2025/028) How do you prove that $\cos20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ =\frac{1}{16}$

Let $A= \cos20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ$

We have as 80 is double of 40 and 40 is double of 20 let us multiply both sides by $2\sin\,20^\circ$ we get 

  $A*2\sin20^\circ = 2\sin20^\circ\cos20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ$

or

$A*2\sin20^\circ = \sin40^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ$ 

or  multiplying by 2 we get 

 $A*4\sin20^\circ = 2\sin40^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ$

or 

$A*4\sin20^\circ = \sin80^\circ \cos 60^\circ \cos 80^\circ$ 

 or $A*8\sin20^\circ = 2\sin80^\circ \cos 60^\circ \cos 80^\circ$

  or $A*8\sin20^\circ = 2\sin80^\circ \cos 80^\circ \cos 60^\circ$

 or $A*8\sin20^\circ = \sin160^\circ \frac{1}{2}$

 or $A*8\sin20^\circ = \sin20^\circ \frac{1}{2}$ (as $\sin20^\circ = \sin 160^\circ$)

or $A * 8 = \frac{1}{2}$

or $ A= \frac{1}{16}$  

Hence proved