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

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已知函数f(x)=cox(2x-π/3)+2sin(x-π/4)sin(x+π/4)
1求函数f(x)的最小正周期和图像的对称轴方程
2求函数f(x)的区间[-π/12,π/2]上的值域

f(x)=cox(2x-π/3)+2sin(x-π/4)sin(x+π/4)
=cox(2x-π/3)+2sin(x-π/4)cos[π/2-(x+π/4)]
=cox(2x-π/3)+2sin(x-π/4)cos(x-π/4)
=cox(2x-π/3)+sin(2x-π/2)
=cox2x*cosπ/3+sin2x*sinπ/3-cos2x
=1/2*cos2x+sin2x*sinπ/3-cos2x
=sin2x*sinπ/3-1/2cos2x
=sin2x*sinπ/3-cosπ/3cos2x
=-cos(2x+π/3)
=cos(2x+4π/3)
1. 最小正周期 2π/2=π；
令2x+4π/3=0,则x=-2π/3；
对称轴 x=-2π/3+kπ,k为自然数；
2.令-π/12≤-2π/3+kπ≤π/2,得 7/12 ≤ k ≤ 7/6,k为自然数,k取1,此时x=π/3；
函数f(x)在区间[-π/12,π/2]上的最大值,最小值在x=-π/12,x=π/2,x=π/3之间产生；
f(-π/12)=cos(7π/6)=-√3/2；
f(π/2)=cos(π+4π/3)=1/2；
f(π/3)=cos(2π/3+4π/3)=1；
函数f(x)的区间[-π/12,π/2]上的值域[-√3/2,1]；
再问：可以解释一下-cos(2x+π/3)怎么化到cos(2x+4π/3) 吗？

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