已知sin A+sinB +siny=0 ,cos A+cos B+cosy=0,求cos(B -y)的值.
问题描述:
已知sin A+sinB +siny=0 ,cos A+cos B+cosy=0,求cos(B -y)的值.
答
sinB+siny=-sinA① cosB+cosy=-cosA② ①+②得 (sinB+siny)+(cosB+cosy)=(-sinA)+(-cosA)=sinA+cosA=1 ∴(sinB+cosB)+(siny+cosy)+2(cosBcosy+sinBsiny)=1 即1+1+2cos(B-y)=1 ∴cos(B-y)=-1/2