Amazing $\prod$ . <2>

From the reference (equation (24)) , we have $\displaystyle \prod_{k=1}^{n-1} \sin \left(\frac {k\pi}n\right) = 2^{1-n} n$ . Then

$\begin{aligned} \prod_{k=1}^{89} \sin \left(\frac {k\pi}{90}\right) & = \frac {90}{2^{89}} \\ \implies \prod_{k=1}^{89} \sin (2k^\circ) & = \frac {45}{2^{88}} \\ \prod_{k=1}^{45} \sin (2k^\circ) \color{#3D99F6} \prod_{k=46}^{89} \sin (2k^\circ) & = \frac {45}{2^{88}} & \small \color{#3D99F6} \text{Since }\sin (180-x)^\circ = \sin x^\circ \\ \prod_{k=1}^{45} \sin (2k^\circ) \color{#3D99F6} \prod_{k=1}^{44} \sin (2k^\circ) & = \frac {45}{2^{88}} & \small \color{#3D99F6} \text{Since }\sin 90^\circ = 1 \\ \prod_{k=1}^{45} \sin (2k^\circ) \color{#3D99F6} \prod_{k=1}^{\color{#D61F06}45} \sin (2k^\circ) & =\frac {45}{2^{88}} \end{aligned}$

$\begin{aligned} \implies \prod_{k=1}^{45} \sin (2k^\circ) & = \sqrt{\frac {45}{2^{88}}} = \frac {\sqrt{45}}{2^{44}} \end{aligned}$

Therefore, $m-n = 45-44 = \boxed{1}$ .

Clearly we should use root of unity .Let $\omega=cos(\frac{2\pi}{90})+isin(\frac{2\pi}{90})$ . Then $\omega^{\frac{k}{2}}=cos(\frac{k\pi}{90})+isin(\frac{k\pi}{90})$ .

$\displaystyle\prod_{k=1}^{45}sin(\frac{k\pi}{90})=\prod_{k=1}^{45}\frac{\omega^{\frac{k}{2}}-\omega^{\frac{-k}{2}}}{2i}=\prod_{k=1}^{45}\frac{\omega^{k}-1}{2i\omega^{\frac{k}{2}}}$ .

Note that $sin(\frac{45\pi}{90})=1$ and $\displaystyle\prod_{k=1}^{44}sin(\frac{k\pi}{90})=\prod_{k=46}^{89}sin(\frac{k\pi}{90})>0$ ,

$\displaystyle (\prod_{k=1}^{45}sin(\frac{k\pi}{90}))^{2}$ $\displaystyle=\prod_{k=1}^{89}sin(\frac{k\pi}{90})$ $\displaystyle=\prod_{k=1}^{89}\frac{\omega^{k}-1}{2i\omega^{\frac{k}{2}}}$

$\displaystyle=2^{-89}\cdot \prod_{k=1}^{89}(1-\omega^{k})=\frac{45}{2^{88}}$ .

The last step relies on the equality $\displaystyle n=\prod_{k=1}^{n-1}(1-\varepsilon_{k})$ ,here $\varepsilon_{k}=exp(\frac{2k\pi i}{n})$ .

Thus $m=45,n=44,m-n=\boxed{1}$ .