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This is a plane Euclidean geometry I did not understand... The problem states::

sin$\frac\pi 7\cdot $ sin$\frac{2\pi} 7\cdot $ sin $\frac{3\pi} 7\cdot$ sin $\frac{4\pi} 7 \cdot $ sin$\frac{5\pi} 7 \cdot $sin $\frac{6\pi} 7 \cdot $ sin $\frac{7\pi} 7 $ = $\frac 7 {64}$

(Source :: )

Need help urgently!!
sin$\frac{7\pi} 7 = 0$, so how come this be proved!!

uestion ta ki tik ase

100% @ confused.... I didn't make any mistake stating the problem....

There was a hint as well. I tried using it but didn't get much out of it!!!
I'll post the hint here.

Hint::

I believe I've posted this problem in the wrong Forum....
It should be moved to the Geometry forum....
If any of the admins could please help me out....

I've also created an AOPS topic for this problem....
Here's the link:::
http://www.artofproblemsolving.com/Foru ... 3#p2358903

maybe its wrong!

u can use this formula
http://www.artofproblemsolving.com/Foru ... hilit=trig
I also think as Nisha said that the given expression isnot valid. whats ur opinion?

This was the theorem I saw::

$n~(\geq 2)$ is a natural number.
$\prod_{k=1}^{n-1}\sin\frac{k\pi}{n}=\frac{n}{2^{n-1}}$

I still believe the problem's invalid!

it will be invalid if the sin 7pi/7=0