Prove that sin20 sin40 sin60 sin80=3/16

 sin20 * sin40 * sin60 * sin80

since sin 60 =  √3/2

√3/2 (sin 20 * sin 40 * sin 80)
√3/2 (sin 20) [sin 40 * sin 80]

Using identity: 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]

Since cos 60 = 1/2:

√3/4 (sin 20) [cos 40 + (1/2)]
√3/4 (sin 20)(cos 40) + √3/8 (sin 20)

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

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

Since sin 60 = √3/2

√3/8 [√3/2 - sin 20] + √3/8 (sin 20)
3/16 - √3/8 sin 20 + √3/8 sin 20

Cancelling out the 2 terms:
3/16

Therefore,  sin20 * sin40 * sin60 * sin80 = 3/16