Product of repeated cosec.

Question

$$P = \prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$$

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?

Answer 1

$$\sin(2n+1)x=(2n+1)\sin x+\cdots+(-1)^n2^{2n}\sin^{2n+1}x$$

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

$\cdots$

$m=22\implies8m+1=177\implies\sin177^\circ=\sin(180-3)^\circ=\sin3^\circ$

$\cdots$

$m=44\implies8m+1=353\implies\sin353^\circ=-\sin7^\circ$

Answer 2

I'm using a non-standard notation inspired by various programming languages in this evaluation because I think it's a bit easier to follow than the traditional big pi product notation. Hopefully, it's self-explanatory. Also, all angles are given in degrees.

This derivation uses the following identities:

$$ \cos x = \sin(90 - x) \\ \sin(180 - x) = \sin x \\ \sin 2x = 2 \sin x \cdot \cos x \\ \text{and hence} \\ \cos x = \frac{\sin 2x}{2 \sin x}\\ $$

$$\begin{align} \text{Let } P & = prod(\csc^2 x: x = \text{1 to 89 by 2}) \\ 1 / P & = prod(\sin x: x = \text{1 to 89 by 2})^2 \\ & = prod(\sin x: x = \text{1 to 89 by 2}) \cdot prod(\cos x: x = \text{1 to 89 by 2}) \\ & = prod(1/2 \sin 2x: x = \text{1 to 89 by 2}) \\ & = 2^{-45} prod(\sin x: x = \text{2 to 178 by 4}) \\ & = 2^{-45} prod(\sin x: x = \text{2 to 86 by 4}) \cdot \sin 90 \cdot prod(\sin x: x = \text{94 to 178 by 4}) \end{align}$$

$$ \text{But } prod(\sin x: x = \text{94 to 178 by 4}) = prod(\sin x: x = \text{2 to 86 by 4}) \\ \text{Let } Q = prod(\sin x: x = \text{2 to 86 by 4}), \text{so } P = \frac{2^{45}}{Q^2} $$

$$\begin{align} Q & = prod(cos x: x = \text{4 to 88 by 4}) \\ & = \frac{prod(sin 2x: x = \text{4 to 88 by 4})}{prod(2 sin x: x = \text{4 to 88 by 4})} \\ & = 2^{-22}\frac{prod(sin x: x = \text{8 to 176 by 8})}{prod(sin x: x = \text{4 to 88 by 4})} \\ & = 2^{-22}\frac{prod(sin x: x = \text{8 to 88 by 8}) \cdot prod(sin x: x = \text{96 to 176 by 8})} {prod(sin x: x = \text{4 to 84 by 8}) \cdot prod(sin x: x = \text{8 to 88 by 8})} \\ \end{align}$$

$\text{But } prod(\sin x: x = \text{96 to 176 by 8}) = prod(\sin x: x = \text{4 to 84 by 8})$

$\text{Hence } Q = 2^{-22} \text{ and } P = 2^{45} / (2^{-22})^2 = 2^{89}$