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已知F x 2sinx sin2x

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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)=1/2cos2x*(2x)]+cosx =cos2x+cosx 显然这是偶函数 cos2x+cosx =2cos²x-1+cosx =2(cosx+1/4)²-9/8 -1

f(x)=2sin²x+2√3sinx×sin(x+π/2) =1-cos2x+2√3sinxcosx =1-cos2x+√3sin2x =2(√3/2*sin2x-1/2*cos2x)+1 =2sin(2x-π/6)+1 最小正周期:T=2π/2=π 行家正解,不明白可以追问!祝您学习进步 满意请点击下面的【选为满意回答】按钮,O(∩_∩)O谢谢

解: ∵f(cosx)=2-sin2x=2-2sinxcosx 设cosx=t ∴sint=√(1-t^2) ∴f(t)=2-2t·√(1-t^2) 再设t=sinx ∴f(sinx)=2-2sinxcosx f(sinx)=2-sin2x

sin2x =sin(x+x) =sinxcosx+cosxsinx =2sinxcosx

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

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/...

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