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Proving that $ \frac{1}{\sin(45°)\sin(46°)}+\frac{1}{\sin(47°)\sin(48°)}+…+\frac{1}{\sin(133°)\sin(134°)}=\frac{1}{\sin(1°)}$

Simplify the following sum

$ \frac{1}{\sin\left(1°\right)\sin\left(2°\right)} + \frac{1}{\sin \left(2°\right) \sin \left(3°\right)} + \frac{1}{\sin\left(3°\right)\sin\left(4°\right)} + ... + \frac{1}{\sin\left(89°\right) \sin \left(90°\right)}$

Can anyone give me a hint?

  • 0
    Nice, thanks for pointing that hint.2012-05-14

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