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fx=2sinx(cosx-sinx)最小正周期
单调增区间呢
人气:177 ℃ 时间:2020-07-01 16:54:21
解答
f(x)=2sinx(cosx-sinx)
=2sinxcosx-2(sinx)^2
=sin2x+1-2(sinx)^2-1
=sin2x+cos2x-1
=√2sin(2x+π/4)-1
∴最小正周期T=2π/2=π
2Kπ-π/2≤2x+π/4≤2Kπ+π/2
则单调增区间为[Kπ-3π/8,Kπ+π/8](K为整数)
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