π－arctan(3/2)－arctan(2/3)=?

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π－arctan(3/2)－arctan(2/3)=?
π－arctan(3/2)－arctan(2/3)=?

π－arctan(3/2)－arctan(2/3)=?
设arctan(3/2)=A,arctan(2/3)=B
所以tanA=3/2>0,tanB=2/3>0
又因为A,B属于[-π/2,π/2]
所以A,B都为锐角
tanA*tanB=1,故A和B互余
A+B=π/2
π－arctan(3/2)－arctan(2/3)=π/2

记x=arctan(3+2√2)，y=arccos[(√6)/3] cosy=√6/3 siny=√3/3 tany=siny/cosy=√2/2 tanx=3+2√2 tan(x-y)=(tanx-tany)/(1+tanxtany)=(3+2√2-√2/2)/(1+(3+2√2)(√2/2))=1 所以arctan(3+2√2)-arccos[(√6)/3]=x-y=π/4
很不错哦，...

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