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已知函数F(x)=sin2x+3sinxsin(x+π2),x∈R.(1...

f(x)=sin2x-2√3sin^2x+√3+1 =sin2x+√3(-2sin^2x+1)+1 =(sin2x+√3cos2x)+1 =(sin2xcos(π/3)+cos2xsin(π/3))*2+1 =2sin(2x+π/3)+1 最小正周期=π -π/2+2kπ

因为这里不便书写,故将我的答案做成图像贴于下方,谨供楼主参考。 (若图像显示过小,点击图片可放大)

(1)f(x)=2sinx(32cosx-12sinx)+3sinxcosx+cos2x=23sinxcosx+cos2x-sin2x=3sin2x+cos2x=2sin(2x+π6)…(6分)∴f(x)的最小正周期T=2π2=π.当2x+π6=2kπ+π2,x=kπ+π6(k∈z)时,f(x)取得最大值2.…(10分)(2)由0≤x≤π2,得π6≤2x+π6≤7π6...

一、使用排除法 A、当x=0,sin2x=0,f(sin2x)=f(0)=sinx =sin0 =0 而当x=π/2时,sin2x=0,f(sin2x)=f(0)=sinx =sin(π/2)=1 出现了f(0)=0和f(0)=1,不符合函数定义,错误。 B、当x=0,sin2x=0,f(sin2x)=f(0)=x^2+x=0 而当x=π/...

f(x)=(sinx)^2+2√3sinxcosx+sin(x+π/4)sin(x-π/4) =1/2(1-cos2x)+√3sin2x+sin(x+π/4)sin[-π/2+(x+π/4)] =1/2-1/2cos2x+√3sin2x-sin(x+π/4)cos(x+π/4) =1/2-1/2cos2x+√3sin2x-1/2sin(2x+π/2) =1/2-1/2cos2x+√3sin2x-1/2cos2x =√3sin2x-cos2x+1/2 ...

(1)∵f(x)=sin2x+23sinxcosx+3cos2x,∴f(x)=1+3sin2x+2cos2x=3sin2x+cos2x+2=2sin(2x+π6)+2,故函数的最小正周期为T=2π2=π,由2kπ-π2≤2x+π6≤2kπ+π2,可得kπ-π3≤x≤kπ+π6,故函数单调递增区间为:[kπ-π3,kπ+π6](k∈Z)(2)∵x∈[-π6,π3],...

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)的最小正周期T=2Pai/2=Pai,初相是:Pai/4 g1(x)=sin(2(x-Pai/8)+Pai/4)=sin2x g(x)=sinx 3.h(x)=|sinx-1/2|sinx =(sinx)^2-1/2sinx,(sinx>=1/2) =(sinx-1/4)^2-1/16 故h(x)max=h(1)=1/2,h(x)min=h(1/2)=0 h(x)=1/2sinx-(sinx)^2,(-1

max=3/2,min=1 解: f(x) =sin²x+√3sinxcosx =(1-cos2x)/2+(√3/2)sin2x =sin(2x-π/6)+1/2 ∵x∈[π/4,π/2] ∴2x-π/6∈[π/3,5π/6] ∴ f(x)_max=1+1/2=3/2 f(x)_min=1/2+1/2=1

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