A geometry problem by Kaushal Diwakar

$\sin20^{\circ}\times{\sin40^{\circ}}\times{\sin60^{\circ}}\times{\sin80^{\circ}}=\frac{\sqrt{3}}{2}\times{\sin20^{\circ}}\times{\cos50^{\circ}}\times{\cos10^{\circ}}\\=\frac{\sqrt{3}}{2}\times{\sin20^{\circ}}\times{\frac{1}{2}}\left(\cos60^{\circ}+\cos40^{\circ}\right)=\frac{\sqrt{3}}{8}\times{\sin20^{\circ}}+\frac{\sqrt{3}}{4}\times{\sin20^{\circ}}\times{\cos40^{\circ}}\\=\frac{\sqrt{3}}{8}\times{\sin20^{\circ}}+\frac{\sqrt{3}}{8}\left(\sin60^{\circ}-\sin20^{\circ}\right) = \frac{3}{16}$

sin(20) sin(40) sin (60) sin (80) sin(60) = √3 /2

√3/2 [ sin(20) sin(40) sin(80) ]

= (√3/2) sin(20) [ sin(40) sin(80) ]

sin A sin B = (1/2) [ cos(A - B) - cos(A + B) ]

= √3/2 sin(20) (1/2)[ cos(40) - cos(120) ]

= √3/4 sin(20) [ cos(40) + cos(60) ]

= √3/4 sin(20) [ cos(40) + 1/2 ]

= √3/4 sin(20)cos(40) + (√3/8) sin(20)

sin A cos B = 1/2 [ sin(A + B) + sin(A - B) ]

= (√3/4)(1/2) [ sin(60) + sin(-20) ]+ (√3/8)sin(20)

= (√3/8) [ (√3 / 2) - sin(20) ]+ (√3/8)sin(20)

= 3/16 - (√3/8)sin(20) + (√3/8)sin(20)

= 3/16