f(x)=cos(π-x)sin(π/2+x)+根号3sinxcosx.求当x属于【0,π/2】是f(x)的最大值和最小值
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f(x)=cos(π-x)sin(π/2+x)+根号3sinxcosx.求当x属于【0,π/2】是f(x)的最大值和最小值
f(x)=cos(π-x)sin(π/2+x)+根号3sinxcosx.求当x属于【0,π/2】是f(x)的最大值和最小值
f(x)=cos(π-x)sin(π/2+x)+根号3sinxcosx.求当x属于【0,π/2】是f(x)的最大值和最小值
f(x)=cos(π-x)sin(π/2-x)+√3sinxcosx
=-(cosx)^2+(√3/2)sin2x
=(√3/2)sin2x-(1/2)cos2x-1/2
=sin2xcosπ/6-cos2xsinπ/6-1/2
=sin(2x-π/6)-1/2
0
f(x)=[-cosx]×[cosx]+√3sinxcosx
=(√3/2)sin2x-cos²x
=(√3/2)sin2x-(1/2)cos2x-(1/2)
=sin(2x-π/6)-(1/2)
因为x∈[0,π/2],则:2x-π/6∈[-π/6,5π/6],则:
sin(2x-π/6)∈[-1/2,1]
...
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f(x)=[-cosx]×[cosx]+√3sinxcosx
=(√3/2)sin2x-cos²x
=(√3/2)sin2x-(1/2)cos2x-(1/2)
=sin(2x-π/6)-(1/2)
因为x∈[0,π/2],则:2x-π/6∈[-π/6,5π/6],则:
sin(2x-π/6)∈[-1/2,1]
则:f(x)∈[-1,1/2]
函数f(x)最小值是-1,最大值是1/2
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f(x)=cos(π-x)sin(π/2+x)+√3sinxcosx
=-cosx²+(√3/2)sin2x
=-(cos2x+1)/2+(√3/2)sin2x
=(√3/2)sin2x-(1/2)cos2x-1/2
=cosπ/6sin2x-sinπ/6cos2x-1/2
=sin(2x-π/6)-1/2
∵x属于【0,π/2】,所以2x...
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f(x)=cos(π-x)sin(π/2+x)+√3sinxcosx
=-cosx²+(√3/2)sin2x
=-(cos2x+1)/2+(√3/2)sin2x
=(√3/2)sin2x-(1/2)cos2x-1/2
=cosπ/6sin2x-sinπ/6cos2x-1/2
=sin(2x-π/6)-1/2
∵x属于【0,π/2】,所以2x-π/6属于【-π/6,5π/6】
sin(2x-π/6)属于【-1/2,1】,则fx∈【-1,1/2】
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