已知sina+sinb+siny=0,cosa+cosb+cosy=0,则cos(a-b)=?
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已知sina+sinb+siny=0,cosa+cosb+cosy=0,则cos(a-b)=?已知sina+sinb+siny=0,cosa+cosb+cosy=0,则cos(a-b)=?已知sina
已知sina+sinb+siny=0,cosa+cosb+cosy=0,则cos(a-b)=?
已知sina+sinb+siny=0,cosa+cosb+cosy=0,则cos(a-b)=?
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
-1/2,在单位圆中做出三个向量,与x正半轴的夹角为a,b,y,,设a>=b>=y,显然当向量做对称的分布时,他们在x轴的分量和在y轴的分量才会同时为0,这样的话,a,b夹角是120,Cos120=-1/2
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