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

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 $2 + 89 = \boxed{91}$ .

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

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