已知函数f x =2倍根号3sin xcosx 2sin^2x-1,x属于R

派,负四分之派到四分之派,再问：可我不知道怎么写过程。

f(x)=-√3sin²x+sinxcosx＝√3/2cos2x+1/2sin2x-1/2=sin(2x+π/3)+1/2T=2π/2=πf(π/6)=sin(π/3+π/3)+1/2=(1

f(x)=√3sin²x+sinxcosx=√3[(1-cos2x)/2]+1/2sin2x=1/2sin2x-√3/2cos2x+√3/2=sin(2x-π/3)+√3/2∵x∈[π/2,

fx=sin²wx+根号3倍的sinwxsin(wx+π/2)=（1-cos2wx)/2+√3sinwxcoswx=1/2-1/2cos2wx+√3/2sin2wx=sin(2wx-π/6)

f(x)=v3sin(π-2x)-2cos^2x+1=v3sin2x-cos2x=2sin(2x-π/6),（1）、f(π/2)=2sin(5π/6)=2*(1/2)=1；（2）、最小正周期T=2π/

f(x)=1/2-1/2cos2x+√3/2sin2x-1/2=sin(2x-π/6)f(-π/12)=sin(-π/3)=-√3/2(2)-π/6

f(x)=2√3sinxcosx+2sin^2x-1=√3sin2x-cos2x=2sin(2x-π/6)最小正周期T=π,单调递增区间:2kπ-π/2

F(X)=2√3sinx/2*cosx/2-(cos²x/2-sin²x/2)=√3sinx-cosx=2(√3/2sinx-1/2cosx)=2(sinxcosπ/6-sinπ/

(1)因为sin²x=（1-cos2x）/2,sin(2x-π/3)=1/2sin2x-根号3/2cos2x所以函数f(x)=2倍根号3sin²x-sin(2x-π/3)=-sin

先化简f(x)=2根号3sinxcosx+2cos^2x-1=根号3sin2x+cos2x=2(根号3/2sin2x+1/2cos2x)=2sin(2x+π/6）则T=2π/ω=2π/2=πy=sin

F(X)=2√3sinx/2*cosx/2-(cos^2x/2-sin^2x/2).=√3sinx-cosx=2(√3/2sinx-1/2cosx)=2sin(x-π/6)当x-π/6=2kπ+π/2

fx=2√3sinxcosx＋2cos^2x-1=√3sin2x＋cos2x=2(√3/2sin2x＋1/2cos2x)=2sin(2x＋π/6)所以最小正周期是π建议你再看看二倍角公式

答：f(x)=2sin(x-π/3)cosx+sinxcosx+√3(sinx)^2=sin(x-π/3+x)+sin(x-π/3-x)+sinxcosx+(√3/2)(1-cos2x)=sin(2x

答：y=f(x)=2√3sinxcosx-2sin²x=√3sin2x+cos2x-1=2*[(√3/2)sin2x+(1/2)cos2x]-1=2sin(2x+π/6)-1y=f(x)关于

第一题A.第二题B

（1）化简可得f(x)=(sin(x/2))^2+((√3)/2)sinx-0.5f'(x)=sin(x/2)cos(x/2)+((√3)/2)cosx=sinx+√3cosx=0√3cosx=-si

解答；f(x)=sin(2x+3分之π)∴sin(2x+π/3)=-3/5∵x∈(0,π/2)∴2x+π/3∈(π/3,4π/3)∵sin(2x+π/3)

解1当2kπ-π/2≤2x+π/3≤2kπ+π/2,k属于Z时,y是增函数即2kπ-5π/6≤2x≤2kπ+π/6,k属于Z时,y是增函数即kπ-5π/12≤x≤kπ+π/12,k属于Z时,y是增函数

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

fx=2根号3sinxcosx+1-2sin^2x=2sin（2x+π/6）周期为π