Question

已知函数y=sinx(sinx十cosx)-1 求(1)函数得最小正周期 (2)函数的递增区间.

Answer 1

y=sinx(sinx十cosx)-1 =sin^2x十sinxcosx-(sin^2x+cos^2x) =1/2sin2x-(cos2x+1)/2 =1/2sin2x-1/2cos2x-1/2 =√2/2sin（x-π/4）-1/2 ①T=2π/1=2π ②递增区间为{x|2kπ-π/2≤x-π/4≤2kπ+π/2,k∈Z} ={x|2kπ-π/4≤x≤2kπ+3π/4,k∈Z}