f(x)=(1 根号2cos(2x-pi 4)) cosx

是“函数f(x)=(1/2)cosx平方+((根号3)/2)sinxcosx-(1/4)”吧f（x)=(cos(2x)+1)/4+(（根号3)sin2x)/4-1/4=1/2(sin(2x+∏/6))

1、f(x)=1/2*sin(2x/3)+√3*(1+cos2x/3)/2=1/2*sin(2x/3)+√3/2*cos(2x/3)+√3/2=sin(2x/3)cosπ/3+cos(2x/3)sin

f(x)=cos^2x-sin^2x+2(根号3)sinxcosx+1=cos2x+(根号3)sin2x+1=2{(1/2)cos2x+[(根号3)/2]sin2x}+1=2sin(2x+派/6)+1

1.f(x)=cos(π/2+2X)COS(π+x)=-sin2x*(-cosx)=sin2xcosxf(-x)=sin(-2x)cos(-x)=-sin2xcosx=-f(x)∴f(x)是奇函数2.

答：f(x)=(1/2)*(cosx)^2+(√3/2)sinxcosx+1=(1/2)*(cos2x+1)/2+(√3/4)sin2x+1=(1/2)[sin2xcosπ/6+cos2xsinπ/6

f(x)=1/2*sin2x+√3/2*cos2x-√3/2=√(1/4+3/4)*sin[2x+arctan√3]-√3/2=2sin(2x+π/3)-√3/2所以T=2π/2=π0

f(x)=2√3sinxcosx-2cos^2x+1=√3*2sinxcosx-(2cos^2x-1)=√3sin(2x)-cos(2x)=2sin(2x-π/6)∴f(x)的最小正周期=2π/2=π

f(x)=2√3sinxcosx+2cos^2x-1=√3sin(2x)+cos(2x)=2[sin(2x)(√3/2)+cos(2x)(1/2)]=2[sin(2x)cos(π/6)+cos(2x)

（1）f(x)=1/2*cos2x*1/2+(根号(3))/2*sin2x*1/2-1/4=1/2*(1/2*cos2x+(根号(3))/2*sin2x)-1/4=1/2*sin(2x+pi/6)-1

一样是先化简的问题f(x)=2cos^2x+√3*sin2x-1＝2(cosx)^2-1+√3*sin2x=cos2x+√3*sin2x=2(1/2*cos2x+√3/2*sin2x)=2(sinπ/

F(X)=2乘根号3sinxcosx+2cos^2X-1=根号3sin2x+cos2x=2cos(2x-派/3)函数最小正周期=2派/2=派单调递减区间[派/6,2派/3]fx0=6/5,cos(2x

f(x)=1+2(√3)sinxcosx+2(cosx)^2f(x)=2(√3)sinxcosx-[1-2(cosx)^2]+2f(x)=(√3)sin2x-cos2x+2f(x)=2{[(√3)/2

f(x)=2＋√3sin2x＋cos2x=2sin(2x＋π/6)＋21、增区间：2kπ－π/2≤2x＋π/6≤2kπ＋π/2kπ－π/3≤x≤kπ＋π/6则：增区间是：[kπ－π/3,kπ＋π/6]

f(x)=2根号3sinxcosx+2cos^2x-1=√3sin2x+cos2x=2sin(2x+π/6)kπ-π/3再问：(cos2x)^2=16/25+2√3/5sin2x+3(sin2x)^2

f(x)=1/2cos2x+根号3/2sin2x-1-cos2x=√3/2×sin2x－1/2×cos2x－1＝sin(2x-π/6)－1当sin(2x-π/6)＝1时f(x)有最大值＝0f(a)=-

1).f(x)=(√3)sin2x+cos2x+2=2sin(2x+π/6)+2单增区间[-π/3+kπ,π/6+kπ](2)f(C)=2sin(2C+π/6)+2=3,故C=π/3a/b=sinA/

（1）f=[根号3]/2sin2x-cos^2(x)-1/2=√3/2*sin2x-1/2(1+cos2x)-1/2=√3/2sin2x-1/2cos2x-1=sin(2x-π/6)-1∵x∈[-π/

f(x)=根号3cos^2x+1/2sin2x=根号3/2*cos2x+1/2sin2x+根号3/2=sin(2x+π/3)+根号3/2y=sinx递减区间[2kπ+π/2,2kπ+3π/2]y=si

f(x)=(根号3/2)sin2x-cos^2x+1/4=(根号3/2)sin2x-(2cos^2x-1)/2+3/4=(根号3/2)sin2x-(cos2x)/2+3/4=sin((2x-π/6))