Question

Упростить

$sin x+ sin 2x+sin 3x+sin 4x+...+sin2017x$

Answer 1

для любого t (по формуле косинуса суммы/разности):
cos(t+x/2) = cos(t) cos(x/2) - sin(t) sin(x/2)
cos(t-x/2) = cos(t) cos(x/2) + sin(t) sin(x/2)
cos(t-x/2) - cos(t+x/2) = 2 sin(t) sin(x/2)
sin(t) = [ cos(t-x/2) - cos(t+x/2) ] / [ 2 sin(x/2) ] (при sin(x/2) <> 0)
-->
t = x: sin(x) = [ cos(x/2) - cos(3x/2) ] / [ 2 sin(x/2) ]
t = 2x: sin(2x) = [ cos(3x/2) - cos(5x/2) ] / [ 2 sin(x/2) ]
...
t = 2017x: sin(2017x) = [cos(4033x/2) - cos(4035x/2) ] / [ 2 sin(x/2) ]
-->
суммируя,
sin(x) + sin(2x) + ... + sin(2017x) =
= [ cos(x/2) - cos(3x/2) + cos(3x/2) - cos(5x/2) + ... + cos(4033x/2) - cos(4035x/2)] / [ 2 sin(x/2) ]

сокращая вошедшие с разным знаком слагаемые (все косинусы, кроме cos(x/2) и cos(4035x/2)),
sin(x) + sin(2x) + ... + sin(2017x) = [ cos(x/2) - cos(4035x/2) ] / [ 2 sin(x/2) ]

Answer 2

3 действия! Спасибо!