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.

Let us use $\sin(1^\circ) = \sin(179^\circ) = \cos(89^\circ)$, etc., to rewrite the product as
$$\prod_{i=1}^{45} \cos^2 \left( \frac{\pi}{180} (2i - 1) \right).$$
Now, $\pm \cos \left( \frac{\pi}{180} (2i-1) \right)$ for $i = 1, \ldots, 45$ are roots of the polynomial $P_{180}(x) + 1$, where $P_n$ is the Chebyshev polynomial such that $P_n(\cos \theta) = \cos (n\theta)$.  In fact, since $-1$ is the minimum possible value of $P_n(x)$ for $-1 \le x \le 1$, they are all double roots; and this accounts for all 180 roots of the polynomial.  On the other hand, $P_{180}(x)$ has the form $2^{179} x^{180} + \cdots + 1$, so $P_{180}(x) + 1$ has the form $2^{179} x^{180} + \cdots + 2$.  Therefore, the square of the product above is equal to the product of roots of this polynomial, which is $\frac{2}{2^{179}} = 2^{-178}$; and the original desired product is $2^{-89}$.
