trigo prove

(1) Since x = p/n, so nx = p, sin nx = sin p = 0 for all positive integer of n.
So sin x + sin 2x + sin 3x + ...... + sin n x + sin (n + 1)x + sin (n + 2) x + .... + sin 2nx = 0 + 0 + 0 + ..... + 0 = 0.
(2) cos nx = cos p = - 1 for all positive integer of n.
So cos x + cos 2x + cos 3x + .... + cos nx = -1 + -1 + -1 + .... = - n.
cos (n + 1) x = cos nx cos x - sin nx sin x = (-1)(-1) - (0)(0) = 1
cos (n + 2) x = cos nx cos 2x - sin nx sin 2x = (-1)(-1) - (0)(0) = 1 and so on.
So cos (n + 1)x + cos (n + 2)x + cos (n + 3)x + ..... cos 2nx = 1 + 1 + 1 + ..... + 1 = n.
So sum of the entire expression = -n + n = 0.



2012-10-15 07:44:34 補充：
(b1) Skipping the details.
cos x + cos 3x + ...... + cos (2k - 1)x + cos [2(k + 1) - ]x = sin 2kx / 2 sin x + cos (2k + 1)x = [sin 2kx + 2 sin x cos (2k + 1)x]/2 sin x = [ sin 2kx + sin (x + 2kx + x) + sin (x - 2kx - x)]/2 sinx = [sin 2kx + sin2(k + 1)x + sin(- 2kx)]/2 sin x = sin 2(k + 1)x/2 sin x

2012-10-15 08:08:49 補充：
(b2) cos 2x + .... + cos 2kx + cos[2(k + 1)x] = sin (2k + 1)x/2 sin x - 1/2 + cos [2(k + 1)x] = - 1/2 + [sin (2k + 1)x + 2 sin x cos (2kx + 2x)]/2 sin x = - 1/2 + [ sin (2k + 1)x + sin ( x + 2kx + 2x) + sin (x - 2kx - 2x)]/2 sin x = - 1/2 + [ sin (2k + 1)x + sin (2kx + 3x) + sin (- 2kx- x) ]/2sin x

2012-10-15 08:10:15 補充：
(b2 cont'd) = sin (2kx + 3x)/s sin x - 1/2 = sin [2(k + 1) + 1]x / 2 sin x.

2012-10-15 08:11:17 補充：
please add back - 1/2 to the last answer. Sorry.