Trigonometry Problem | AIME I, 2015 | Question 13

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).

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 \(2+89=91\).

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