Question

Show cos 20 * cos 40 * cos 80 =1/8

Answer 1

From Mathematica Home Edition:\[\text{Cos}[x]\text{Cos}[2x]\text{Cos}[4x]=\frac{1}{8}\text{ }\frac{\text{Sin}[8 x]}{\text{Sin}[x]}=\frac{1}{8}\text{ }\frac{\text{Sin}\left[\frac{8 \pi }{9}\right]}{\text{Sin}\left[\frac{\pi }{9}\right]}=\frac{1}{8}\text{ }\frac{\text{Sin}\left[\frac{\pi }{9}\right]}{\text{Sin}\left[\frac{\pi }{9}\right]}=\frac{1}{8} \]

Answer 2

I'll give you a clue, you have to find a way to turn 20's into 40's etc (use sin(2a) = 2sin a cos a).

Answer 3

Mathematica came up with the initial identity after some fiddling around with it. The identity appears to be valid after using some nontrivial angle values for verification. I agree that is not a proof, however, I think I'll let it go at that.

Thank you for your response.

Answer 4

First step, multiply by 2 sin 20/2 sin 20

Answer 5

My final Mathematica solution to this problem is attached.

Answer 6

Multiply top and bottom by 2 sin 20

cos 20.cos 40.cos 80 = 2 sin 20 cos 20.cos 40.cos 80 / 2 sin 20

-> sin 40 cos 40 cos 80 / 2 sin 20 then multiply top and bottom by 2

2 sin 40 cos 40 cos 80 / 4 sin 20

-> sin 80 cos 80/ 4 sin 20 and by 2 again

2 sin 80 cos 80 /8 sin 20

-> sin 160 / 8 sin 20

-> sin ( 180 -20) / 8 sin 20

-> sin 20/8 sin 20 = 1/8