sinA+sinB+sinC=cosA+cosB+cosC=0,则cos（B-C）的值为?

sinA+sinB+sinC=0
sinA=-(sinB+sinC)=-2sin(B+C)/2cos(B-C)/2
cosA+cosB+cosC=0
cosA=-(cosB+cosC)=-2cos(B+C)/2cos(B-C)/2
1=sin^2A+cos^2A
=4cos^2 (B-C)/2 (sin^2 (B+C)/2 + cos^2 (B+C)/2)
=4cos^2 (B-C)/2
2cos^2 (B-C)/2=1/2
cos（B-C）=2cos^2 (B-C)/2 -1
=1/2-1
=-1/2

cos（B-C）=cosBcosC+sinBsinC
(sinA)^2=(sinB+sinC)^2=(sinB)^2+2sinB*sinC+(sinC)^2,
同理(cosA)^2=(cosB)^2+2cosB*cosC+(cosC)^2,
所以相加得1=1+2(cosB*cosC+sinB*sinC)+1,
公式cos(B-C)=cosB*cosC+sinB*sinC
∴所以相加得1=1+2cos(B-C)+1
得到cos（B-C）=-1/2