sinA+sinB+sinY=0 ,cosA+cosB+cosY=0,求cos(B-Y)的值.

sinA+sinB+sinY=0
-sinA=sinB+sinY
两边平方：
(sinA)^2=(sinB)^2+2sinBsinY+(sinY)^2--------1

cosA+cosB+cosY=0
-cosA=cosB+cosY
两边平方:
(cosA)^2=(cosB)^2+2cosBcosY+(cosY)^2--------2

1+2:
2(sinBsinY+cosBcosY)=-1
2cos(B-Y)=-1
cos(B-Y)=-1/2

sinA+sinB+sinY=0
sinB+sinY=-sinA
两边平方
sin^2B+2sinBsinY+sin^2Y=sin^2A (1)
cosA+cosB+cosY=0
cosB+cosY=-cosA
两边平方
cos^2B+2cosBcosY+cos^2Y=cos^2A (2)
(1)+(2)
sinBsinY+cosBcosY=-1/2
cos(B-Y)=-1/2

sinB+sinY=-sinA
cosB+cosY=-cosA
从而有（sinB+sinY）^2+(cosB+cosY)^2=1
即sinB^2+sinY^2+2sinBsinY+cosB^2+cosY^2+2cosBcosY=1
从而2sinBsinY+2cosBcosY=-1
即cos(B-Y)=cosBcosY+sinBsinY=-0.5

sinα=-sinβ-sinγ
cosα=-cosα-cosγ
两个平方相加
1=2+2sinβsinγ+2cosβcosγ
sinβsinγ+cosβcosγ=-0.5
cos(β-γ)=cosβcosγ+sinβsinγ=-0.5

cos(B-Y)=cosBcosY+sinBsinY
sinA+sinB+sinY=0 =>sinA=-(sinB+sinY)
cosA+cosB+cosY=0 =>cosa=-(cosB+cosY)
(sinA)^2+(cosA)^2=1
=>(sinB)^2+2sinBsinY+(sinY)^2+(cosB)^2+2cosBcosY+(cosY)^2=1
=>2+2cos(B-Y)=1
=>cos(B-Y)=-1/2