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问题: 已知tanb,tan2b,tana成等差数列,求证:tan(a-b)=sin2b.

解答:

证明:∵tanb,tan2b,tana成等差数列,
∴tana-tan2b=tan2b-tanb,
即sina/cosa-sin2b/cos2b=sin2b/cos2b-sinb/cosb.
两边都乘以cosacos2bcosb,得
cosb(sinacos2b-cosasin2b)=cosa(sin2bcosb-cos2bsinb),
即cosbsin(a-2b)=cosasin(2b-b)=cosasinb.
∴2sin(a-2b)cosb=2cosasinb,
即sin(a-b)+sin(a-3b)=sin(a+b)-sin(a-b).
∴2sin(a-b)=sin(a+b)-sin(a-3b),
即2sin(a-b)=2cos(a-b)sin2b.
∴sin(a-b)/cos(a-b)=sin2b,
即tan(a-b)=sin2b.