AIME Problem 6

Note first that $S = \displaystyle\prod_{k=1}^{45} \sin^{2}(2k - 1)^{\circ} = \prod_{k=1}^{90} \sin(2k - 1)^{\circ}.$

So then $S = \displaystyle\dfrac{\prod_{k=1}^{179} \sin(k)^{\circ}}{\prod_{k=1}^{89} \sin(2k)^{\circ}} = \dfrac{\prod_{k=1}^{179} \sin(k)^{\circ}}{2^{89} \prod_{k=1}^{89} \sin(k)^{\circ}\cos(k)^{\circ}} =$

$\displaystyle\dfrac{\prod_{k=90}^{179} \sin(k)^{\circ}}{2^{89} \prod_{k=1}^{89} \cos(k)^{\circ}} = \dfrac{1}{2^{89}}$ since $\sin(90 + x)^{\circ} = \cos(x)^{\circ}.$

The given product is then just $\dfrac{1}{S} = 2^{89},$ and so $m + n = 2 + 89 = \boxed{91}.$

Let $T_n$ be the nth Chebyshev Polynomial where $T_n(\cos(x))=\cos(nx)$

Using the identity that $\cos(x)=-\cos(180-x)$

Looking at all the terms of $\displaystyle \prod_{1}^{45}\cos^2(2k-1)$ (we can do this because the identity $\sin(x)=\cos(90-x)$ ) let each of these terms be a root of the polynomial $f(x)$

Thus we have $(x-\cos^2(1))(x-\cos^2(3))(x-\cos^2(5))...(x-\cos^2(89))$

Let $y^2=x$ . Then since $y^2-\cos^2(t)=(y-\cos(t))(y+\cos(t))$ we have

$(y-\cos(1))(y+\cos(1))(y-\cos(3))(y+\cos(3))...(y-\cos(89))(y+\cos(89))$

Using the identity $\cos(t)=-\cos(180-t)$

$(y-\cos(1))(y-\cos(179))(y-\cos(3))(y-\cos(177))...(y-\cos(89))(y-\cos(91))$

Resubstituting for x

$(\sqrt{x}-\cos(1))(\sqrt{x}-\cos(179))(\sqrt{x}-\cos(3))(\sqrt{x}-\cos(177))...(\sqrt{x}-\cos(89))(\sqrt{x}-\cos(91))$

This polynomial is equivalent to $T_{90}(\sqrt{x})=0$

We need to look at the constant term of the polynomial divided by lead coefficient as by vietas, this will yield the product of the roots which is what we're looking for.

Since 90 is not divisible by 4, our constant term will be $-1$ . Also, the lead coefficient will be $2^{89}$ .

Thus our result is $\frac{1}{2^{89}}$ . However, since this is the product of $\cos$ , we need to invert this to get our product of $\csc$ so our answer here is $2^89$ and $m+n=91$