设函数f(x)=1-sin^2(x+π/4)+2sin(x+π/4)cos(x+π/4) (1)函数f(x)的最小正周期 （2)函数f(x）的值域

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设函数f(x)=1-sin^2(x+π/4)+2sin(x+π/4)cos(x+π/4) (1)函数f(x)的最小正周期 （2)函数f(x）的值域

设函数f(x)=1-sin^2(x+π/4)+2sin(x+π/4)cos(x+π/4) (1)函数f(x)的最小正周期 （2)函数f(x）的值域
设函数f(x)=1-sin^2(x+π/4)+2sin(x+π/4)cos(x+π/4) (1)函数f(x)的最小正周期 （2)函数f(x）的值域

设函数f(x)=1-sin^2(x+π/4)+2sin(x+π/4)cos(x+π/4) (1)函数f(x)的最小正周期 （2)函数f(x）的值域
f(x)=[cos(x+π/4)]^2+sin[2(x+π/4)]
=(1/2){cos[2(x+π/4)]+1}+sin[2(x+π/4)]
=(1/2){cos[2(x+π/4)]+2sin[2(x+π/4)]}+1/2
=(√5/2){(1/√5)cos[2(x+π/4)]+(2/√5)sin[2(x+π/4)]}+1/2
=(√5/2)sin[2(x+π/4)+ξ]}+1/2
=(√5/2)sin(2x+π/2+ξ)+1/2
=(√5/2)cos(2x+π/2+ξ)+1/2
其中,sinξ=1/√5,cosξ=2/√5
所以,最小正周期为2π/2=π；
值域为[(1-√5)/2,(1+√5)/2].

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