Find the value of $k$ in the equation $\sin {1^\circ}\sin {3^\circ}.......\sin {179^\circ} = \frac{1}{{{2^k}}}$ Find the value of 'k' in the equation $\sin {1^\circ}\sin {3^\circ}.......\sin {179^\circ} = \frac{1}{{{2^k}}}$
My approach is as follow
$\sin {1^\circ} = \sin {179^\circ}$
$T = {\sin ^2}{1^\circ}{\sin ^2}{3^\circ}..{\sin ^2}{89^\circ}$
$\left( {\frac{{1 - \cos {2^\circ}}}{2}} \right) = {\sin ^2}{1^\circ}$
$T = \left( {\frac{{1 - \cos {2^\circ}}}{2}} \right)\left( {\frac{{1 - \cos {6^\circ}}}{2}} \right)\left( {\frac{{1 - \cos {{10}^\circ}}}{2}} \right)....\left( {\frac{{1 - \cos {{178^\circ}}}}{2}} \right)$
$2 + \left( {n - 1} \right)4 = 178 \Rightarrow n = 45$
$T = \frac{1}{{{2^{45}}}}\left( {1 - \cos {2^\circ}} \right)\left( {1 - \cos {6^\circ}} \right)\left( {1 - \cos {{10}^\circ}} \right)....\left( {1 - \cos {{178^\circ}}} \right)$
$\left( {1 - \cos {{178}^\circ}} \right) = \left( {1 + \cos {2^\circ}} \right)$
$T = \frac{1}{{{2^{45}}}}\left( {1 - {{\cos }^2}{2^\circ}} \right)\left( {1 - {{\cos }^2}{6^\circ}} \right)\left( {1 - {{\cos }^2}{10^\circ}} \right)....\left( {1 - {{\cos }^2}{{86}^\circ}} \right)\left( {1 - \cos {{90}^\circ}} \right)$
$T = \frac{{{{\sin }^2}{2^\circ}.{{\sin }^2}{6^\circ}.{{\sin }^2}{{10}^\circ}......{{\sin }^2}{{86}^\circ}}}{{{2^{45}}}}$
Not able to proceed from here

The connection of these kind of products with Chybshev polynomials in my opinion is the most important. Because, these polynomials show up everywhere like a unification of many things. Some nice properties of them are listed here. I know so few of them!: https://en.wikipedia.org/wiki/Chebyshev_polynomials
$T_n(x)=2^{n-1}\prod_{k=0}^{n-1}(x-\cos(\frac{2k+1}{2n}\pi))$ and $T_n(0)=(-1)^{\frac{n}{2}}$ when $n$ is even. So we get $2^{n-1}\prod_{k=0}^{n-1}\cos(\frac{2k+1}{2n}\pi)=(-1)^{\frac{n}{2}}$ when $n$ is even. Now let $n=90$. Then $2^{89}\prod_{k=0}^{89}\cos(\frac{2k+1}{180}\pi)=-1$ and thus $2^{89}\prod_{k=0}^{44}\cos^2(\frac{2k+1}{180}\pi)=2^{89}\prod_{k=0}^{44}\sin^2(\frac{2k+1}{180}\pi)=2^{89}\prod_{k=0}^{89}\sin(\frac{2k+1}{180}\pi)=1$.
