设 x=arctan（1/2) y=arctan(-2）
得到 cosx=2sinx
cosy=-1/2siny
则 cos[arctan(1/2) - arctan(-2)]= cos(x-y)
=cosxcosy+sinxsiny
=0
cos[arctan(1/2) - arctan(-2)]=?
设 x=arctan（1/2) y=arctan(-2）
得到 cosx=2sinx
cosy=-1/2siny
则 cos[arctan(1/2) - arctan(-2)]= cos(x-y)
=cosxcosy+sinxsiny
=0