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当0<x<π/4时,函数F(x)=Cos^2x/Cosxsinx%sin^2x的...

分子分母同除以 cos^2 x, 得 f(x) = 1/(tanx -tan^2 x) 0= 2√ 1/u(1-u) √ 1/u(1-u) >=2 1/u(1-u) >= 4 f(x) =cos^2x/(cosxsinx-sin^2x) = 1/(tanx -tan^2 x) = 1/u(1-u) >=4 最小值是4

∵0<x<π4,∴0<tanx<1,cosx≠0.∴f(x)=2cos2xsinxcosx?sin2x=2tanx?tan2x=2?(tanx?12)2+14,由0<tanx<1,∴分母>0,∴当且仅当tanx=12,f(x)取得最小值8.故选D.

待续

f(x)=2sinxcosx-sin²x+cos²x =sin2x+cos2x =√2sin(2x+π/4) 在[0, π]上,2x+π/4∈[π/4, 2π+π/4] 其中单调增区间为: 2x+π/4 ∈[π/4, π/2]U[3π/2,9π/4] 单调减区间: 2x+π/4 ∈[π/2, 3π/2] 即f(x)的单调增区间为:[0, π/8]U[5π/8, π] f(x)...

f(x)=(cosx)^4-2sinxcosx-(sinx)^4 =[(cosx)^2+(sinx)^2][(cosx)^2-(sinx)^2]-2sinxcosx =(cosx)^2-(sinx)^2-2sinxcosx =cos(2x)-sin(2x) = √2*[cos(2x)*√2/2-sin(2x)*√2/2] = √2*[cos(2x)cos(π/4)-sin(2x)sin(π/4)] = √2cos(2x+π/4) 。 (2)因...

f(x)=sin2x+√2(cosxcosπ/4-sinxsinπ/4) =2sinxcosx+(cosx-sinx) 令a=cosx-sinx=√2cos(x+π/4) 则-√2

1、 f(x)=sinx(sinxcosπ/6+cosxsinπ/6) =√3/2*sin²x+1/2*sinxcosx =√3/2*(1-cos2x)/2+1/2*(sin2x)/2 =√3/4+1/2*(1/2*sin2x-√3/2*cos2x) =√3/4+(sin2xcosπ/3-cos2xsinπ/3) =√3/4+sin(2x-π/3) 所以T=2π/2=π 2、 0≤x≤π/2 -π/3≤2x-π/3≤2π/3 -√3...

解: (1) f(x)=cos²x-√3sinxcosx+½ =½[1+cos(2x)]-(√3/2)sin(2x)+½ =½cos(2x)-(√3/2)sin(2x)+1 =cos(2x+π/3)+1 最小正周期T=2π/2=π cos(2x+π/3)=1时,f(x)取得最大值f(x)max=1+1=2 cos(2x+π/3)=-1时,f(x)取得最小值f(...

f(x) = 4sinx(cosx-sinx)+3 = 4sinxcosx-4sin²x+3 = 2sin2x-2(1-cos2x)+3 = 2sin2x+2cos2x+1 = 2√2(sin2xcosπ/4+cos2xsinπ/4) + 1 = 2√2sin(2x+π/4) + 1 x∈(0,π) 2x∈(0,2π) 2x+π/4∈(π/4,9π/4) 2x+π/4∈(π/2,3π/2)时单调递减 此...

1.f(x)=(sinx+sin3x)/(cosx+cos3x) =(sinx+3sinx-4sin^3 x)/(cosx+4cos^3 x-3cosx) =(4sinx-4sin^3 x)/(4cos^3 x-2cosx) =[4sinx(1-sin^2 x)]/[2cosx(cos^2 x-1)] =(4sinxcos^ x)/(2cosxcos2x) =(2sin2xcosx)/(2cosxcos2x) =tan2x T=π/2 2.f(x)=...

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