设△ABC的内角A、B、C所对应的边分别为a、b、c,cos(A-C)+cosB=3/2,b2=ac,则B=_.

问题描述:

设△ABC的内角A、B、C所对应的边分别为a、b、c,cos(A-C)+cosB=

3
2
,b2=ac,则B=______.

∵B=π-(A+C),∴已知等式变形得:cos(A-C)-cos(A+C)=32,即cosAcosC+sinAsinC-cosAcosC+sinAsinC=2sinAsinC=32,∴sinAsinC=34,将b2=ac利用正弦定理化简得:sin2B=sinAsinC=34,∴sinB=32或sinB=-32(舍去)...