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已知函数F(x)=√3sin(2φx%pAi/3)+B,且该函数图像的...

最小距离为pai/4说明周期为pai…所以φ为1…最大值为1 所以b=1-√3 解析式带入即可 当x=0时,函数取最小值 x=pai/3时,取最大值 最小带入-3那边 最大带入+3那边 求出m范围

f(x)=√3sin2x+cos2x =2(√3/2sin2x+1/xcos2x) =2sin(2x+π/6) f(x-φ)=2sin[2(x-φ)+π/6] =2sin(2x+π/6-2φ) ∵f(x-φ)图像关于y轴对称 ∴x=0时,函数取得最值 即π/6-2φ=kπ+π/2,k∈Z φ=-kπ/2-π/6,k∈Z ∵0

已知函数fx=2cosxsin(x+pai/3)-根号3sin^2x+sinxcosx,求振幅,周期,初相,单调区间 解析:fx=2cosxsin(x+π/3)-√3sin^2x+sinxcosx=2cosx(sinx*1/2+√3/2*cosx)-√3sin^2x+1/2sin2x =sin2x+√3(cos^2x-sin^2x)=sin2x+√3cos2x=2sin(2x+π/3) 所以...

f(x)=cos(2x-π/3)+2sin(x-π/4)cos[π/2-(x+π/4)] =cos(2x-π/3)+2sin(x-π/4)cos(π/4-x) =cos(2x-π/3)+2sin(x-π/4)cos(x-π/4) =cos(2x-π/3)+sin(2x-π/2) =cos(2x-π/3)-cos2x =(1/2)cos2x+(√3/2)sin2x-cos2x =(√3/2)sin2x-(1/2)cos2x =sin(2x-π/6) ...

令wx=a f(x)=sin²a+√3sinacosa=(1-cos2a)/2+√3/2sin2a =1/2+√3/2sin2a-1/2cos2a =sin(2a-π/6)+1/2 =sin(2wx-π/6)+1/2 根据题意 2π/2w=π w=1

f(x)=根号3sin2x+cos2x+1+m=2sin(2x+Pai/6)+m+1 0

f(x)=cos(2x+π/3)+sin²x =cos2xcosπ/3-sin2xsinπ/3+[1-cos(2x)]/2 =1/2cos2x-√3/2sin2x+1/2-1/2cos2x =-√3/2sin2x+1/2 当sin2x=-1时,最大值=(1+√3)/2 最小正周期=2π/2=π f(C/2)=-√3/2sinC+1/2=-1/4 sinC=√3/2 ∵C为锐角 C=π/3 cosC=1/2 ∵cos...

可以先采纳嘛,做出来了

f(x)=2√3sinxcosx+2cos²x+m=√3sin2x+1+cos2x +m=2sin(2x+π/6) +m+1。 0

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