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

In your approach, there's a shorter way to arrive at T, but anyway taking off from there :
$$ T = \frac{{{{\sin }^2}{2^\circ}.{{\sin }^2}{6^\circ}.{{\sin }^2}{{10}^\circ}......{{\sin }^2}{{86}^\circ}}}{{{2^{45}}}} \\
= \frac{({\sin }{2^\circ}.{\sin }{6^\circ}.{\sin }{{10}^\circ}......{\sin }{{86}^\circ})^2}{{{2^{45}}}} \\
= \frac{({\cos }{4^\circ}.{\cos }{8^\circ}....{\cos }{{84}^\circ}.{\cos }{{88}^\circ})^2}{{{2^{45}}}} $$
Now let $p= {\cos }{4^\circ}.{\cos }{8^\circ}....{\cos }{{84}^\circ}.{\cos }{{88}^\circ} \\ $
And let  $q = {\sin }{4^\circ}.{\sin }{8^\circ}....{\sin }{{84}^\circ}.{\sin }{{88}^\circ}$
So $pq = {(1/2^{22})}.{{\sin }{8^\circ}.{\sin }{16^\circ}...{\sin }{{176}^\circ}}\\
= {(1/2^{22})}.{\sin }{4^\circ}.{\sin }{8^\circ}....{\sin }{{84}^\circ}.{\sin }{{88}^\circ} {(\because \sin(180-x)=\sin x)}\\
=  {(1/2^{22})}q $
So $p={(1/2^{22})}$
Therefore $T={(1/2^{89})}$
So $k=89$
