f(x)=2cos2(x-

π

4

)-[sin(x+

π

4

)+cos(x+

π

4

)] 2

=1+cos(2x-

π

2

)-[

2

sin(x+

π

2

) ] 2

=1+sin2x-2cos²x=sin2x-cos2x  

=

2

sin(2x-

π

4

)

（1）T=

2π

2

=π

（2）当 f（x）取最大值时，sin(2x-

π

4

)=1，即2x-

π

4

=

π

2

+2kπ⇒{x|x=kπ+

3π

8

，k∈Z}