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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.

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up vote 16 down vote accepted

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

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(1‌​34°))$, 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

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