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已知函数Fx=Cos2x%sin2x+4sinx·Cosx,求F(x)的最小...

f(x)=cosxsin(x+π/6)-cos2x-1/4, =cosx(√3/2sinx+1/2cosx)-cos2x-1/4, =√3/2sinxcosx+1/2(cosx)^2-cos2x-1/4, =√3/4sin2x+1/4(1+cos2x)-cos2x-1/4, =√3/4sin2x-3/4cos2x =√3/2(1/2sin2x-√3/2cos2x) =√3/2sin(2x-π/3) 2x-π/3∈[2kπ-π/2,2kπ+π/2]单...

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...

解1f(x)=cos2x+√3×2sinxcosx =cos2x+√3sin2x =2(1/2cos2x+√3/2sin2x) =2sin(2x+π/6) 故T=2π/2=π 当2kπ-π/2≤2x+π/6≤2kπ+π/2,k属于Z时,y是增函数。 即2kπ-2π/3≤2x≤2kπ+π/3,k属于Z时,y是增函数。 即kπ-π/3≤x≤kπ+π/6,k属于Z时,y是增函数。 故函数...

(Ⅰ)f(x)=cos2x+3sin2x=2sin(2x+π6),(6分)∴f(π12)=2sin(π6+π6)=2sinπ3=3.(8分)(Ⅱ)由(Ⅰ)可知f(x)=2sin(2x+π6),∴函数f(x)的最小正周期T=2π2=π.(11分)函数f(x)的最大值为2.(13分)

f'(x)=1/2cos2x*(2x)]+cosx =cos2x+cosx 显然这是偶函数 cos2x+cosx =2cos²x-1+cosx =2(cosx+1/4)²-9/8 -1

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)=(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)=√3sinxcosx-cos²x-1/2 =(√3/2)sin2x-(1+cos2x)/2-1/2 =(√3/2)sin2x-(1/2)cos2x-1 =sin(2x-π/6)-1. 早小正周期T=2π/2=π. sin(2x-π/6)=-1, 即x=kπ-π/6时, 最小值f(x)|min=-2。

(1)f(x)=√3sin2x+cos2x=2(√3/2sin2x+1/2cos2x)=2sin(2x+π/6) 所以T=2π/2=π 因为x∈[0,π/2] ==>2x+π/6∈[π/6,7π/6] 根据正弦曲线的单调性 f(x)max=2×1 当2x+π/6=π/2时取得 f(x)max=2×(-1/2)=-1 当2x+π/6=7π/6时取得 (2)由题意可得2sin(2x1+π/6)=...

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