证明：∵2sinx2cosnx=sin(x2+nx)+sin(x2−nx)．∴2sinx2（cosx+cos2x+…+cosnx）=（sin3x2−sinx2）+(sin5x2−3x2)+…+(sin1+2n2x−sin1−2n2x)=sin1+2n2x−sinx2=2cosn+12xsinn2x．∴cos+cos2x+…+cosnx=cosn+12x•...