Y=根号3sinxcosx-cos²x 1

y=2√3sinxcosx+2cos²x=√3sin2x+cos2x+1=2(√3/2sin2x+1/2*cos2x)+1=2(sin2xcosπ/6+cos2xsinπ/6)+1=2sin

y=√3cos2x+sin2x=√[1²+(√3)²]sin(2x+z)=2sin(2x+z)其中tanz=√3/1=√3所以最大=2,最小=-2T=2π/2=π

函数y=sinxcosx+(√3)cos²x-(√3)/2的最小正周期y=(1/2)sin2x+(√3)(1+cos2x)/2-(√3)/2=(1/2)sin2x+(√3/2)cos2x=s

Y=√3SINXCOSX-COS²X=√3/2sin2x-(1+cos2x)/2=√3/2sin2x-1/2-cos2x/2=sin(2x-∏/6)-1/2ymax=1/2

y=1/2*cos²x+√3/2*sinxcosx+1=1/4*(cos2x-1)+√3/4*sin2x+1=1/2*(1/2*cos2x+√3/2*sin2x)+3/4=1/2*(sin3

y=√3cos²x+sinxcosx=(√3/2)[cos2x+1]+(1/2)sin2x=sin(2x+π/3)+√3/2所以值域为[-1+√3/2,1+√3/2]最小正周期为2π/2=π

y=1-4(sinx)^2-2倍根号3sin（2x）=1-2（1-cos2x）-2倍根号3sin（2x）=4sin（π/6-2x）-1当x∈[0,π/2],π/6-2x属于[-5π/6,π/6]所以最

y=1/2cos²x+√3/2sinxcosx+1=1/2(1/2+1/2cos2x+√3/2sin2x)+1=1/2(sin2xcosπ/6+cos2xsinπ/6)+1+1/4=1/2s

y=2cosxsin(x+π/3)-根号3*(sin^2)x+sinxcosx,后两项先提出一个sinx,然后括号内部分用叠加原理,得到y=2cosxsin(x+π/3)+2sinxcos(x+π/3

y=2根号3sinxcosx-2cos^2x+1＝根号3sin2x-cos2x=2sin(2x-π/6)T=2π/2=π2x-π/6=2kπ+π/2当x=kπ+π/3时,函数的最大值=22x-π/6=

y=sinxcosx+√3cos²x-√3/2=1/2*(2sinxcosx)+√3/2(2cos²x-1)=1/2*sin2x+√3/2cos2x.正弦余弦的二倍角公式=sin(

-3

y=sinxcosx+根号3cos^2x＝(1/2)sin2x＋根号3×(1＋cos2x)／2＝(1/2)sin2x＋(根号3／2)cos2x＋根号3／2＝sin(2x＋60°)＋根号3／2当2x＋6

y=-(cos²x-sin²x）-√3sinxcosx=-cos2x-(√3/2)sin2x=-sin(2x+π/6)或者y=-(cos²x-sin²x）-√3

y=sin²x+√3sinxcosx-1=[1-cos(2x)]/2+(√3/2)sin(2x)-1=(√3/2)sin(2x)-(1/2)cos(2x)-1/2=sin(2x-π/6)-1

由题知,已知函数y=cos²x+(√3)sinxcosx+1,x∈R,所以,化简y=cos²x+(√3)sinxcosx+1=(cos(2x)+1)/2+(√3/2)sin(2x)

y=cos²x+2√3sinxcosx-sin²x=cos²x-sin²x+2√3sinxcosx=cos2x+2√3sinxcosx=cos2x+√3sin2

函数y=√3cos^2x+sinxcosx=√3/2*(1+cos2x)+1/2*sin2x=√3/2+√3/2*cos2x+1/2*sin2x=√3/2+sin(2x+π/3)周期T=π最大值√3/

y=√3cos^2x-sinxcosx=√3/2*cos2x-1/2*sin2x+√3/2=cos(2x+30°）+√3/2最大值1+√3/2