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

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,...

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...

n=1时公式成立; 现在假设对n-1公式成立 那么sinx+sin2x+sin3x+……+sinnx=sinx+sin2x+sin3x+……+sin(n-1)x+sinnx =[sin((n-1)x/2)sin(nx/2)]/sin(x/2)+sinnx =[sin((n-1)x/2)sin(nx/2)+sinnxsin(x/2)]/sin(x/2) =sin(nx/2)[sin((nx/2-x/2)+2cos(nx...

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...

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)=sin2x+√2(cosxcosπ/4-sinxsinπ/4) =2sinxcosx+(cosx-sinx) 令a=cosx-sinx=√2cos(x+π/4) 则-√2

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)=(sin2x-2sin²x)/sinx =(2sinxcosⅹ-2sin²x)/sinx =2(cosx-sinx) =2√2[cosxcos(π/4)-sinxsin(π/4)] =2√2cos(x+π/4) 显然,定义域为x∈R,即(-∞,+∞). cos(x+π/4)=1, 即x=2kπ-π/4时, 最大值f(x)|max=2√2. 0

f(x)=2sinxcosx+2√3sin²x-√3 =2sinxcosx+√3sin²x+√3(sin²x-1) =2sinxcosx+√3sin²x-√3cos²x =2sinxcosx+√3(sin²x-cos²x) =sin2x-√3cos2x =2(1/2 sin2x-√3/2 cos2x) =2sin(2x-π/3) sinx的单调递减区间为[π/2+2...

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