Question

f(x)=cos(2X+π/3)+sin(x-π/4)sin(x+π/4)的最小正周期和图像的对称轴方程是什么？

f(x)=cos(2X+π/3)+sin(x-π/4)sin(x+π/4)的最小正周期和图像的对称轴方程是什么？在【-π/12,π/2】的值域是多少？

越快越好，谢谢帮助，30分吧~
原函数应当是f(x)=cos(2X-π/3)+sin(x-π/4)sin(x+π/4)

Answer 1

解：f(x)=cos(2x+π/3)+sin(x-π/4)sin(x+π/4)

        =cos(2x+π/3)+sin(x-π/4)sin(x+π/4-π/2+π/2)

        =cos(2x+π/3)+sin(x-π/4)sin(x-π/4+π/2)

        =cos(2x+π/3)+sin(x-π/4)cos(x-π/4)

        =cos(2x+π/3)+1/2sin(2x-π/2) 

        =cos(2x+π/3)-1/2cos(2x) 

        =1/2cos(2x)-√3/2sin(2x)-1/2cos(2x)

        =-√3/2sin(2x)

最小正周期T=π

图像的对称轴方程：x=π/4+或-kπ/2(k为整数)

 当x属于[-π/12,π/2]

sin2x属于[-1/2,1]

所以值域为[-√3/2,√3/4]

Answer 2

f(x)=cos(2x-π/3)+sin(x-π/4)sin(x+π/4)
= cos(2x-π/3)-sin(π/4-x)sin(x+π/4)
= cos(2x-π/3)-cos(x+π/4)sin(x+π/4)
=(1/2)cos2x+(√3/2)sin2x-(1/2)sin(2x+π/2)
=(1/2)cos2x+(√3/2)sin2x-(1/2)cos2x
= (√3/2)sin2x，

∴最小正周期T=π，

令2x=kπ+π/2，k∈Z，
得x=kπ/2+π/4，k∈Z，
∴图象的对称轴方程为x=kπ/2+π/4，k∈Z，

当x∈[-π/12，π/2]时，2x∈[-π/6，π]，sin2x∈[-1/2，1]，
∴值域为[-(√3/4)，√3/2].