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已知函数F(x)=2Cosxsin(x+π/3)+sinxCosx%√3sin²x...

f(x)=2cosxsin(x+π/3)-√3sinx+sinxcosx=2cosx(1/2*sinx+√3/2*cosx) -√3sinx+sinxcosx= sinxcosx+√3cosx-√3sinx+sinxcosx=2 sinxcosx+√3(cosx-sinx)=sin2x+√3cos2x=2sin(2x+π/3)(1)最小正周期=2π/2=π(2)x∈[0,5π/12]2x+π/3∈[π/3,7π/6]sin(2x+π/3)∈[-1/2,1]2sin(2x+π/3)∈[-1,2]f(x)>m成立m 评论0 0 0

解∵f(x)=2cosx[sinxcos(π/3)+cosxsin(π/3)]-√3(sinx)^2+sinxcosx =sinxcosx+√3(cosx)^2-√3(sinx)^2+sinxcosx =sin2x+√3cos2x =2[(1/2)sin2x+(√3/2)cos2x] =2[sin2xcos(π/3)+cos2xsin(π/3)] =2sin[2x+(π/3)] ∴当2x+(π/3)=π/2时,函数f(x)有最大值且f(

f(x)=2cosx*[sinx*(1/2)+cosx*(√3/2)]-√3sin^2 x+sinxcosx=sinxcosx+√3cos^2 x-√3sin^2 x+sinxcosx=2sinxcox+√3(cos^2 x-sin^2 x)=sin2x+√3cos2x=2sin[2x+(π/3)]所以:①最小正周期T=2π/2=π②最大值=2,最小值=-2③递增区间为2x+(π/3)∈[2kπ

f(x)=[2(sinx*1/2+cosx*√3/2)+sinx]cosx-√3sinx =(2sinx+√3cosx)cosx-√3sinx =2sinxcosx+√3(cosx-sinx) =sin2x-√3cos2x =2sin(2x-π/3) T=2π/2=π

f(x)=2sin(x-π/3)cosx+sinxcosx+√3sin^2x =2(sinxcosπ/3-cosxsinπ/3)cosx+sinxcosx+√3sin^2x =sinxcosx-√3cos^2x+sinxcosx+√3sin^2x =2sinxcosx-√3(cos^2x-sin^2x) =sin2x-√3cos2x =2(1/2sin2x-√3/2cos2x) =2sin(2x-π/3)由2kπ-π/2≤2x-π/3≤2kπ+π/2,k∈Z得kπ-π/12≤x≤kπ+5π/12,k∈Z∴f(x)递增区间为[kπ-π/12,kπ+5π/12],k∈Z

f(x)=2cosx*(1/2*sinx+√3/2*cosx)-√3sinx+sinxcosx =sinxcosx+√3*cosx-√3sinx+sinxcosx =2sinxcosx+√3(cosx-sinx) =sin2x+√3cos2x =2(1/2*sin2x+√3/2*cos2x) =2sin(2x+π/3)(1)令2kπ+π/2≤2x+π/3≤2kπ+3π/2解得:kπ+π/12≤x≤2kπ+7π/12所以f(x)的单调递减区间集合为[kπ+π/12,2kπ+7π/12] (k∈Z)

最大值2,最小值-2.通过化简,化成一个三角函数式,即可知道结果. 请参考图片.

f(x)=2cosxsin(x+π/3)+sinx(cosx-√3sinx)=2cosxsinxcosπ/3+2cos^2xsinπ/3+sinxcosx-√3sin^2x=sin2x*1/2+(cos2x-1)*√3/2+1/2sin2x-√3/2(1-cos2x)=sin2x+√3cos2x-√3=2(1/2sin2x+√3/2cos2x)-√3=2sin(2x+π/3)-√3T=2π/2=π

1.f(x)=2cosx*sin(x+π/3)-√3sin^2x+sinx*cosx= 2cosx*sin(x+π/3)- 2sinx*[(√3/2)sinx-(1/2)cosx]= 2cosx*sin(x+π/3)- 2sinx*[sin(π/3)sinx-cos(π/3)cosx)]= 2cosx*sin(x+π/3)+ 2sinxcos(x+π/3)= 2sin(2x+π/3)f(x)的单调区间 2kπ - π/2 [2kπ - π/2 - π/3]/2 2sin[2(x-

(1). f(x)=2cosx*sin(x+π/3)-√3sin^2x+sinx*cosx=2cosx(1/2sinx+√3/2cosx)-√3sin^2x+sinxcosx=2sinxcosx+√3cos^2x-√3sin^2x=sin2x+√3cos2x=2(1/2sin2x+√3/2cos2x)=2sin(2x+π/3)令π/2+2kπ<=2x+π/3<=3π/2+2kπ,即π/12+kπ<=x<=7π/12+kπ,即f

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