求函数f(x)=sin(x+π/3)+根号3sin(x-π/6)的单调区间函数y=cosx+cos(x+π/3)的最大值
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求函数f(x)=sin(x+π/3)+根号3sin(x-π/6)的单调区间
函数y=cosx+cos(x+π/3)的最大值
增区间[2kπ-7π/6,2kπ-π/6] (k∈Z),减区间[2kπ-π/6,2kπ+5π/6] (k∈Z),
y=cosx+cos(x+π/3)=cosx+cosxcosπ/3-sinxsinπ/3=(3/2)cosx-(√3/2)sinx=√3[(√3/2)cosx-(1/2)sinx]=√3[sinπ/3cosx-cosπ/3sinx]=√3sin(π/3-x)
∴函数y=cosx+cos(x+π/3)的最大值为√3
增区间[2kπ-7π/6,2kπ-π/6] (k∈Z),减区间[2kπ-π/6,2kπ+5π/6] (k∈Z), y=cosx+cos(x+π/3)=cosx+cosxcosπ/3-sinxsinπ/3=(3/2)cosx-(√3/2)sinx=√3[(√3/2)cosx-(1/2)sinx]=√3[sinπ/3cosx-co...
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增区间[2kπ-7π/6,2kπ-π/6] (k∈Z),减区间[2kπ-π/6,2kπ+5π/6] (k∈Z), y=cosx+cos(x+π/3)=cosx+cosxcosπ/3-sinxsinπ/3=(3/2)cosx-(√3/2)sinx=√3[(√3/2)cosx-(1/2)sinx]=√3[sinπ/3cosx-cosπ/3sinx]=√3sin(π/3-x) ∴函数y=cosx+cos(x+π/3)的最大值为√3
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