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已知函数F(x)4sin²(π/4+x)%2√3Cos2x%1,x∈...

∵f(x)=4sin^(π/4+x)-2√3cos2x-1 ∴f(x)=2[1-cos(π/2+2x)]-2√3cos2x-1 =2sin2x-2√3cos2x+1 =4sin(2x-π/6)+1 又π/2≤2x≤π,π/3≤2x-π/6≤5/6π ∴f(x)max=5 f(x)min=3 若q|f(x)-m|

f(x)=2sin²(π/4+x)-根号3cos2x =1-cos(π/2+2x)-√3cos2x =sin2x-√3cos2x+1 =2sin(2x-π/3)+1 ∵ x∈[π/4,π/2] ∴ 2x-π/3∈[π/6,2π/3] ∴ sin(2x-π/3)∈[1/2,1] ∴ 2x-π/3=π/6时,f(x)有最小值2 2x-π/3=π/2时,f(x)有最大值3

解答:(1)f(x)=2sin²(π/4+x)-根号3cos2x=1-cos(π/2+2x)-√3cos2x=sin2x-√3cos2x+1=2sin(2x-π/3)+1∵x∈[π/4,π/2]∴2x-π/3∈[π/6,2π/3]∴sin(2x-π/3)∈[1/2,1]∴2x-π/3=π/6时,f(x)有最小值22x-π/3=π/2时,f(x)有最大值3(2)|f(x)-m|m-2且f(x)m-2...

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...

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...

(1)由题意知,f(x)=2sin2(π4+x)?3cos2x?1=1?cos(π2+2x)?3cos2x?1=sin2x?3cos2x=2sin(2x?π3),∴h(x)=f(x+t)=2sin(2x+2t?π3),∴h(x)的图象的对称中心为(kπ2+π6?t,0),k∈Z,又∵已知点(?π6,0)为h(x)的图象的一个对称中心,∴t=kπ2+...

f(x)=2sin平方(π/4+x)-根号3cos2x,x∈(π/4,π/2) =[1-cos(2x+π/2)]-√3cos2x =1+sin2x-√3cos2x =1+2sin(2x-π/3) f(X)的最大值3和最小值1 x∈(π/4,π/2),y∈(1/2,1) │f(x)-m│<2 f(x)-2

(1)∵2sin2(π4+x)=2×1? cos(π2+2x) 2=1-cos(π2+2x)=1+sin2x,∴f(x)=2sin2(π4+x)+3cos2x?1=sin2x+3cos2x=2sin(2x+π3)所以f(x)的最小正周期T=2π2=π,由?π2+2kπ≤2x+ π3≤π2+2kπ,解得kπ?5π12≤x≤kπ+π12∴函数f(x)的单增区间为[kπ?5π12,kπ+π12]...

(1)∵f(x)=2sin2(π4+x)-3cos2x-1=-cos2(x+π4)-3cos2x=sin2x-3cos2x=2sin(2x-π3).则函数的最大值为2,最小值为-2,函数的周期T=2π2=π.(2)∵f(x)=2sin(2x-π3),∴h(x)=f(x+t)=2sin(2x+2t-π3),∵h(x)=f(x+t)的图象关于点...

f(x)=2sin2(π4+x)?3cos2x?1=2×1? cos(π2+2x) 2-3cos2x-1=1-cos(π2+2x)-3cos2x-1=sin2x-3cos2x=2sin(2x-π3),(5分)(1)f(π8)=sinπ4-3cosπ4=2?

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