函数问题！！在线等

解：f（x〕=cos(2x-∏/3〕+2sin(x-∏/4〕sin(x+∏/4〕
=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(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]