Question

已知函数f(X)=(cos(2x-π/4)+根号2/2)²+cos²(2x+π/4+nπ)-3/2(n∈Z)

1.求函数f（X）的最小周期T
2，当x∈【π/4,3π/4】时，求函数f（x）的最大值和最小值

Answer 1

函数f(X)=(cos(2x-π/4)+根号2/2)²+cos²(2x+π/4+nπ)-3/2

=cos²(2x-π/4)+根号2*cos(2x-π/4)+1/2+ cos²(2x+π/4+nπ) -3/2
又cos(2x+π/4+π)=-cos(2x+π/4)
cos²(2x+π/4+nπ)=cos²(2x+π/4)
所以f(X)=cos²(2x-π/4)+根号2*cos(2x-π/4)+ cos²(2x+π/4) -1
又cos(x)=sin(x+π/2) 所以cos(2x-π/4)=sin(2x+π/4)
cos²(2x-π/4)+cos²(2x+π/4)=sin²(2x+π/4)+cos²(2x+π/4)=1

所以f(X)=根号2*cos(2x-π/4)

f(X)的最小周期T为π

当x∈【π/4,3π/4】时 ，2x-π/4∈【π/4,5π/4】

cos(2x-π/4)∈【-1,根号2/2】 ，f(X)∈【-2,根号2】