Demostrando que $ \frac{1}{\sin(45°)\sin(46°)}+\frac{1}{\sin(47°)\sin(48°)}+...+\frac{1}{\sin(133°)\sin(134°)}=\frac{1}{\sin(1°)}$

Question

Me gustaría demostrar que la siguiente suma trigonométrica

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

telescopios para $$\frac{1}{\sin(1°)}$$

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

Así que la suma es:

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

Aunque no creo que esta nueva expresión simplifique el problema.

Answer 1

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

Súmalos y tendrás tu suma telescópica ;)