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2000 AIME II Problems/Problem 15

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Problem

Find the least positive integer n such that
\frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ}+\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}=\frac 1{...

Solution

We apply the identity

\begin{align*}\frac{1}{\sin n \sin (n+1)} &= \frac{1}{\sin 1} \cdot \frac{\sin (n+1) \cos n - \sin n \cos (n+1)}{\sin n \...

The motivation for this identity arises from the need to decompose those fractions, possibly into telescoping series.

Thus our summation becomes

\sum_{k=23}^{67} \frac{1}{\sin (2k-1) \sin 2k} = \frac{1}{\sin 1} \left(\cot 45 - \cot 46 + \cot 47 - \cdots + \cot 133 - \co...

Since \cot (180 - x) = - \cot x, the summation simply reduces to \frac{1}{\sin 1} \cdot \left( \cot 45 - \cot 90 \right) = \frac{1 - 0}{\sin 1} = \frac{1}{\sin 1^{\circ}}. Therefore, the answer is \boxed{001}.

See also

2000 AIME II (ProblemsResources)
Preceded by
Problem 14
Followed by
Last Question
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