Evaluate `cos alpha cos 2 alpha cos 3 alpha..........cos 999alpha`, where `alpha=(2pi)/(1999)`

Explanation

Let
`P=cos alpha cos 2 alpha cos 3 alpha........cos 999alpha`.
`Q=sin alpha sin 2 alpha sin 3 alpha.........sin 999alpha`.
Then `2^(999)PQ=(2sin alpha cos alpha)(2 sin 2 alpha cos 2 alpha)`......
`(2sin 999alpha cos 999alpha)`
`=sin 2alpha sin 4 alpha......sin 1998alpha`
`(sin alpa sin 4 alpha.......sin 998alpha)[-sin 999alpha cos 999alpha]`
`[-sin (2pi-1002alpha)]....[-sin(2pi-1998alpha)]`
`(because 2pi=199alpha)`
`=sin 2alpha sin 4 alpha.......sin 998alpha sin 997alpha` ...... `sin alpha=Q`.
It is easy to see that `Q ne0`, hence the desired product is `P=(1)/(2^(999))`.