Question

求函数f(x)=sin^2x-sinxcosx的单调区间

Answer 1

f(x)=sin^2x-sinxcosx
= (1-cos2x)/2-(1/2)sin2x
=(-1/2)(sin2x+cos2x)+1/2
=-(√2/2)*[sin2x*(√2/2)+cos2x*(√2/2)]+1/2
=(-√2/2)*[sin2x*cos(π/4)+cos2x*cos(π/4)]+1/2
=(-√2/2)*sin(2x+π/4)+1/2
（1）增区间,即y=sin(2x+π/4)的减区间
∴ 2kπ+π/2≤2x+π/4≤2kπ+3π/2,k∈Z
∴ 2kπ+π/4≤2x≤2kπ+5π/4,k∈Z
∴ kπ+π/8≤x≤kπ+5π/8,k∈Z
∴ 增区间是 [kπ+π/8,kπ+5π/8],k∈Z
（1）减区间,即y=sin(2x+π/4)的增区间
∴ 2kπ-π/2≤2x+π/4≤2kπ+π/2,k∈Z
∴ 2kπ-3π/4≤2x≤2kπ+π/4,k∈Z
∴ kπ-3π/8≤x≤kπ+π/8,k∈Z
∴ 减区间是 [kπ-3π/8,kπ+π/8],k∈Z