Question

Prove that `cos((2pi)/9)*cos((4pi)/9)*cos((8pi)/9)=-1/8`?

Answer 1

We use formula `2cosAcosB=cos(A+B)+cos(A-B)`. For details see explanation.

Explanation:

`cos((2pi)/9)*cos((4pi)/9)*cos((8pi)/9)`

= `1/2cos((4pi)/9)[2cos((8pi)/9)cos((2pi)/9)]`

= `1/2cos((4pi)/9)[cos((8pi)/9+(2pi)/9)+cos((8pi)/9-(2pi)/9)]`

= `1/2cos((4pi)/9)[cos((10pi)/9)+cos((6pi)/9)]`

= `1/2cos((4pi)/9)[cos(pi+pi/9)+cos((2pi)/3)]`

= `1/2cos((4pi)/9)[-cos(pi/9)-1/2]`

= `-1/4cos((4pi)/9)-1/4[2cos((4pi)/9)cos(pi/9)]`

= `-1/4cos((4pi)/9)-1/4[cos((4pi)/9+pi/9)+cos((4pi)/9-pi/9)]`

= `-1/4cos((4pi)/9)-1/4[cos((5pi)/9)+cos((3pi)/9)]`

= `-1/4cos((4pi)/9)-1/4[cos(pi-(4pi)/9)+cos(pi/3)]`

= `-1/4cos((4pi)/9)-1/4[-cos((4pi)/9)+1/2]`

= `-1/4cos((4pi)/9)+1/4cos((4pi)/9)-1/8]`

= `-1/8`