cos[arctan(1/2) - arctan(-2)]=?
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cos[arctan(1/2) - arctan(-2)]=?
cos[arctan(1/2) - arctan(-2)]=?
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
cos[arctan(1/2) - arctan(-2)]=?
cos[2arctan(-1/2)]=?
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