已知f(x)=根3cosx的四次方

f(x)=sinx+cosxf'(x)=cosx-sinx=√2((1/√2)cosx-(1/√2)sinx)=√2(cos(x+π/4))f'(x)的最小正周期=2πy-f'(x)=sinx+cos

（1）已知函数f（x）=sinx+cosx,则f′（x）=sinx-cosx．代入F（x）=f（x）f′（x）+[f（x）]2易得F(x)=cos2x+sin2x+1=2sin(2x+π4)+1当2x

f(x)=sin2x+√3cosx=2((1/2)sin2x+(√3/2)cosx)=2sin(2x+π/3)单调递增区间2kπ-π/2

f(x)=2cosx(cosx+根号3sinx）=2cos^2x+2*根号3sinx*cosx=cos2x+根号3sin2x+1当cos2x=1/2sin2x=根号3/2有最大值3当cos2x=-1/

f'(x)=cosx-sinx=3(sinx+cosx)=>4sinx=-2cosx=>cosx=-2sinxsin^2x+cos^2x=1=>sin^2x=1/5cos^2x=4/5=>(sin^2

向量a=(sinx,根号3cosx),向量b=(cosx,cosx),f(x)=向量a*向量b=sinxcosx+√3cos²x=1/2sin2x+√3/2(1+cos2x)=1/2sin2

f(x)=sinxcosx+√3cos²x=(1/2)sin2x+(√3/2)(cos2x+1)=(1/2)sin2x+(√3/2)cos2x+√3/2=sin(2x+π/3)+√3/2周期

令x-2=t,则x=t+2f(t)=3(t+2)-5=3t+1t换成x,f(x)=3x+1f(1-cosx)=sin²x=1-cos²x=(1+cosx)(1-cosx)=(-1+

f（x）=-1+cos方x-cosx+9/4=cos方x-cosx+5/4=(cosx-1/2)^2+1cosx=1/2,f(x)最小=1cosx=-1,f（x）最大=9/4+1=13/4有问题留言

f(x)=√3/2*(cos(2x)-1)–1/2*sin2x=cos(2x+π/6)–√3/2(1)Tmin=π(2)-π/3≤x≤π/6,-π/2≤2x+π/6≤π/2值域[–√3/2,1–√3/

f(x)=sqrt(3)sinxcosx+cos^2x=sqrt(3)/2sin2x+1/2+1/2cos2x=sin(2x+pi/6)+1/2

f（x）=sinx+√3cosx=2(1/2sinx+√3/2cosx)=2(cosπ/3sinx+sinπ/3cosx)=2sin(x+π/3)所以最小正周期为：2π振幅为2再答：请采纳哦，谢谢再答

f(x)=2(sinxcosπ/6-cosxsinπ/6)=2sin(x-π/6)-1≤sin(x-π/6)≤1-2≤f(x)≤2值域是[-2,2]

|x|≤π/4f(x)=(cosx)^2+sinx=1-(sinx)^2+sinx=-[(sinx)^2-sinx+1/4]+5/4=-(sinx-1/2)^2+5/4当x=-π/4时,sinx=-√

f(x)=2(sinx*√3/2-cosx*1/2)=2(sinxcosπ/6-cosxsinπ/6)=2sin(x-π/6)、1/2表示2分之1=====sinx=4/5,题目应该有x∈〔-π/2,

1)f(x)=2cos^2x+2√3sinxcosx+m=cos2x+√3sin2x+m+1=2sin(2x+π/6)+m+1最小正周期T=2π/2=π2)f(x)在[-π/6,π/6]上时2x+π/

F(x)=3sinxcosx-cos²x=(3/2)sin2x-(1+cos2x)/2=(3/2)sin2x-(1/2)cos2x-1/2=(√10/2)sin(2x-∅)-1/

cosαsinβ=[sin(α+β)-sin(α-β)]/2f(x)=(√3cosx-sinx)sin2x/2cosx+1/2=(√3cosx-sinx)sinx+1/2=2cos(x+π/4)sin

f(π/3-x)=cos(π/3-x)/cos(π/6-(π/3-x))=cos(π/3-x)/cos(x-π/6)=cos(π/3-x)/cos(π/6-x)注：cosx=cos(-x)所以：f(x

你的问题呢?再问：求m的值，快一点，谢谢了、再答：是（sinx-cosx）的3次方？你条件和问题不清楚啊再问：已知函数,f(x)=(sinx-cosx)的平方乘m,x属于R再答：(sinx-cosx)