Product of 126 sines

First, we are going to state the following formula : $1-e^{2\theta i}=2 \sin{ \theta} e^{i(\theta+3\pi/2)}\quad\quad\quad(*)$

The proof of this formula is very simple. We just have to use the fact that $\sin{\theta}=\frac{e^{i\theta}-e^{-i\theta}}{2i},$ and the fact that $e^{i(3\pi/2)}=-i.$

Now, for any natural number $n$ and any angle $\theta$ we can define the polynomial $p(z)=z^n-e^{2n\theta i},$ with zeros $e^{i(2\theta+\frac{2k\pi}{n})},$ where $k$ is any integer number satisfying $0\leq k \leq n-1.$ Then $z^n-e^{2n\theta i}=\prod_{k=0}^{n-1}(z-e^{i(2\theta+\frac{2k\pi}{n})}).$ Evaluating at $z=1,$ we get $1-e^{2n\theta i}=\prod_{k=0}^{n-1}(1-e^{i(2\theta+\frac{2k\pi}{n})}).$ Now we use the formula $(*)$ in both sides, so we get $2 \sin {(n\theta)} e^{i(n\theta+3\pi/2)} = \prod_{k=0}^{n-1}(2 \sin{(\theta+\frac{k\pi}{n})} e^{i(\theta+\frac{k\pi}{n}+{3\pi}/{2})})$ The right side of which can be transform like this: $2^n* \prod_{k=0}^{n-1} \sin{(\theta+\frac{k\pi}{n})} \prod_{k=0}^{n-1} e^{i(\theta+\frac{k\pi}{n}+{3\pi}/{2})} =2^n* e^{i(n\theta+\frac{(n-1)\pi}{2}+{3n\pi}/{2})} \prod_{k=0}^{n-1} \sin{(\theta+\frac{k\pi}{n})} =$ $=2^n* e^{i(n\theta+2n \pi-{\pi}/{2})} \prod_{k=0}^{n-1} \sin{(\theta+\frac{k\pi}{n})}=2^n* e^{i(n\theta-{\pi}/{2})} \prod_{k=0}^{n-1} \sin{(\theta+\frac{k\pi}{n})}.$

Now, dividing both sides of the formula $2 \sin {(n\theta)} e^{i(n\theta+3\pi/2)}=2^n* e^{i(n\theta-{\pi}/{2})} \prod_{k=0}^{n-1} \sin{(\theta+\frac{k\pi}{n})}$ by $2* e^{i(n\theta-{\pi}/{2})} \sin{\theta},$ we obtain that $2^{n-1}\prod_{k=1}^{n-1} \sin{(\theta+\frac{k\pi}{n})}=\frac{\sin n\theta}{\sin \theta}.$ Making $n=127,$ in this general formula, we obtain $2^{126}\prod_{k=1}^{126} \sin{(\theta+\frac{k\pi}{127})}=\frac{\sin 127\theta}{\sin \theta}.$ Therefore $A=127$ and $B=1,$ and the answer is $AB=\boxed{127}.$