问题补充:
设三角形ABC的内角A,B,C的对边分别为a,b,c,cos(A-—C)+cosB=3/2,b²=ac,求B
答案:
因为b²=ac,所以由正弦定理得sin^2B=sinAsinC,所以 cos(A-C)=cosAcosC+sinAsinC
=cosAcosC-sinAsinC+2sinAsinC=cos(A+C)+2sin^2B=-cosB+2sin^2B
因为cos(A-—C)+cosB=3/2,所以 2sin^2B =3/2,所以 sinB=根号3/2 (舍负)
所以 B=派/3 (舍B=2派/3)