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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.

1 Answers 1

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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 ;)

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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.2011-12-31
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    It gives $ \frac{1}{\sin(1°)}(\cot(45°)-\cot(90°))=\frac{1}{\sin(1°)}$, thanks!2011-12-31
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    could you guys like explain this again ;-; how did he get sin 1 in the numerator smh2018-06-28
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    @AayushPaurana Multiply the top and bottom by $\sin(1^\circ)$:$$\frac{1}{\sin(45°)\sin(46°)}+\frac{1}{\sin(47°)\sin(48°)}+\cdots+\frac{1}{\sin(133°)\sin(134°)}=\frac{1}{\sin(1^\circ)}\left(\frac{\sin(1^\circ)}{\sin(45°)\sin(46°)}+\frac{\sin(1^\circ)}{\sin(47°)\sin(48°)}+\cdots+\frac{\sin(1^\circ)}{\sin(133°)\sin(134°)}\right)$$2018-06-28
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    @N.S. Thanks a lot!2018-06-30