A B 为锐角tan(A-B)=sin(2B),求证2tan(2B)=tanA+tanB
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A B 为锐角tan(A-B)=sin(2B),求证2tan(2B)=tanA+tanB
A B 为锐角tan(A-B)=sin(2B),求证2tan(2B)=tanA+tanB
A B 为锐角tan(A-B)=sin(2B),求证2tan(2B)=tanA+tanB
tan(A-B) = (tanA-tanB)/(1+tanAtanB) =sin(2B)=2sinBcosB
tanA-tanB=2sinBcosB(1+tanAtanB)= 2sinBcosB+2tanAsin²B
tanA(1-2sin²B)= 2sinBcosB+tanB
tanA=(2sinBcosB+tanB)/cos(2B)
tanA+tanB={sin(2B)+tanB[1+cos(2B)]}/cos(2B)= {sin(2B)+ sin(2B)}/cos(2B)
= 2tan(2B)
证明:
tan(A-B)=(tanA-tanB)/(1+tanAtanB)
sin2B=2tanB/(1+tan^2B)
tan(A-B)=sin2B
(tanA-tanB)/(1+tanAtanB) =2tanB/(1+tan^2B);
化简得tanA=tanB(3+tan^2B)/(1-tan^2B)
tanA+tanB
=tanB(3+tan^2B)/(1-tan^2B)+tanB
=4tanB/((1-tan^2B)
=2tan2B
证毕
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