热心网友
求证：tg(3x/2)-tg(x/2)=2sinx/(cosx+cos2x)证明：tg(3x/2)-tg(x/2)=[sin(3x/2)/cos(3x/2)]-[sin(x/2)/cos(x/2)]=[sin(3x/2)*cos(x/2)-sin(x/2)*cos(3x/2)]/cos(3x/2)*cos(x/2) =sin[(3x/2)-(x/2)]/(1/2){[cos[(3x/2)+(x/2)]+cos[(3x/2)-(x/2)]} =sinx/[(1/2)(cos2x+cosx)]=2sinx/(cos2x+cosx)得证