y=sin²(x+π/2) - sin²(x-π/4)
=cos²x + [1-2sin²(x-π/4)-1]/2
=(2cos²x-1+1)/2 + [cos2(x-π/4)-1]/2
=1/2cos2x+1/2 + 1/2cos(2x-π/2)-1/2
=1/2cos2x + 1/2sin2x
=√2/2(√2/2cos2x+√2/2sin2x)
=√2/2sin(2x+π/4)
所以最小正周期T=2π/W=2π/2=π
因为-1≤sin(2x+π/4)≤1
所以-√2/2≤√2/2sin(2x+π/4)≤√2/2
即值域为[-√2/2,√2/2]