0.5+cosx+cos2x+cos3x…………cosnx,把这个式子化简成分子和分母的形式

把原式乘以sinx后再除以sinx原式={1/2(sinx)+1/2(sin2x-sin0)+1/2(sin3x-sinx)+1/2(sin4x-sin2x)+………+1/2[sin(n+1)x-sin(n-1)x]}/sinx=1/2[sin(nx)+sin(n+1)x]/sinx=sin{[(2n+1)/2]x}cos(x/2)/sinx