在锐角三角形ABC中，sinA=2根号2/3，求sin[(B+C)/2]的平方+cos(3pai-2A)的值

sinA=2√2/3,A为锐角,cosA=1/3,cos2A=-7/9
sin(B+C)/2 = sin(π-A)/2 = cos(A/2),
sin^2((B+C)/2)+cos(3π-2A)
=cos^2(A/2)+cos(π-2A)
=1/2(1+cosA)-cos2A
=1/2(1 + 1/3)+ 7/9
=13/9

B+C=π-A
sin[(B+C)/2]的平方+cos(3pai-2A)
=(sinA/2)^2-cos2A
=（1-cos2A）/2-cos2A
=1/2-3cos2A/2
cos2A=1-2sina^2=1-2*8/3
=-13/3

sin[(B+C)/2]的平方+cos(3pai-2A)=1/2+13/2=7

2根号2/3，???