原式=(2cos3&+sin2&+cos&-3)/(2+2cos2&+cos&)=(2cos3&+1-cos2&+cos&-3)/(2+2cos2&+cos&)=(2cos3&-2+cos&-cos2&)/(2+2cos2&+cos&)=[2(cos&-1)(cos2&+cos&+1)-cos&(cos&-1)]/(2+2cos2&+cos&)=(cos&-1)(2cos2&+2cos&+2-cos&)/(2+2cos2&+cos&)=cos&-1把f=(π/3)代入cos&-1,得:cos(π/3)-1=0.5