已知函数f(x)=1-2sin²(x+π/8)+2sin(x+π/8)cos(x+π/8),求f(x)的最小正周期,求f(x)的单调增区
问题描述:
已知函数f(x)=1-2sin²(x+π/8)+2sin(x+π/8)cos(x+π/8),求f(x)的最小正周期,求f(x)的单调增区
答
f(x)=1-2sin²(x+π/8)+2sin(x+π/8)cos(x+π/8)=cos(2x+π/4)+sin(2x+π/4)=√2sin(2x+π/2)=√2cos2x因此f(x)的最小正周期是2π/2=π2kπ≤2x≤2kπ+π单减,2kπ-π≤2x≤2kπ单增即kπ≤x≤kπ+π/2...