Verified answer

as 80 =2 * 40 and 160 = 2 * 80

multiply and divide LHS by sin 40

we have LHS = 1/ sin 40( sin 40 cos 40 cos 80 cos 160)

= 1/ sin 40 ( 1/2 ( sin 80) cos 80 cos 160) as sin 40 cos 40 = 1/2 ( 2 sin 40 cos 40) = 1/2( sin 80)

= ( sin 80 cos 80 cos 160)/ ( 2 sin 40)

= ( (sin 160)/2 cos 160) / ( 2 sin 40) as sin 80 cos 80 = (sin 160)/2

= ( sin 160 cos 160)/ ( 4 sin 40)

= (sin 320)/ ( 8 sin 40)

= sin (360- 40)/ ( 8 sin 40)

= - sin 40/( 8 sin 40) = - 1/8

All angles are in degrees in the solution given below.

i) cos(160) = cos(180 - 20) = -cos(20)

ii) Thus the given left side is:P = -cos(20)*cos(40)*cos(80)

iii) Multiply and divide by 8*sin(20):

==> P = {-1/8*sin(20)}*[8*sin(20)cos(20)cos(40)cos(80)]

= {-1/8*sin(20)}*[{2*sin(20)cos(20)}{2cos(40)}{2cos(80)}]

= [-1/{8*sin(20)}]*[{2sin(40)cos(40)}{2cos(80)}] [Since 2sin(20)cos(20) = sin(40); by the application of multiple angle identity of sin(2A)]

Applying this multiple angle identity of sin(2A), successively for the products in bracket,

P = [-1/{8*sin(20)}*[sin(160)

= [-1/{8*sin(20)}*[sin(180 - 20)]

= [-1/{8*sin(20)}*[sin(20)] = -1/8

Thus it is proved that cos(40)*cos(80)*cos(160) = -1/8

Too easy. Check out M.L. Agarwal.

I proved it on paper. You should try using paper, or maybe a calculator.