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已知函数Fx等于2sin²x+2根号3sinx×sin(x+π/2)...

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

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)

f(x)=1/2•(cos²x-sin²x)-2asinx+b =-sin²x-2asinx+b-1/2 =-(sinx-a)²+b-1/2+a²。 u=sinx, -1≤u≤1, f(x)=-(u-a)²+b-1/2+a² 转化为二次函数在动轴定区间上的值域问题。分类讨论01. 注意抛物线y=g(u)开...

解由f(x)=√3/2sin2x+1/2(1-cos2x)+1/2 =sin(2x-π/3)+1 由x属于[-π/12,5π/12] 则2x属于[-π/6,5π/6] 则2x-π/3属于[-π/2,π/2] 则sin(2x-π/3)属于[-1,1] 则sin(2x-π/3)+1属于[-2,2] 故函数的值域为[-2,2]

已知函数f(x)=2(√3)sinxsin(π/2-x)+2cos²x+2,求f(x)的最小正周期与递减区间。 解:f(x)=2(√3)sinxsin(π/2-x)+2cos²x+2=2(√3)sinxcosx+(1+cos2x)+2 =(√3)sin2x+cos2x+3=2[(√3/2)sin2x+(1/2)cos2x]+3=2[sin2xcos(π/6)+cos2xsin(π/6)]+3...

f(x)=sinx^2+2√3sinxcosx+3cosx^2 =sinx^2+cosx^2+2√3sinxcosx+2cosx^2 =2√3sinxcosx+2cosx^2-1+2 =√3sin2x+cos2x+2 =2(√3/2sin2x+1/2cos2x)+2 =2sin(2x+π/6)+2 剩下就简单了

解: f(x)=2√3sinxcosx+1 -2sin²x =√3sin(2x)+cos(2x) =2[(√3/2)sin(2x)+(1/2)cos(2x)] =2sin(2x+π/6) -1≤sin(2x+π/6)≤1 -2≤2sin(2x+π/6)≤2 -2≤f(x)≤2 函数的值域为[-2,2]

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...

f(x)=sin(x/2)cos(x/2)+√3*sin²(x/2)+√3/2 =1/2*sinx+√3/2*(1-cosx)+√3/2 =1/2*sinx-√3/2*cosx+√3 =sin(x-π/3)+√3 ∴f(x)的最小正周期T=2π 令-π/2+2kπ≤x-π/3≤π/2+2kπ 得:-π/6+2kπ≤x≤5π/6+2kπ ∴f(x)的单调递增区间为[-π/6+2kπ,5π/6+2kπ] (k∈...

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