sin(B+C)=sinA=2√2/3cos(B+C)=-cosA=-1/3cos[(B+C)/2]=√[(1-1/3)/2]=√3/3sin[(B+C)/2]=√6/3cos2A=2(cosA)^2-1=-7/9所以[sin(B+C)/2]^2+cos2A=2/3-7/9=-1/9用正弦定理求出c=3√3/4,b=√6/2+√3/4