2000 AIME II Problems/Problem 15

Problem

Find the least positive integer $n$ such that

$\frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ}+\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}=\frac 1{\sin n^\circ}.$

Solution 1

We apply the identity

$\begin{align*} \frac{1}{\sin n \sin (n+1)} &= \frac{1}{\sin 1} \cdot \frac{\sin (n+1) \cos n - \sin n \cos (n+1)}{\sin n \sin (n+1)} \\ &= \frac{1}{\sin 1} \cdot \left(\frac{\cos n}{\sin n} - \frac{\cos (n+1)}{\sin (n+1)}\right) \\ &= \frac{1}{\sin 1} \cdot \left(\cot n - \cot (n+1)\right). \end{align*}$

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

Thus our summation becomes

$\sum_{k=23}^{67} \frac{1}{\sin (2k-1) \sin 2k} = \frac{1}{\sin 1} \left(\cot 45 - \cot 46 + \cot 47 - \cdots + \cot 133 - \cot 134 \right).$

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 $\boxed{001}$ .

Solution 2

We can make an approximation by observing the following points:

The average term is around the 60's which gives 4/3.

There are 45 terms so the approximate sum is 60.

Therefore the entire thing equals approximately $\frac{1}{60}$ .

Recall that the approximation of sinx in radians is x if x is close to zero. In this case x is close to zero. Converting to radians we see that sin1 in degrees is about sin $\frac{1}{57}$ in radians, or is about $\frac{1}{57}$ because of the approximation. What we want is apparently close to that so we make the guess that n is equal to 1 degree. Basically, it boils down to the approximation of sin1= $\frac{1}{60}$ in degrees, convert to radians and use the small angle approximation sinx=x.

-jasonhu4

2000 AIME II (Problems • Answer Key • Resources)
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2000 AIME II Problems/Problem 15

Contents

Problem

Solution 1

Solution 2

See also