1.求函数y=2cos(x+π/4)cos(x-π/4)+根号三×xin2x的值域和最小周期?2.设函数f(x)=根号2/2×cos(2x+π/4)+sin²x,求函数f（x）的最小正周期?

Question

1.求函数y=2cos(x+π/4)cos(x-π/4)+根号三×xin2x的值域和最小周期?
2.设函数f(x)=根号2/2×cos(2x+π/4)+sin²x,求函数f（x）的最小正周期?

Answer 1

1、y=2(cosx·√2/2-sinx·√2/2)(cosx·√2/2+sinx·√2/2)+√3sin2x
=2(cos²x/2-sin²/2)+√3sin2x
=cos²x-sin²x+√3sin2x
=cos2x+√3sin2x
=2sin(2x+π/6).
值域为[-2,2],周期为π.
2、f(x)=√2/2(cos2x·√2/2-sin2x·√2/2)+sin²x
=cos2x/2-sin2x/2+(1-cos2x)/2
=-1/2sin2x+1/2.
T=π.