f(x)=(sinx+cosx)^2-2sin^2x
=sinx^2+cos^2+2sinxcosx-2sin^2x
=cos^2x-sin^2x+sin2x
=cos2x+sin2x
=√2sin(π/4+2x)
单调递减区间
2Kπ+π/2≤π/4+2x≤(2K+1)π+π/2
2Kπ+π/4≤2x≤2Kπ+5π/4
Kπ+π/8≤x≤Kπ+5π/8
f′(x)=2(sinx+cosx)*(cosx-sinx)-4sinxcosx=2(cos^2x-sin^2x)-2sin2x=2cos2x-2sin2x=2√2sin(2x-π/4)当f′(x)≥0时,函数单调递增sin(2x-π/4)≥02Kπ≤2x-π/4≤(2K+1)π2Kπ+π/4≤2x≤2Kπ+5π/4Kπ+π/8≤x≤Kπ+5π/8