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

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...