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已知函数F(x)=sin2x+3sinxsin(x+π2),x∈R.(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

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...

(I)f(x)=1?cos2x2+32sin2x+(1+cos2x)=32sin2x+12cos2x+32=sin(2x+π6)+32.∴f(x)的最小正周期T=2π2=π.由题意得2kπ?π2≤2x+π6≤2kπ+π2,k∈Z,即kπ?π3≤x≤kπ+π6,k∈Z.∴f(x)的单调增区间为[kπ?π3,kπ+π6],k∈Z.(II)先把y=sin2x图象上所有...

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)=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...

因为f(x)=[2sin(x+π3)+sinx]cosx?3sin2x=[2(sinxcosπ3+sinπ3cosx)+sinx]cosx-3sin2x=sin2x+3cos2x=2sin(2x+π3)于是(I)函数f(x)的最小正周期T=2π2=π.(II)因为x∈[0,π4]∴π3≤2x+π3≤5π6∴12≤sin(2x+π3)≤ 1即:1≤y≤2∴f(x)max=2,f(x)min=1

f(x)=cos(x)-2sin2x+1,当x=0时,最大,f(x)=1-0+1=2;第二问找出x0/2和cos2x0的关系即可。希望可以帮到你

由cos2x=1-2sin2x,整理得f(x)=sinx+2sinx(0<x<π).令t=sinx,0<t≤1,则函数y=t+2t在t=1时有最小值3.故选B.

(1)函数y=sinx在(0,π2)上单调递增,但并不是在第一象限单调递增,故(1)错误;(2)∵f(x)=sin(2x3+7π2)=sin(23x-π2)=-cos23x,∴f(-x)=-cos(-23x)=-cos23x=-f(x),∴函数f(x)=sin(2x3+7π2)是偶函数,正确;(3)∵f(t+π3)=...

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