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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
    perhaps if you write this like $$\frac{1}{\sin(90-89)\sin(90-88)}+\cdots+\frac{1}{\sin(90-1)\sin(90-0)}$$2012-05-14
  • 0
    and how to proceed next?2012-05-14
  • 0
    I don't know... i'm asking myself the same, hehe. I realized this could not help anyway...2012-05-14
  • 2
    I just found a similar problem here http://math.stackexchange.com/questions/95291/proving-that-frac1-sin45-sin46-frac1-sin47-sin48-fr I can solve it now. Thanks.2012-05-14
  • 0
    Nice, thanks for pointing that hint.2012-05-14

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