问几道高一数学题!再三角形ABC中,角A角B角C所对的边分别为abc,且cosA=1/3（1）求sin^2[(B+C)/

1.(1)A+B+C=π
sin^2[(B+C)/2]=sin^2[(π-A)/2]=sin^2(π/2-A/2)=sin^2(A/2)=1/2-1/2cosA
sin^2[(B+C)/2]+=1/2-1/2cosA+cos2A=-4/9
(2)a^2=b^2+c^2-2bccosA a=根号3 cosA=1/3
3=b^2+c^2-2/3bc由
因b>0,c>0,由上式可知b2+c2-3>0
由均值不等式可得,b2+c2>=2bc
代入得1/3=（b2+c2-3）/2bc>=(2bc-3)/2bc
解得bc

1、sin^2[(B+C)/2]+cos2A=sin^2(π/2-A/2)+2cosA-1
=cos^2A/2-7/9=1/2(cosA+1)-7/9=-1/9
cosA=(b^2+c^2-a^2)/(2bc)=(b^2+c^2-3)/2bc=1/3
>=(2bc-3)/2bc=1-3/(2bc) [当且仅当b=c取等]
bc的...

sin^2[(B+C)/2]+cos2A=(1+cosA)/2+2cos^2A=8/9