已知函数f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)

2sin(x-π/4)sin(x+π/4)=cos(x-π/4-x-π/4)-cos(x-π/4+x+π/4)=-cos2x
f(x)=cos(2x-π/3)-cos2x=cos(2x-π/6-π/6)-cos(2x-π/6+π/6)=2sin(2x-π/6)sin(π/6)=sin(2x-π/6)
最小正周期是2π/2=π,图象对称轴很容易知道是π/3+nπ/2
把[-π/12,π/2]代入2x-π/6,发现sin里的是[-π/3,7π/6]
那么一画图,值域是[-1/2,1]