函数f(x)=5根号3cosx的平方加根号3sinx的平方减4sinxcosx

已知:函数f(x)=2sinxcosx+2√3cos²x-√3求:(1)单调增区间和最小正周期；(2)当x∈[-π/4,π/4]时求最值.f(x)=2sinxcosx+2√3cos²

f(x)=2（sin兀/3cosx-cos兀/3sinx）=2sin（兀/3-x）0

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/

1.ab={5√3cosx,cosx}*{sinx,2cosx}=5√3cosx*sinx+cosx*2cosx=5√3sinxcosx+2cos^x=(5√3/2)sin2x+2cos^xb^=si

ab=5根号3sinxcosx+2cos^2xb^2=sin^2x+4cos^2x+4sinxcosxf(x)=5根号3sinxcosx+2cos^2x+sin^2x+4cos^2x+4sinxcos

首先把cosx乘进去,得到根号3+sinx;然后用三角公式得到2*sin(x+60).其中2是根号3的平方加上sin前面那个1的平方再开根号所得.所以答案就是2.

f(x)=5根号3*sinxcosx+2(cosx)²+(sinx)²+4(cosx)²=5根号3sinxcosx+5(cosx)²+1=5*根号3/2*sin

（1）f(x)＝.2sinx3(sinx−cosx)sinx+cosxcosx.＝sin2x+3cos2x＝2sin(2x+π3)…（3分）所以函数f（x）的最小正周期为π…（3分）（2）

f(x)=sinx+根号3cosx=2*sin(x+pi/3)1.T=2pi2.x用x-pi/3代替:y=sinx单调增区间：【0,pi/2】

万能代换：设sinx=2k/(1+k^2),cosx=(1-k^2)/(1+k^2).代入得y=(3k^2+2k+3)/√(5k^2+5+8k+3-3k^2)=(3k^2+2k+3)/√2(k^2+4

1a·b=√3sinxcosx+cosx^2=√3sin(2x)/2+(1+cos(2x))/2=sin(2x+π/6)+1/2故：f(x)=a·b-1/2=sin(2x+π/6)最小正周期：T=2π

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]

f(x)=cos²x+√3sinxcosx=1/2(2cos²x-1)+√3sinxcosx+1/2=1/2cos2x+√3/2sin2x+1/2=cos(2x-π/3)+1/2当

1、f(x)=(√3/2)sin2x+(1/2)cos2x=sin(2x+π/6)周期T=2π/2=π当x∈【0,π/2】时,2x+π/6∈【π/6,7π/6】则sin(2x+π/6)∈【-1/2,1

f(x)=sinx+根号3cosx=2sin(x+π/3),即最小正周期为2π得到的g(x)=2sin(x+π/3-π/3)=2sinx,即在(O,π/2】上单调递增,在【π/2,π）上单调递减

函数y=sinx/根号(5+4cosx)[0

f(x)=√3sinxcosx+(cosx)^2+m=sin(2x+π/6)+m+1/21.最小正周期T=2π/2=π2.x∈[-π/6,π/3]2x+π/6∈[-π/6,5π/6]可见x=-π/6时

1、由x范围则cosx>0sin²x+cos²x=1所以cosx=3/5所以f(x)=(4√3-3)/52、f(x)=2(sinx*√3/2-cosx*1/2)=2(sinx

f(x)=√3sinx-cosx=2(√3/2sinx-1/2cosx)=2(sinxcosπ/6-cosxsinπ/6)=2sin(x-π/6)所以最小值是-2