Question

已知函数f(x)=sin(x+7π/4)+cos(x-3π/4),x∈R (1)求f(x)的最小正周期和最小值

已知函数f(x)=sin(x+7π/4)+cos(x-3π/4),x∈R (1)求f(x)的最小正周期和最小值

Answer 1

f(x)=sin(x+7π/4)+cos(x-3π/4
=-sin（x+π/4）-cos（x-π/4）
=-（sinxcosπ/4+cosxsinπ/4）-（cosxcosπ/4+sinxsinπ/4）
=根号2（cosx+sinx）
=2sin（x+π/4）
所以最小正周期2π，最小值-2

Answer 2

f(x)=sin(x+7π/4)+cos(x-3π/4)
=sin(x+2π-π/4)+cos(x-π+π/4)
=sin(x-π/4)+cos[-(π-x-π/4)]
=sin(x-π/4)+cos(π-x-π/4)
=sin(x-π/4)-cos(x+π/4)
=sin(x-π/4)-cos(x+π/2-π/4)
=sin(x-π/4)-cos(π/2+x-π/4)
=sin(x-π/4)+sin(x-π/4)
=2sin(x-π/4)

T=2π/1=2π

-2<=2sin(x-π/4)<=2
所以函数f(x)为：-2