已知函数f(x )=(sinx-cosx)sin2x /sinx (1)求f(x )的定义域及最小正周期 (2)求f(x)的单调增区间
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解答:
(1)sinx≠0, x≠kπ,k∈Z
f(x)=(sinx-cosx)sin2x/sinx
=(sinx-cosx)*2cosx
=2sinxcosx-2cos²x
=sin2x-cos2x-1
=√2sin(2x-π/4)-1
T=2π/2=π
(2)增区间为
2kπ-π/2≤2x-π/4<2kπ或 2kπ<2x-π/4≤2kπ+π/2
2kπ-π/4≤2x<2kπ+π/4或 2kπ+π/4<2x≤2kπ+3π/4
kπ-π/8≤x<kπ+π/8或 kπ+π/8<x≤kπ+3π/8
增区间【 kπ-π/8,kπ+π/8)和( kπ+π/8,kπ+3π/8】,k∈Z
(1)sinx≠0, x≠kπ,k∈Z
f(x)=(sinx-cosx)sin2x/sinx
=(sinx-cosx)*2cosx
=2sinxcosx-2cos²x
=sin2x-cos2x-1
=√2sin(2x-π/4)-1
T=2π/2=π
(2)增区间为
2kπ-π/2≤2x-π/4<2kπ或 2kπ<2x-π/4≤2kπ+π/2
2kπ-π/4≤2x<2kπ+π/4或 2kπ+π/4<2x≤2kπ+3π/4
kπ-π/8≤x<kπ+π/8或 kπ+π/8<x≤kπ+3π/8
增区间【 kπ-π/8,kπ+π/8)和( kπ+π/8,kπ+3π/8】,k∈Z
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