \tan\left(\left[\frac{175}{2}\right]^\circ\right)
$$
A: By geometric series,
$$T=\sin 5°+\sin10°+\sin 15°+...+\sin175° = \\
\operatorname{Im} (\sum_{n=1}^{35}\exp(i n 5 \pi/180)) =\\
\operatorname{Im} \exp(i 5 \pi/180) \frac{\exp(i 35 \cdot 5 \pi/180)-1}{\exp(i 5 \pi/180)-1} =\\
\operatorname{Im} \exp(i 5 \pi/180) \exp(i 34 \cdot 5 \pi/360) \frac{\sin( 35 \cdot 5 \pi/360)}{\sin( 5 \pi/360)} =
\\
\operatorname{Im}  \exp(i 36 \cdot 5 \pi/360) \frac{\sin( 35 \cdot 5 \pi/360)}{\sin( 5 \pi/360)}  =
\\ \frac{\sin( 35 \cdot 5 \pi/360)}{\sin( 5 \pi/360)} \\
 $$
and
$$
\arctan(\frac{\sin( 35 \cdot 5 \pi/360)}{\sin( 5 \pi/360)} ) = \\
\arctan(\frac{\sin( 35 \cdot 5 \pi/360)}{\cos(  \frac{\pi}{2}  - 5 \pi/360)} ) = \\
\arctan(\frac{\sin( 35 \cdot 5  \pi/360)}{\cos( 35 \cdot 5 \pi/360)} ) = 
\frac{35 \pi}{72}
$$
or, in degrees, $\frac{35 \cdot 180}{72} =\frac{175}{2} = 87.5 $
