Question

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

Answer 1

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