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_其他解答

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

人气：468 ℃　时间：2019-10-19 13:44:37

解答

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]}+...