三角形ABC中,cos(A-C)+cosB=3/2 b^2=ac 求B

cos(A-C)+cosB=cos(A-C)-cos(A+C)=cosAcosC+sinAsinC-cosAcosC+sinAsinC=2sinAsinC=3/2即sinAsinC=3/4根据正弦定理,a/sinA=b/sinB=c/sinC=2Rb^2=sin^B*4R^2 a=sinA*2R c=sinC*2R所以,sin^B=sinA*sinC=3/4因为B...