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Difference between revisions of "2000 AIME II Problems/Problem 15"
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Since <math>\cot (180 - x) = - \cot x</math>, the summation simply reduces to <math>\frac{1}{\sin 1} \cdot \left( \cot 45 - \cot 90 \right) = \frac{1 - 0}{\sin 1} = \frac{1}{\sin 1^{\circ}}</math>. Therefore, the answer is <math>\boxed{001}</math>.
 
Since <math>\cot (180 - x) = - \cot x</math>, the summation simply reduces to <math>\frac{1}{\sin 1} \cdot \left( \cot 45 - \cot 90 \right) = \frac{1 - 0}{\sin 1} = \frac{1}{\sin 1^{\circ}}</math>. Therefore, the answer is <math>\boxed{001}</math>.
 +
 +
== Solution 2(Not Rigorous) ==
 +
We can make an approximation by observing the following points:
 +
Each of the terms is greater than 1, so the overall sum will be greater than 1.
 +
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 <math>\frac{1}{60}</math>.
 +
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<math>\frac{1}{57}</math> in radians, or is about <math>\frac{1}{57}</math> because of the approximation. What we want is apparently close to that so we make the guess that n is equal to 1 degree.
 +
-jasonhu4
  
 
== See also ==
 
== See also ==

Revision as of 19:53, 16 April 2015

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

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 series.

Thus our summation becomes

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

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(Not Rigorous)

We can make an approximation by observing the following points: Each of the terms is greater than 1, so the overall sum will be greater than 1. 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. -jasonhu4

See also

2000 AIME II (Problems • Answer Key • Resources)
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