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

(Ⅰ)化简可得f(x)=sin2x+3sinxsin(x+π2)=1?cos2x2+3sinxcosx=32sin2x-12cos2x+12=sin(2x-π6)+12,可得周期T=2π2=π;(Ⅱ)由?π2+2kπ≤2x?π6≤π2+2kπ得?π6+kπ≤x≤π3+kπ,k∈z∴函数f(x)的单调递增区间是[?π6+kπ,π3+kπ],k∈z;(Ⅲ)由x∈[0,2π...

解fx=2sinxsin(x+π/6) =cos[x-(x+π/6)]-cos[x+(x+π/6)] =-cos(2x+π/6)+√3/2 故函数的周期T=2π/2=π 当2kπ≤2x+π/6≤2kπ+π,k属于Z函数是增函数 故函数的增区间是[kπ-π/12,kπ+5π/12],k属于Z 由x属于【0,π/2】 知2x属于【0,π】 即2x+π/6属于【π/6,...

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)=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、 f(x)=sinx(sinxcosπ/6+cosxsinπ/6) =√3/2*sin²x+1/2*sinxcosx =√3/2*(1-cos2x)/2+1/2*(sin2x)/2 =√3/4+1/2*(1/2*sin2x-√3/2*cos2x) =√3/4+(sin2xcosπ/3-cos2xsinπ/3) =√3/4+sin(2x-π/3) 所以T=2π/2=π 2、 0≤x≤π/2 -π/3≤2x-π/3≤2π/3 -√3...

解:f(x)=½sin(2x)+(√3/2)(cosx-sinx)=½sin(2x)+(√6/2)cos(x+π/4)cos(x+π/4)的最小正周期T₁=2π/1=2πf(x+2π)=½sin[2(x+2π)]+(√6/2)cos(x+2π+π/4)=½sin(2x+4π)+(√6/2)cos(x+2π+π/4)=½sin(2x)+(√6/2)cos(x+π/4)=f(x...

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

f(x)=sinx(sinxcosπ/6+cosxsinπ/6) =sinx(sinx*√3/2+cosx*1/2) =sin²x*√3/2+sinxcosx*1/2 =√3/2*(1-cos2x)/2+1/2*(sin2x)/2 =√3/4-√3/4*cos2x+1/4*sin2x =√3/4-1/2*(√3/2*cos2x+1/2*sin2x) =√3/4-1/2*(sinπ/3cos2x+cosπ/3sin2x) =√3/4-1/2*...

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(2x+π/3)-cos2x+(√3)sin2x =cos(2x+π/3)-2[(1/2)cos2x-(√3/2)sin2x] =cos(2x+π/3)-2[cos2xcos(π/3)-sin2xsin(π/3)] =cos(2x+π/3)-2cos(2x+π/3)=-cos(2x+π/3) ①.单调区间: 由 2kπ≤2x+π/3≤2kπ+π,得单增区间为:[kπ-π/6,kπ+π/3]; 由 2kπ...

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