sinv=-(sina+sinb)
sin^2v=(sina+sinb)^2=sin^2a+sin^2b+2sinasinb
cosv=-(cosa+cosb)
cos^2v=(cosa+cosb)^2=cos^2a+cos^2b+2cosacosb
于是 sin^2v+cos^2v=sin^2a+sin^2b+2sinasinb+cos^2a+cos^2b+2cosacosb
=2+2(sinasinb+cosacosb)
=2+2cos(b-a)
=1
即cos(b-a)=-1/2
又0≤α＜β＜γ＜π
于是 b-a=2π /3