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先求导的那种方法怎么做？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