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求函数的单调区间:(1)y=sin(π/4-3x),(2)f(x)=sinx(sinx-cosx)
人气:163 ℃ 时间:2020-01-19 22:35:09
解答
(1)y=sin(π/4-3x)
递增2kπ-π/2<π/4-3x<2kπ+π/2
2kπ-3π/4<3x<2kπ+π/4
2kπ/3-π/4递减2kπ+π/2<π/4-3x<2kπ+3π/2
2kπ+π/4<-3x<2kπ+5π/4
-2kπ/3-5π/12(2)f(x)=sinx(sinx-cosx)
=sin^2x-sinxcosx
=(1-cos2x)/2-sin2x/2
=1-(sin2x+cos2x)/2
=1-根号2sin(2x+π/4)/2
递增2kπ+π/2<2x+π/4<2kπ+3π/2
2kπ+π/4<2x<2kπ+5/4π
kπ+π/8递减2kπ-π/2<2x+π/4<2kπ+π/2
2kπ-3π/4<2x<2kπ+π/4
kπ-3π/8
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