What is the value of $\sin 1 ^\circ \sin3^\circ\sin5^\circ \sin 7^\circ \sin 9^\circ \cdots \sin 179^\circ $? 
What is the value of
  $\sin 1 ^\circ \sin3^\circ\sin5^\circ \sin 7^\circ \sin 9^\circ \cdots \sin 179^\circ $ ?

The question is indeed intriguing. We could start by condensing it using $\sin \theta = \sin (180-\theta)$, This reduces the problem as the products till $89^\circ$.  But that doesn't help proceed.  
Thanks in advanced.



The product of the chords in this figure is
$(2\cos \frac {2\pi}{5})(2\cos \frac {\pi}{5})(2\cos 0)(2 \cos -\frac {\pi}{5})(2\cos -\frac {2\pi}{5}) = 2^5\prod_\limits {n=-2}^2 \cos \frac {n\pi}{5}$
If we map this figure to the complex plane the product of those lengths = $|(1+e^{\frac {\pi i}{5}})(1+e^{\frac {3\pi i}{5}})(1+e^{\frac {5\pi i}{5}})(1+e^{\frac {7\pi i}{5}})(1+e^{\frac {7\pi i}{5}})|$
Note: $(z+e^{\frac {\pi i}{5}})(z+e^{\frac {3\pi i}{5}})(z+e^{\frac {5\pi i}{5}})(z+e^{\frac {7\pi i}{5}})(z+e^{\frac {7\pi i}{5}}) = z^5 + 1$
Evaluated at $z= 1$
$2^5\prod_\limits {n=-2}^2 \cos \frac {n\pi}{5} = 2\\
\prod_\limits {n=-2}^2 \cos \frac {n\pi}{5} = 2^{-4}$
And this generalizes:
$\prod_\limits {n=1}^k \cos \frac {(2n-1)\pi}{2k} = 2^{-(k-1)}$
