Let k = 1o. Show that 2 sin 2k + 4 sin 4k + 6 sin 6k + ... + 180 sin 180k = 90 cot k.
Solution
Multiply the expression by sin k. We have 2 sin 2nk sin k = cos(2n-1)k - cos(2n+1)k. So 2n sin 2nk sin k = n cos(2n-1)k - n cos(2n+1)k. Adding these equations for n = 1, 2, ... , 90 gives: 2 sin 2k + 4 sin 4k + 6 sin 6k + ... + 180 sin 180k = cos k + (2 - 1) cos 3k + (3 - 2) cos 5k + ... + (90 - 89) cos 179k - 90 cos 181k. Now cos 181k = - cos k, so the last term is + 90 cos k. The other terms sum to zero in pairs: cos k + cos 179k = 0, cos 3k + cos 177k = 0, ... , cos 89k + cos 91k = 0. Hence result.
© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002