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Difference between revisions of "2015 AIME I Problems/Problem 13"

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With all angles measured in degrees, the product <math>\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n</math>, where <math>m</math> and <math>n</math> are integers greater than 1. Find <math>m+n</math>.
 
With all angles measured in degrees, the product <math>\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n</math>, where <math>m</math> and <math>n</math> are integers greater than 1. Find <math>m+n</math>.
  
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==Solution==
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==Solution 1==
 
Let <math>x = \cos 1^\circ + i \sin 1^\circ</math>. Then from the identity
 
Let <math>x = \cos 1^\circ + i \sin 1^\circ</math>. Then from the identity
 
<cmath>\sin x = \frac{x - \frac{1}{x}}{2i} = \frac{x^2 - 1}{2 i x},</cmath>
 
<cmath>\sin x = \frac{x - \frac{1}{x}}{2i} = \frac{x^2 - 1}{2 i x},</cmath>
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<cmath>\frac{1}{M} = \dfrac{1}{2^{90}}|1 - x^2| |1 - x^6| \dots |1 - x^{358}| = \dfrac{1}{2^{90}} |1^{90} + 1| = \dfrac{1}{2^{89}}.</cmath>
 
<cmath>\frac{1}{M} = \dfrac{1}{2^{90}}|1 - x^2| |1 - x^6| \dots |1 - x^{358}| = \dfrac{1}{2^{90}} |1^{90} + 1| = \dfrac{1}{2^{89}}.</cmath>
 
It is easy to see that <math>M = 2^{89}</math> and that our answer is <math>2 + 89 = \boxed{91}</math>.
 
It is easy to see that <math>M = 2^{89}</math> and that our answer is <math>2 + 89 = \boxed{91}</math>.
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==Solution 2==
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Let <math>p=\sin1\sin3\sin5...\sin89</math>
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 +
<cmath>=\sqrt{\sin1\sin3\sin5...\sin177\sin179}</cmath>
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<cmath>=\sqrt{\frac{\sin1\sin2\sin3\sin4...\sin177\sin178\sin179}{\sin2\sin4\sin6\sin8...\sin176\sin178}}</cmath>
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 +
<cmath>=\sqrt{\frac{\sin1\sin2\sin3\sin4...\sin177\sin178\sin179}{(2\sin1\cos1)\cdot(2\sin2\cos2)\cdot(2\sin3\cos3)\cdot....\cdot(2\sin89\cos89)}}</cmath>
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<cmath>=\sqrt{\frac{1}{2^{89}}\frac{\sin90\sin91\sin92\sin93...\sin177\sin178\sin179}{\cos1\cos2\cos3\cos4...\cos89}}</cmath>
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<math>=\sqrt{\frac{1}{2^{89}}}</math> because of the identity <math>\sin(90+x)=\cos(x)</math>
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we want <math>\frac{1}{p^2}=2^{89}</math>
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Thus the answer is <math>2+89=091</math>
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2015|n=I|num-b=12|num-a=14}}
 
{{AIME box|year=2015|n=I|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 10:39, 21 March 2015

Problem

With all angles measured in degrees, the product $\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$, where $m$ and $n$ are integers greater than 1. Find $m+n$.

Solution 1

Let $x = \cos 1^\circ + i \sin 1^\circ$. Then from the identity \[\sin x = \frac{x - \frac{1}{x}}{2i} = \frac{x^2 - 1}{2 i x},\] we deduce that (taking absolute values and noticing $|x| = 1$) \[|2\sin x| = |x^2 - 1|.\] But because $\csc$ is the reciprocal of $\sin$ and because $\sin z = \sin (180^\circ - z)$, if we let our product be $M$ then \[\frac{1}{M} = \sin 1^\circ \sin 3^\circ \sin 5^\circ \dots \sin 177^\circ \sin 179^\circ\] \[= \frac{1}{2^{90}} |x^2 - 1| |x^6 - 1| |x^{10} - 1| \dots |x^{354} - 1| |x^{358} - 1|\] because $\sin$ is positive in the first and second quadrants. Now, notice that $x^2, x^6, x^{10}, \dots, x^{358}$ are the roots of $z^{90} + 1 = 0.$ Hence, we can write $(z - x^2)(z - x^6)\dots (z - x^{358}) = z^{90} + 1$, and so \[\frac{1}{M} = \dfrac{1}{2^{90}}|1 - x^2| |1 - x^6| \dots |1 - x^{358}| = \dfrac{1}{2^{90}} |1^{90} + 1| = \dfrac{1}{2^{89}}.\] It is easy to see that $M = 2^{89}$ and that our answer is $2 + 89 = \boxed{91}$.

Solution 2

Let $p=\sin1\sin3\sin5...\sin89$

\[=\sqrt{\sin1\sin3\sin5...\sin177\sin179}\]

\[=\sqrt{\frac{\sin1\sin2\sin3\sin4...\sin177\sin178\sin179}{\sin2\sin4\sin6\sin8...\sin176\sin178}}\]

\[=\sqrt{\frac{\sin1\sin2\sin3\sin4...\sin177\sin178\sin179}{(2\sin1\cos1)\cdot(2\sin2\cos2)\cdot(2\sin3\cos3)\cdot....\cdot(2\sin89\cos89)}}\]

\[=\sqrt{\frac{1}{2^{89}}\frac{\sin90\sin91\sin92\sin93...\sin177\sin178\sin179}{\cos1\cos2\cos3\cos4...\cos89}}\]

$=\sqrt{\frac{1}{2^{89}}}$ because of the identity $\sin(90+x)=\cos(x)$

we want $\frac{1}{p^2}=2^{89}$

Thus the answer is $2+89=091$

See Also

2015 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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