已知f[x]=[sinx+cosx]的平方+cos2x

f(x)=(sinx+cosx)²+cos2xf(x)=1+sin2x+cos2xf(x)=1+√2·sin(2x+π/4)f(x)的最小正周期=2π/2=π2x+π/4=π/2，即：x=π/8时，sin(2x+π/4)=1，f(x)有最大值1+√22x+π/4=3π/2，即：x=5π/8时，sin(2x+π/4)=-1，f(x)有最小值1-√2