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Math Help - Today's Putnam Problem of the Day

  1. #1
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    Today's Putnam Problem of the Day

    Evaluate the sum

    <br />
S = \frac{1}{ \cos{0^o} \cos{1^o} } + \frac{1}{ \cos{1^o} \cos{2^o} } + ... + \frac{1}{ \cos{88^o} \cos{89^o} }<br />

    Source : http://www.math.harvard.edu/putnam/
    Last edited by simplependulum; May 19th 2010 at 12:28 AM.
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  2. #2
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    Quote Originally Posted by simplependulum View Post
    Evaluate the sum

    <br />
S = \frac{1}{ \cos{0^o} \cos{1^o} } + \frac{1}{ \cos{1^o} \cos{2^o} } + ... + \frac{1}{ \cos{88^o} \cos{89^o} }<br />

    Source : Harvard Mathematics Department : Putnam Competition
    we have \frac{1}{\cos n \cos (n+1)} = \frac{1}{\sin 1} \left(\frac{\sin(n+1)}{\cos(n+1)} - \frac{\sin n}{\cos n} \right) and so (\sin 1)S is just a simple telescoping sum: (\sin 1)S= \sum_{n=0}^{88} \left(\frac{\sin(n+1)}{\cos(n+1)} - \frac{\sin n}{\cos n} \right)=\frac{\sin 89}{\cos 89}=\frac{\cos 1}{\sin 1}. thus S=\frac{\cos 1}{\sin^21}.
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