向量a=(cosx,sinx)b=(cosy,siny)为什么它们的夹角是y-xa=(cosx,sinx)b=(cosy,siny)θ=y-xa+b=(cosx+cosy,sinx+siny)|a+b|=√(2+2cos(x-y))|a-b|=√(2-2cos(x-y))
向量a=(cosx,sinx)b=(cosy,siny)为什么它们的夹角是y-x
a=(cosx,sinx)
b=(cosy,siny)
θ=y-x
a+b=(cosx+cosy,sinx+siny)
|a+b|=√(2+2cos(x-y))
|a-b|=√(2-2cos(x-y))
设夹角z,
cosz=a∙b/(|a||b|)-----------------------------------------------(1)
|a|=√[(cos(x))^2+(sin(x))^2]=1---------------------------------(2)
|b|=√[(cos(y))^2+(sin(y))^2]=1---------------------------------(3)
a∙b=cosxcosy+sinxsiny=cos(x-y)=cos(y-x)------------(4)
从(1),(2),(3),(4),cosz=cos(y-x)--------------------------(5)
z=y-x
|a+b|=√(a+b)∙(a+b)=√{a∙a)+2(a∙b)+b∙b}---------------(6)
从(2),(3),(4),
|a+b|=√(2+2cos(x-y))
同理,|a-b|=√(2-2cos(x-y))
