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已知函数F(x)=2sin2(π4+x)?3Cos2x?1(x∈R)(1)求F...

(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+...

(1)∵f(x)=2sin2(π4+x)-3cos2x=[1-cos(π2+2x)]-3cos2x=1+sin2x-3cos2x=2sin(2x-π3)+1∴函数f(x)的最小正周期T=π(2)2x-π3∈[2kπ+π2,2kπ+3π2],k∈Z 解得:x∈[kπ+5π12,kπ+11π12],k∈Z 函数f(x)的单调递减区间[kπ+5π12,kπ+11π12],k...

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

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)=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...

下图继续……

(1)f【4/π】=-(a+1)sinθ=0, ∵θ∈(0,π). ∴sinθ≠0, ∴a+1=0,即a=-1 ∵f(x)为奇函数, ∴f(0)=(a+2)cosθ=0, ∴cosθ=0,θ=2/π。 (2)由(1)知f(x)=(-1+2cos2x)cos(2x+2/π)=cos2x•(-sin2x)=-2/1sin4x, ∴f(4/a)=-2/1 ...

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

(I)f(x)=sin2x-3cos2x+1=2sin(2x-π3)+1,∵ω=2,∴T=π;(II)∵x∈[π4,π2],2x-π3∈[π6,2π3],∴sin(2x-π3)∈[12,1],即2sin(2x-π3)+1∈[2,3],则函数f(x)的最大值为3,最小值为2.

sin方转换成两倍角,然后和化积。

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