cosx + cos(x+d) + cos(x+2d) + …… + cos(x+nd)＝

来源：百度知道 编辑：UC知道 时间：2024/05/07 22:45:10
解出答案，并给出详细的证明。
d 不是周期，为任意的实数

cosx+cos(x+d)+cos(x+2d)...+cos(x+nd)
=[sin(d/2)*cos(x)+sin(d/2)*cos(x+d)...+sin(d/2)*cos(x+nd)]/(sin(d/2))
=1/2*[sin(x+d/2)+sin(-x+d/2)+sin(x+3d/2)+sin(-x-d/2)+...+sin(x+nd+d/2)+sin(-x-nd+d/2)]/(sin(d/2))
=1/2*[-sin(x-d/2)+sin(x+d/2)-sin(x+d/2)...-sin(x+nd-d/2)+sin(x+nd+d/2)]/(sind/2)
=1/2*[sin(x+nd+d/2)-sin(x-d/2)]/(sin(d/2))
=cos(x+nd/2)*sin[(n+1)d/2]/(sin(d/2))
积化和差， 和差化积

ncosx
d为周期