2015 AIME I Problems/Problem 13

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 $\sin 1 = \frac{x - \frac{1}{x}}{2i} = \frac{x^2 - 1}{2 i x},$ we deduce that (taking absolute values and noticing $|x| = 1$ ) $|2\sin 1| = |x^2 - 1|.$ 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 $p=\sin1\sin3\sin5...\sin89$

$p=\sqrt{\sin1\sin3\sin5...\sin177\sin179}$

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

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

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

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

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

Thus the answer is $2+89=091$

Solution 3

Similar to Solution $2$ , so we use $\sin{2\theta}=2\sin\theta\cos\theta$ and we find that: $\begin{align*}\sin4\sin8\sin12\sin16\cdots\sin84\sin88&=(2\sin2\cos2)(2\sin4\cos4)(2\sin6\cos6)(2\sin8\cos8)\cdots(2\sin42\cos42)(2\sin44\cos44)\\ &=(2\sin2\sin88)(2\sin4\sin86)(2\sin6\sin84)(2\sin8\cos82)\cdots(2\sin42\sin48)(2\sin44\sin46)\\ &=2^{22}(\sin2\sin88\sin4\sin86\sin6\sin84\sin8\sin82\cdots\sin42\sin48\sin44\sin46)\\ &=2^{22}(\sin2\sin4\sin6\sin8\cdots\sin82\sin84\sin86\sin88)\end{align*}$ Now we can cancel the sines of the multiples of $4$ : $1=2^{22}(\sin2\sin6\sin10\sin14\cdots\sin82\sin86)$ So $\sin2\sin6\sin10\sin14\cdots\sin82\sin86=2^{-22}$ and we can apply the double-angle formula again: $\begin{align*}2^{-22}&=\sin2\sin6\sin10\sin14\cdots\sin82\sin86\\ &=(2\sin1\cos1)(2\sin3\cos3)(2\sin5\cos5)(2\sin7\cos7)\cdots(2\sin41\cos41)(2\sin43\cos43)\\ &=(2\sin1\sin89)(2\sin3\sin87)(2\sin5\sin85)(2\sin7\sin87)\cdots(2\sin41\sin49)(2\sin43\sin47)\\ &=2^{22}(\sin1\sin89\sin3\sin87\sin5\sin85\sin7\sin83\cdots\sin41\sin49\sin43\sin47)\\ &=2^{22}(\sin1\sin3\sin5\sin7\cdots\sin41\sin43)(\sin47\sin49\cdots\sin83\sin85\sin87\sin89)\end{align*}$ Of course, $\sin45=2^{-\frac{1}{2}}$ is missing, so we multiply it to both sides: $2^{-22}\sin45=2^{22}(\sin1\sin3\sin5\sin7\cdots\sin41\sin43)(\sin45)(\sin47\sin49\cdots\sin83\sin85\sin87\sin89)$ $(2^{-22})(2^{-\frac{1}{2}})=2^{22}(\sin1\sin3\sin5\sin7\cdots\sin83\sin85\sin87\sin89)$ $2^{-\frac{45}{2}}=2^{22}(\sin1\sin3\sin5\sin7\cdots\sin83\sin85\sin87\sin89)$ Now isolate the product of the sines: $\sin1\sin3\sin5\sin7\cdots\sin83\sin85\sin87\sin89=2^{-\frac{89}{2}}$ And the product of the squares of the cosecants as asked for by the problem is the square of the inverse of this number: $\csc^21\csc^23\csc^25\csc^27\cdots\csc^283\csc^285\csc^287\csc^289=(\frac{1}{2^{\frac{89}{2}}})^2=(2^{\frac{89}{2}})^2=2^{89}$ The answer is therefore $m+n=(2)+(89)=\boxed{091}$ .

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
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2015 AIME I Problems/Problem 13

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also