Question

已知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