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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...

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2 Answers 2

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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 $$ \frac{\sin(5^\circ)}{1 - \cos(5^\circ)} = \tan\left(90^\circ - \frac{5^\circ}{2}\right) = \tan(87.5^\circ) = \tan\left(\left[\frac{175}{2}\right]^\circ\right) $$

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  • $\begingroup$ That looks negative. $\endgroup$
    – Angina Seng
    Jan 2, 2019 at 20:20
  • $\begingroup$ @LordSharktheUnknown Found the sign error, fixing now $\endgroup$
    – Ben Grossmann
    Jan 2, 2019 at 20:21
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    $\begingroup$ $(\sin x)/(1-\cos x)=\cot (x/2)$. $\endgroup$
    – Angina Seng
    Jan 2, 2019 at 20:22
  • $\begingroup$ @LordSharktheUnknown Well, that certainly explains it $\endgroup$
    – Ben Grossmann
    Jan 2, 2019 at 20:23
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    $\begingroup$ @Beginner you should see this post linked above. It looks to me like you were at least trying the right kind of trick $\endgroup$
    – Ben Grossmann
    Jan 2, 2019 at 20:32
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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 $

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  • $\begingroup$ Is there a quick or clever way to compute $$ \arctan(\frac{\sin( 35 \cdot 5 \pi/360)}{\sin( 5 \pi/360)} ) = \frac{35 \pi}{72} ? $$ $\endgroup$
    – Ben Grossmann
    Jan 2, 2019 at 20:43
  • $\begingroup$ Never mind, I see now that the expression in the parentheses can immediately be rewritten as $\sin(87.5^\circ)/\cos(87.5^\circ)$. $\endgroup$
    – Ben Grossmann
    Jan 2, 2019 at 20:47
  • $\begingroup$ @Omnomnomnom I put it in the main text $\endgroup$
    – Andreas
    Jan 2, 2019 at 20:53

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