己知f(x)=cos(2x-2/3)+2sin(x-兀/4)Sin(x+兀/4)①f(x)单调区间
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解决时间 2021-04-01 09:16
- 提问者网友：自食苦果
- 2021-04-01 01:45
己知f(x)=cos(2x-2/3)+2sin(x-兀/4)Sin(x+兀/4)①f(x)单调区间
最佳答案
- 五星知识达人网友：人间朝暮
- 2021-04-01 01:58
f(x)=cos(2x-⅔π)+2sin(x-π/4)sin(x+π/4)?
=cos(2x-⅔π)-2cos[π/2+(x-π/4)]sin(x+π/4)
=cos(2x-⅔π)-2cos(x+π/4)sin(x+π/4)
=cos(2x-⅔π)-sin[2(x+π/4)]
=cos(2x-⅔π)-cos2x
=-½cos(2x)+√3/2sin2x-cos2x
=-√3cos(2x+π/6)
=√3sin(2x-π/3)
将2x-π/3看成整体：
单调递增区间2x-π/3∈(2kπ-π/2, 2kπ+π/2)→x∈(kπ-π/12,kπ+5π/12)
单调递减区间2x-π/3∈(2kπ+π/2, 2kπ+3π/2)→x∈(kπ+5π/12,kπ+11π/12)
=cos(2x-⅔π)-2cos[π/2+(x-π/4)]sin(x+π/4)
=cos(2x-⅔π)-2cos(x+π/4)sin(x+π/4)
=cos(2x-⅔π)-sin[2(x+π/4)]
=cos(2x-⅔π)-cos2x
=-½cos(2x)+√3/2sin2x-cos2x
=-√3cos(2x+π/6)
=√3sin(2x-π/3)
将2x-π/3看成整体：
单调递增区间2x-π/3∈(2kπ-π/2, 2kπ+π/2)→x∈(kπ-π/12,kπ+5π/12)
单调递减区间2x-π/3∈(2kπ+π/2, 2kπ+3π/2)→x∈(kπ+5π/12,kπ+11π/12)
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