三角形ABC中,角ABC所对的边分别为a,b,c,且cosA=1/3.求[sin(B+C)/2]^2+cos2A

题目

三角形ABC中,角ABC所对的边分别为a,b,c,且cosA=1/3.求[sin(B+C)/2]^2+cos2A

答案

[sin(B+C)/2]^2+cos2A
=sin^2A/2+cos2A
=(1-cosA)/2+cos2A
=2cos^2A+(1-cosA)/2
=2*1/3^2+(1-1/3)/2
=2/9+4/3
=14/9