在△ABC中,角A,B,C的对边分别为a,b,c,且cos(A-B)cosB-sin(A-B)sin(A+C)=-3/5（1）求sinA的值

(1)cos(A-B)cosB-sin(A-B)sin(A+C)=-3/5∵A+C=180º-B∴sin(A+C)=sinB∴cos(A-B)cosB-sin(A-B)sinB=-3/5∴cos(A-B+B)=-3/5即cosA=-3/5∵A为三角形内角∴sinA=4/5(2)a=4√2,b=5根据正弦定理a/sinA=b/sinB∴sinB=b...