sina+sinb+sinc=0,cosa+cosb+cosc=0,求cos(B-C)的值?

(sina)^2=(sinb+sinc)^2=(sinb)^2+2sinb*sinc+(sinc)^2,
(cosa)^2=(cosb+cosc)^2=(cosb)^2+2cosb*cosc+(cosc)^2,
两式相加得：1=1+2(cosb*cosc+sinb*sinc)+1,
所以 cos(b-c)=-1/2.