Question

cosx+cos2x+cos3x+....+cosnx=

要具体过程

Answer 1

具体回答如下：

cosx+cos2x+cos3x+....+cosnx

=sin(x/2)*[ cosx+cos2x+cos3x+....+cosnx] / sin(x/2) 

= 【sin[x(2n+1)/2] - sin(x/2) 】/ [2sin(x/2)]

和角公式：

sin ( α ± β ) = sinα · cosβ ± cosα · sinβ

sin ( α + β + γ ) = sinα · cosβ · cosγ + cosα · sinβ · cosγ + cosα · cosβ · sinγ - sinα · sinβ · sinγ

cos ( α ± β ) = cosα cosβ ∓ sinβ sinα

tan ( α ± β ) = ( tanα ± tanβ ) / ( 1 ∓ tanα tanβ )

Answer 2

cosx+cos2x+cos3x+....+cosnx
=sin(x/2)*[ cosx+cos2x+cos3x+....+cosnx] / sin(x/2) （ 将sin(x/2) 移入方括号里并化简）
= {sin[x(2n+1)/2] - sin(x/2) }/ [2sin(x/2)]