已知sina+sinb+sinc=0且cosa+cosb+cosc=0 求cos(a-b)的值

问题描述:

已知sina+sinb+sinc=0且cosa+cosb+cosc=0 求cos(a-b)的值

sina+sinb=-sinc;
cosa+cosb=-cosc;
两式平方再相加,化简
得cosa*cosb+sina*sinb=-1/2;
∴cos(a-b)=cosa*cosb+sina*sinb=-1/2