若x∈【-π/2,0】,求函数f(x)=cos(X+π/6)-cos(X-π/6)+ 根号三*cosx 的最大值和最小值.
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若x∈【-π/2,0】,求函数f(x)=cos(X+π/6)-cos(X-π/6)+ 根号三*cosx 的最大值和最小值.
若x∈【-π/2,0】,求函数f(x)=cos(X+π/6)-cos(X-π/6)+ 根号三*cosx 的最大值和最小值.
若x∈【-π/2,0】,求函数f(x)=cos(X+π/6)-cos(X-π/6)+ 根号三*cosx 的最大值和最小值.
f(x)=cos(X+π/6)-cos(X-π/6)+ 根号三*cosx
=cosxcosπ/6-sinxsinπ/6-(cosxcosπ/6+sinxsinπ/6)+√3cosx
=-sinx+√3cosx
=-2(1/2sinx-√3/2cosx)
=-2sin(x-π/6)
因为x∈【-π/2,0】
所以x-π/6∈【-π/2-π/6,0-π/6】=【-2π/3,-π/6】
所以最大值=2;
最小值=2*1/2=1.
f(x)=cos(x+π/6)-cos(x-π/6)+√3cosx
=(cosπ/6cosx-sinπ/6sinx)-(cosπ/6cosx+sinπ/6sinx)+√3cosx
=cosπ/6cosx-sinπ/6sinx-cosπ/6cosx-sinπ/6sinx+√3cosx
=-2sinπ/6sinx+√3cosx
=-2*1/2sinx+√3cosx
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f(x)=cos(x+π/6)-cos(x-π/6)+√3cosx
=(cosπ/6cosx-sinπ/6sinx)-(cosπ/6cosx+sinπ/6sinx)+√3cosx
=cosπ/6cosx-sinπ/6sinx-cosπ/6cosx-sinπ/6sinx+√3cosx
=-2sinπ/6sinx+√3cosx
=-2*1/2sinx+√3cosx
=-sinx+√3cosx
=2(√3/2cosx-1/2sinx)
=2cos(x+π/6)
x∈[-π/2,0]
x+π/6∈[-π/3,π/6]
cos(x+π/3)∈[1/2,√3/2]
2cos(x+π/3)∈[1,√3]
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f(x)=cos(X+π/6)-cos(X-π/6)+ 根号三*cosx
=-2sinxsin(π/6)+√3cosx
=√3cosx-sinx
=2[(√3/2)cosx-(1/2)sinx]
=2sin(π/3-x)
因x∈【-π/2,0】, π/3-x∈【π/3, 2π/3】
所以f(x)最大=f(-π/6)=2
f(x)最小=f(0)=2*(√3/2)=√3