I happen to remember this from trig class....
Note that we can write
sin 40 = 2sin20cos 20 rearrange
cos20 = sin40/ [2 sin20]
And....likewise
sin80 = 2sin40cos40
cos40 = sin80 /[2sin40]
And
sin160 = 2sin80cos80
cos80 = sin160/[2sin80]
So
cos20*cos40*cos80 =
sin40/ [ 2sin20] * sin80/ [2sin40] * sin160 / [2sin80] =
sin160 / [8 sin20] =
(1/8) sin160/sin20
Note that sin160 = sin(180 - 20) = sin180cos20 - sin20cos180 = sin20
So
(1/8)sin20 / sin20 =
(1/8)
