由于a/2+B/2=(a-B/2)-(a/2-B),故
cos(a/2+B/2)=cos[(a-B/2)-(a/2-B)]
=cos(a-B/2)cos(a/2-B)+sin(a-B/2)sin(a/2-B)
由π/2又cos(a-B/2)=-1/9,sin(a/2-B)=2/3,故sin(a-B/2)=(4倍根号5)/9,cos(a/2-B)=(根号5)/3
因此cos(a/2+B/2)=(7倍根号5)/27