TITLE: Proving that $ \frac{1}{\sin(45°)\sin(46°)}+\frac{1}{\sin(47°)\sin(48°)}+...+\frac{1}{\sin(133°)\sin(134°)}=\frac{1}{\sin(1°)}$
QUESTION [13 upvotes]: 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.

REPLY [19 votes]: $$\frac{\sin(1^\circ)}{\sin(x^\circ) \sin(x+1)^\circ}=\frac{\sin((x+1)^\circ-x^\circ)}{\sin(x^\circ) \sin(x+1)^\circ}=$$
$$\frac{\sin((x+1)^\circ) \cos (x^\circ)}{\sin(x^\circ) \sin(x+1)^\circ}-\frac{\sin(x^\circ) \cos(x+1)^\circ}{\sin(x^\circ) \sin(x+1)^\circ}= \cot(x^\circ)-\cot(x+1)^\circ$$
Add them and you get your telescopic sum ;)