根据倍角公式：
cos2x = 1-2sin²x
2sin²x = 1-cos2x
∴
f(x)=1-cos2x+sin2x
=sin2x-cos2x+1
=√2[(√2/2)sin2x+(-√2/2)cos2x]+1
=√2sin[2x-(π/4)]+1
T = 2π/2 = π
显然,sin[2x-(π/4)] ≤ 1
因此,f(x)的最大值：√2+1