<s>[INST] Product of repeated cosec. 
$$P = \prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$$

I realize that there must be some sort of trick in this.
$$P = \csc^2(1)\csc^2(3).....\csc^2(89) = \frac{1}{\sin^2(1)\sin^2(3)....\sin^2(89)}$$
I noticed that: $\sin(90 + x) = \cos(x)$ hence,
$$\sin(89) = \cos(-1) = \cos(359)$$
$$\sin(1) = \cos(-89) = \cos(271)$$
$$\cdots$$
$$P \cdot P = \frac{\cos^2(-1)\cos^2(-3)....}{\sin^2(1)\sin^2(3)....\sin^2(89)}$$
But that doesnt help?
 [/INST] $$\sin(2n+1)x=(2n+1)\sin x+\cdots+(-1)^n2^{2n}\sin^{2n+1}x$$
If we set $2n+1=45,2^{44}\sin^{45}x-\cdots+45\sin x-\sin45x=0$
If we set $\sin45x=\sin45^\circ,45x=360^\circ m+45^\circ$ where $m$ is any integer
$\implies x=8^\circ m+1^\circ$ where $0\le m\le44$
$\implies Q=\prod_{m=0}^{44}\sin(8^\circ m+1^\circ)=\dfrac1{\sqrt2}\cdot\dfrac1{2^{44}}$
$Q^2=\dfrac1{2\cdot2^{88}}$
Clearly, $\{\sin^2(8^\circ m+1^\circ);0\le m\le44\}=\{\sin^2(2r-1)^\circ,1\le r\le45\}$
as $x=1^\circ,9^\circ,17^\circ,25^\circ,33^\circ,41^\circ,49^\circ,57^\circ,65^\circ,73^\circ,81^\circ,89^\circ,$
$97^\circ[\sin97^\circ=\sin(180-97)^\circ=\sin83^\circ],$
$105^\circ[\sin75^\circ]$ and so on
$\cdots$
$m=22\implies8m+1=177\implies\sin177^\circ=\sin(180-3)^\circ=\sin3^\circ$
$\cdots$
$m=44\implies8m+1=353\implies\sin353^\circ=-\sin7^\circ$
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