1.向量a=(2cos(x/2),tan(x/2+π/4)),向量b=((√2)sin(x/2+π/4),tan(x/2-π/4),令f(x)=a•b,则
f(x)max=————,T=————.
f(x)=a•b=2(√2)cos(x/2)sin(x/2+π/4)+tan(x/2+π/4)tan(x/2-π/4)
=2cos(x/2)[sin(x/2)+cos(x/2)]+[1+tan(x/2)][1-tan(x/2)]/{[1-tan(x/2)][1+tan(x/2)]}
=sinx+2cos²(x/2)+1=sinx+cosx+2=(√2/2)sin(x+π/4)+2
故maxf(x)=2+(√2)/2,minf(x)=2-(√2)/2；最小正周期Tmin=2π
2.已知向量m=（sinA,cosA）,向量n=（√3,-1）,向量m•n=1,A为锐角,求①A=——
②f（x）=cos2x+4cosAsinx（x∈R）的值域.
① m•n=(√3)sinA-cosA=2[(√3/2)sinA-(1/2)cosA]=2[sinAcos(π/6)-cosAsin(π/6)]
=2sin(A-π/6)=1,故sin(A-π/6)=1/2,∴A-π/6=π/6,故A=π/3.
②f(x)=cos2x+4cos(π/3)sinx=cos2x+2sinx=1-2sin²x+2sinx=-2(sin²x-sinx)+1=-2[(sinx-1/2)²-1/4]+1
=-2(sinx-1/2)²+3/2
由于0≦(sinx-1/2)²≦9/4,(当sinx=1/2时获得最小值0；当sinx=-1时获得最大值9/4)；
故-9/2≦-2(sinx-1/2)²≦0；于是得 -3≦-2(sinx-1/2)+3/2≦3/2,即f(x)的值域为[-3,3/2]. 1.向量a=(2cos(x/2),tan(x/2+π/4)),向量b=((√2)sin(x/2+π/4),tan(x/2-π/4),令f(x)=a•b,则
f(x)max=————,T=————.
f(x)=a•b=2(√2)cos(x/2)sin(x/2+π/4)+tan(x/2+π/4)tan(x/2-π/4)
=2cos(x/2)[sin(x/2)+cos(x/2)]+[1+tan(x/2)][1-tan(x/2)]/{[1-tan(x/2)][1+tan(x/2)]}
=sinx+2cos²(x/2)+1=sinx+cosx+2=(√2/2)sin(x+π/4)+2
故maxf(x)=2+(√2)/2,minf(x)=2-(√2)/2；最小正周期Tmin=2π
2.已知向量m=（sinA,cosA）,向量n=（√3,-1）,向量m•n=1,A为锐角,求①A=——
②f（x）=cos2x+4cosAsinx（x∈R）的值域.
① m•n=(√3)sinA-cosA=2[(√3/2)sinA-(1/2)cosA]=2[sinAcos(π/6)-cosAsin(π/6)]
=2sin(A-π/6)=1,故sin(A-π/6)=1/2,∴A-π/6=π/6,故A=π/3.
②f(x)=cos2x+4cos(π/3)sinx=cos2x+2sinx=1-2sin²x+2sinx=-2(sin²x-sinx)+1=-2[(sinx-1/2)²-1/4]+1
=-2(sinx-1/2)²+3/2
由于0≦(sinx-1/2)²≦9/4,(当sinx=1/2时获得最小值0；当sinx=-1时获得最大值9/4)；
故-9/2≦-2(sinx-1/2)²≦0；于是得 -3≦-2(sinx-1/2)+3/2≦3/2,即f(x)的值域为[-3,3/2].