已知sina+sinb+siny=0,cosa+cosb+cosy=0,则cos(a-b)=?

问题描述:

已知sina+sinb+siny=0,cosa+cosb+cosy=0,则cos(a-b)=?

sina+sinb=-siny①
cosa+cosb=-cosy②
①^2+②^2
sinasinb+cosacosb=-1/2
所以cos(a-b)=(-1)/2