π/3<B<5π/6,
0<B-A<π/2,
sin(B-A)>0
由于 cos(B-A)=4√3/7
则 sin(B-A)=1/7
cosB=cos(B-A+A)
=cos(B-A)cosA-sin(B-A)sinA
=4√3/7*1/2-1/7*√3/2
=3√3/14
π/3<B<5π/6,
0<B-A<π/2,
sin(B-A)>0
由于 cos(B-A)=4√3/7
则 sin(B-A)=1/7
cosB=cos(B-A+A)
=cos(B-A)cosA-sin(B-A)sinA
=4√3/7*1/2-1/7*√3/2
=3√3/14