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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

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]

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...

已知函数f(x)=2sin²x+2根号3sinxcosx+1,求函数的单调递增区间 函数在区间【0,п/2】上的最值 解析:∵函数f(x)=2sin²x+2√3sinxcosx+1=-cos2x+√3sin2x+2 =2sin(2x-π/6)+2 其单调增区间为: 2kπ-π/2

亲,网友,您说的是不是下面的问题: 已知函数f x=sin(兀/2-x)sinx-根号3cos^2x,求周期、最值。 f(x)=1/2 sin2x-√3/2(cos2x+1) =sin(2x-π/3)-√3/2 T=2π/2=π。 f max=1-√3/2, f min=-1-√3/2. 送您 2015 夏祺 凉快

(1)f(x)=(1-cos2x)/2+√3/2*sin2x+1+cos2x=√3/2*sin2x+1/2*cos2x+3/2=sin(2x+π/6)+3/2所以f(x)的最小正周期为π在(kπ-π/3,kπ+π/6)上单调递增,在(kπ+π/6,kπ+2π/3)上单调递减(2)f(x)可以由函数y=sin2x的图像向左平移π/12,向上平移3/2得到

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]

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)

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