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已知函数F(x)=4sinxsin²(π/4+x/2)+Cos2x%1

f(x)=4sinxsin²(π/4+x/2)+cos2x-1 =4sinxsin²[(π/2+x)/2]+cos2x-1 =4sinx[1-cos(π/2+x]/2+cos2x-1 =2sinx+2sin²x+1-2sin²x-1 =2sinx ∴f(ωx)=2sinωx x∈[-π/2,2π/3]是增函数 f'(ωx)=2ωcosωx>0 ∵ω>0 ∴cosωx>0 ω·2π/3≤π/2→ω≤3/4...

(1)函数f(x)=4sinx?sin 2 ( π 4 + x 2 )+cos2x=2sinx[1-cos( π 2 +x )]+cos2x=2sinx+2sin 2 x+cos 2 x-sin 2 x=1+2sinx…(4分)由题意需[- π 2 , 2π 3 ] ?[- π 2ω , π 2ω ] 得ω ∈(0, 3 4 ] …(6分)(2)由题意当 π 6 ≤x≤ 2π 3 时,...

(1)∵f(x)=2sin2x-23cos2x+1=4sin(2x-π3)+1.又∵π4≤x≤π2,∴π6≤2x-π3≤2π3,即3≤4sin(2x-π3)+1≤5∴f(x)max=5,f(x)min=3(2)∵|f(x)-m|<2,∴m-2<f(x)<m+2又p是q的充分不必要条件∴m?2<3m+2>5,∴3<m<5.∴m的取值范围为(3,5)

解: (1) tanx有意义,x≠kπ+ π/2,(k∈Z) 函数定义域为{x|x≠kπ+ π/2,k∈Z} f(x)=4tanxsin(π/2 -x)cos(x- π/3) -√3 =4tanxcosxcos(x-π/3)-√3 =4sinx[cosxcos(π/3)+sinxsin(π/3)] -√3 =4sinx[(1/2)cosx+(√3/2)sinx] -√3 =2sinxcosx+2√3sin²x-√...

(1)f(x)=4sinx[sin(π/4+x/2)]^2+2(cosx)^2+1+a =2sinx[1-cos(π/2+x)]+2+cos2x+a =2sinx(1+sinx)+cos2x+2+a =2sinx+2(sinx)^2+cos2x+2+a =2sinx+3+a, x∈R是一个奇函数, ∴f(0)=3+a=0,a=-3. (2)f(x)=2sinx, f(2x)=2sin2x>=-√3, ∴sin2x>=-√3/2, ∴...

f(x)=1+2cos2x?12cosx+sinx+a2sin(x+π4)=cosx+sinx+a2sin(x+π4)=2sin(x+π4)+a2sin(x+π4)=(2+a2)sin(x+π4).依题意有2+a2=2+3,∴a=±3.故答案为:±3

f(x)=4sinxcos(x+π/3)+根号3 =4sinx(1/2cosx-√3/2sinx)+√3 =2sinxcosx-2√3sin²x+√3 =sin2x-2√3sin²x+√3(sin²x+cos²x) =sin2x+√3cos²x-√3sin²x =sin2x+√3(cos²x-sin²x) =sin2x+√3con2x =2sin(2x+π/3) ...

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...

已知函数f(x)=-(√2)sin(2x+π/4)+6sinxcosx-2cos²x+1,x∈R; (1).求f(x)的最小正周期;(2).求f(x)在区间[0,π/2]上的最大值和最小值. 解:(1)。f(x)=-(sin2x+cos2x)+3sin2x+cos2x=2sin2x,故最小正周期T=π; (2)。在区间[0,π/2]上的最大...

(1)f(x)=sin2x-cos2x=2sin(2x-π4),∴T=2π2=π.(2)∵x∈[0,π2],∴2x-π4∈[-π4,3π4],∴sin(2x-π4)∈[-22,1],∴函数的值域为[-1,2].

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