题目没有打对,至少少了半边括号已知函数f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4）f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4） =cos(2x-π/3)+cos[(x-π/4)-(x+π/4)]-cos[(x-π/4)+(x+π/4)] =cos(2x-π/3)-cos2x+cos(-π/2) =cos(2x-π/3)-cos2x =-2sin(2x-π/6)sin(-π/6) =sin(2x-π/6)故最小正周期为π令2x-π/6=kπ（k∈Z），解得：x=(k/2+1/12)π这就是图像的对称轴方程由x∈[-π/12,π/2]得2x-π/6∈[-π/3,5π/6]由π/2∈[-π/3,5π/6]函数最大值为1最小值为sin(-π/3)=-√3/2