Evaluate the sum
$\displaystyle
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} }
$
Source : http://www.math.harvard.edu/putnam/
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Evaluate the sum
$\displaystyle
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} }
$
Source : http://www.math.harvard.edu/putnam/
we have $\displaystyle \frac{1}{\cos n \cos (n+1)} = \frac{1}{\sin 1} \left(\frac{\sin(n+1)}{\cos(n+1)} - \frac{\sin n}{\cos n} \right)$ and so $\displaystyle (\sin 1)S$ is just a simple telescoping sum: $\displaystyle (\sin 1)S= \sum_{n=0}^{88} \left(\frac{\sin(n+1)}{\cos(n+1)} - \frac{\sin n}{\cos n} \right)=\frac{\sin 89}{\cos 89}=\frac{\cos 1}{\sin 1}.$ thus $\displaystyle S=\frac{\cos 1}{\sin^21}.$Quote:
Originally Posted by simplependulum
