prove that $\displaystyle A\cdot B= 2^{-6}\bigg(\csc \frac{\pi}{22}-1\bigg)$ 
If $\displaystyle A= \prod^{5}_{k=1}\cos \left(\frac{k\pi}{11}\right)$ and $\displaystyle B= \sum^{5}_{k=1}\cos \left(\frac{k\pi}{11}\right)$ then show that $\displaystyle A\cdot B= 2^{-6}\bigg(\csc \frac{\pi}{22}-1\bigg)$

Attempt: $\displaystyle B= \sum^{5}_{k=1}\cos \left(\frac{k\pi}{11}\right)=\cos \frac{\pi}{11}+\cos \frac{2\pi}{11}+\cos \frac{3\pi}{11}+\cos \frac{4\pi}{11}+\cos \frac{5\pi}{11}$
$\displaystyle =\frac{\sin \frac{5\pi}{22}}{\sin \frac{\pi}{22}}\bigg[\cos \bigg(\frac{\pi}{11}+4\frac{\pi}{22}\bigg)\bigg] =\frac{\sin \frac{5\pi}{22}}{\sin \frac{\pi}{22}}\bigg[\cos \frac{6\pi}{22}\bigg]=\frac{1}{2}\frac{-\sin(\frac{11\pi}{22})+\sin\frac{\pi}{11}}{\sin \frac{\pi}{22}}=\frac{1}{2}\bigg[-\csc \frac{\pi}{22}+1\bigg]$
could some help me how to solve it, thanks

COMMENT.- This is not properly a hint but a (nice) remark that I can not avoid mentioning. Given the value $x=\dfrac{\pi}{11}$ there are triangles $\triangle{ABC}$ like that of the attached figure intimately linked for which there must be a counterpart of the proposed problem in angles with the sides and bisectors $BB'$and $CC'$ of the triangle (in particular $\triangle{ACC'}$ is isosceles and $\triangle{ABB'}\sim \triangle{ABC}$).
One has all the five angles, $x,2x,3x,4x,5x$ of the problem in  $\triangle{ABC}$ because of $5x+4x+2x=\pi$.

