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I would like to show that the following trigonometric sum

$$ \frac{1}{\sin(45°)\sin(46°)}+\frac{1}{\sin(47°)\sin(48°)}+\cdots+\frac{1}{\sin(133°)\sin(134°)}$$

telescopes to $$\frac{1}{\sin(1°)}$$

We have: $$\begin{align} \sin(45°)\sin(46°)&=\frac{1}{2}(\cos(1°)+\sin(1°))\\ \sin(47°)\sin(48°)&=\frac{1}{2}(\cos(1°)+\sin(5°))\\ \sin(49°)\sin(50°)&=\frac{1}{2}(\cos(1°)+\sin(9°))\\ &\ \vdots\\ \sin(133°)\sin(134°)&=\frac{1}{2}(\cos(1°)+\sin(177°)) \end{align}$$

So the sum is:

$$\begin{align} \sum_{k=0}^{44} &\frac{2}{\cos(1°)+\sin(1+4k)} =\frac{2}{\cos(1°)+\sin(1°)}+\frac{2}{\cos(1°)+\sin(5°)}+\\ &\kern2.5in +\frac{2}{\cos(1°)+\sin(9°)}+\cdots+\frac{2}{\cos(1°)+\sin(177°)}. \end{align}$$

Although I don't think this new expression simplifies the problem.

J. M. ain't a mathematician
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Chon
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  • https://math.stackexchange.com/questions/425966/finite-series-reciprocals-of-sines and https://math.stackexchange.com/questions/464031/find-the-sum-frac1-cos0-circ-cos1-circ-frac1-cos1-circ-cos2-cir – lab bhattacharjee Mar 22 '20 at 10:27

1 Answers1

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$$\frac{\sin(1^\circ)}{\sin(x^\circ) \sin(x+1)^\circ}=\frac{\sin((x+1)^\circ-x^\circ)}{\sin(x^\circ) \sin(x+1)^\circ}=$$ $$\frac{\sin((x+1)^\circ) \cos (x^\circ)}{\sin(x^\circ) \sin(x+1)^\circ}-\frac{\sin(x^\circ) \cos(x+1)^\circ}{\sin(x^\circ) \sin(x+1)^\circ}= \cot(x^\circ)-\cot(x+1)^\circ$$

Add them and you get your telescopic sum ;)

N. S.
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  • So the sum is: $ \frac{1}{\sin(1°)}(\cot(45°)-\cot(46°)+\cot(47°)-\cot(48°)+...+\cot(133°)-\cot(134°))$, but I can't see how it telescopes. – Chon Dec 31 '11 at 10:17
  • It gives $ \frac{1}{\sin(1°)}(\cot(45°)-\cot(90°))=\frac{1}{\sin(1°)}$, thanks! – Chon Dec 31 '11 at 10:24
  • could you guys like explain this again ;-; how did he get sin 1 in the numerator smh – ibuprofen Jun 28 '18 at 15:00
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    @AayushPaurana Multiply the top and bottom by $\sin(1^\circ)$:$$\frac{1}{\sin(45°)\sin(46°)}+\frac{1}{\sin(47°)\sin(48°)}+\cdots+\frac{1}{\sin(133°)\sin(134°)}=\frac{1}{\sin(1^\circ)}\left(\frac{\sin(1^\circ)}{\sin(45°)\sin(46°)}+\frac{\sin(1^\circ)}{\sin(47°)\sin(48°)}+\cdots+\frac{\sin(1^\circ)}{\sin(133°)\sin(134°)}\right)$$ – N. S. Jun 28 '18 at 19:06
  • @N.S. Thanks a lot! – ibuprofen Jun 30 '18 at 07:58