cosx+cos2x+.cos(nx)=sin(nx/2)cos((n+1)x/2)/sin(x/2)
sinx+sin2x+.sin(nx)=sin(nx/2)sin((n+1)x/2)/sin(x/2)
方法：
cosx+cos2x+.cos(nx)+i[sinx+sin2x+.sin(nx)]=e^(ix)+e^(i2x)+...e^(inx)
=e^(ix)*(1-e^(inx))/(1-e^(ix))
=e^(ix)*(sin(nx/2)/sin(x/2))e^(i(n-1)x/2)
整理分別取實部和虛部即得最后一步怎么來的？(1-e^(inx))/(1-e^(ix))上下乘以1-e^(-ix)即可得或者畫圖可得1-e^(ix)=sin(x/2)e^(i*(x-pi)/2)再有復(fù)數(shù)除法規(guī)則可得