Question

已知函数y=sin^2 x+2sinxcosx-cox^2 x,(x∈R)求最小正周期.递减区间

Answer 1

y=sin^2 x+2sinxcosx-cox^2 x = sin(2x) - cos(2x) = √2 sin(2x - π/4)

周期 π ,2kπ + π/2 < 2x - π/4 < 2kπ + 3π/2 时单调递减

递减区间：（kπ + 3π/8,kπ + 7π/8 ）