Question

已知函数f(x)=(sinx+cos)^2-2sin^2x求单调递减区间

帮我解答一下

Answer 1

你好

f(x)=(sinx+cosx)^2-2sin^2x
=sinx^2+cos^2+2sinxcosx-2sin^2x
=cos^2x-sin^2x+sin2x
=cos2x+sin2x
=√拦粗败2sin（π/简颤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

数凳辩学辅导团为您解答，不理解请追问，理解请及时选为满意回答！(*^__^*)谢谢！

追问

先求导的那种方法怎么做？

追答

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）≥0
2Kπ≤2x-π/4≤（2K+1）π
2Kπ+π/4≤2x≤2Kπ+5π/4
Kπ+π/8≤x≤Kπ+5π/8

Answer 2

f(x)=(sinx+cosx)^2-2sin^2x

=1+2sinxcosx-2(1-cos2x)/笑银团2
=2sin2x+2cos2x
=2√2sin(2x+π/搏团4)
当2kπ+ π/2≤2x+π/4≤2kπ+3π/2，即kπ+ π/8≤x≤kπ+5π/8 时，
f(x)单调递减
=>f(x)的单调递减区间为：碰橘【kπ+ π/8，kπ+5π/8】