Finding $\sin {1^\circ}\sin {3^\circ}\sin {5^\circ}......\sin {179^\circ} = \frac{1}{{{2^\lambda }}}$

Question

Evaluate $\sin {1^\circ}\sin {3^\circ}\sin {5^\circ}......\sin {179^\circ} = \frac{1}{{{2^\lambda }}}$

My approach is as follow

After grouping $\left( {\sin {{179}^\circ}\sin {1^\circ}} \right)\left( {\sin {{177}^\circ}\sin {3^\circ}} \right).....\left( {\sin {{91}^\circ}\sin {{89}^\circ}} \right)$

Number of terms $91 + \left( {n - 1} \right)2 = 179 \Rightarrow n = 45$

$$\begin{gathered}\frac{{\left( {2\sin {{179}^\circ}\sin {1^\circ}} \right)\left( {2\sin {{177}^\circ}\sin {3^\circ}} \right)\cdots\left( {2\sin {{93}^\circ}\sin {{87}^\circ}} \right)\left( {2\sin {{91}^\circ}\sin {{89}^\circ}} \right)}}{{{2^{45}}}}\\ =\frac{{\left( {\cos {{178}^\circ} - \cos {{180}^\circ}} \right)\left( {\cos {{174}^\circ} - \cos {{180}^\circ}} \right)\cdots\left( {\cos {6^\circ} - \cos {{180}^\circ}} \right)\left( {\cos {2^\circ} - \cos {{180}^\circ}} \right)}}{{{2^{45}}}}\\ =\frac{{\overbrace {\left( {\cos {{178}^\circ} + 1} \right)\left( {\cos {{174}^\circ} + 1} \right)\cdots\left( {\cos {{94}^\circ} + 1} \right)}^{\text{22 terms}}\left( {\cos {{90}^\circ} + 1} \right)\overbrace {\left( {\cos {{86}^\circ} + 1} \right)\cdots\left( {\cos {6^\circ} + 1} \right)\left( {\cos {2^\circ} + 1} \right)}^{\text{22 terms}}}}{{{2^{45}}}}\\ =\frac{{\left( { - \cos {2^\circ} + 1} \right)\left( { - \cos {6^\circ} + 1} \right)\cdots\left( { - \cos {{86}^\circ} + 1} \right)\left( {\cos {{90}^\circ} + 1} \right)\left( {\cos {{86}^\circ} + 1} \right)\cdots\left( {\cos {6^\circ} + 1} \right)\left( {\cos {2^\circ} + 1} \right)}}{{{2^{45}}}}\\ =\frac{{\left( {1 - {{\cos }^2}{2^\circ}} \right)\left( {1 - {{\cos }^2}{6^\circ}} \right)\cdots\left( {1 - {{\cos }^2}{{86}^\circ}} \right)}}{{{2^{45}}}} = \frac{{{2^{22}}\,{{\sin }^2}{1^\circ}\,{{\sin }^2}{3^\circ}\cdots{{\sin }^2}43^\circ}}{{{2^{45}}}}\end{gathered}$$

How do I proceed further