已知sinα +sinβ+ siny=0,cosα +cosβ +cosy=0,求cos(α-β)
问题描述：
已知sinα +sinβ+ siny=0,cosα +cosβ +cosy=0,求cos(α-β)
答
sinr=-(sina+sinb) cosr=-(cosa+cosb) 1=sin^2r+cos^2r =(sina+sinb)^2+(cosa+cosb)^2 =sin^2a+2sinasinb+sin^b+cos^2a+2cosacosb+cos^2b =2+2(sinasinb+cosacosb) =2+2cos(a-b) 2cos(a-b)=-1 cos(a-b)=-1/2