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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+...

∵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|

(1)f(x)=1-cos(π/2+2x)-√3cos2x =1+sin2x-√3cos2x =2sin(2x-π/3)+1 (2)sin(2x-π/3)=1时取得最大值3, sin(2x-π/3)=-1时取得最小值-1 最小正周期T=2π/2=π.

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...

sin方转换成两倍角,然后和化积。

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)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)=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)=[cos(x+π/12)]^2=[1+cos{2(2x+π/6)}]÷2 (x=5π/12对称轴)代入g(x)=1+1/2sin2x=5/4(2)h(x)=[cos(x+π/12)]^2+1+1/2sin2x=3/2+1/4sin2x+√3/4cos2x=2/3+1/2sin(π/3+2x)至于值域自己会球吧

(1)∵f(x)=2cos(2x+π4)=1,∴cos(2x+π4)=12;又x∈[-π4,3π4],∴2x+π4∈[-π4,7π4],∴2x+π4=π3或2x+π4=5π3,解得:x=π24或x=17π24,∴x取值的集合为{π24,17π24};(2)∵2cos(2x+π4)≤-2,∴cos(2x+π4)≤-22,又2x+π4∈[-π4,7π4],∴3π4≤2x+π...

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