已知2sinβ=sin(2α+β)求证tan(α+β)=3tanα

题目

已知2sinβ=sin(2α+β)求证tan(α+β)=3tanα

答案

∵tan(α+β)-tanα
=[sin(α+β)]/[cos(α+β)]-(sinα/cosα)
={[sin(α+β)]cosα-[cos(α+β)]sinα}/{[cos(α+β)](cosα)}
=[sin(α+β-α)]/{[cos(α+β)](cosα)}
=(sinβ)/{[cos(α+β)](cosα)}
又∵ 2sinβ=sin(2α+β)
∴tan(α+β)-tanα
=(1/2)[sin(2α+β)]/{[cos(α+β)](cosα)}
=(1/2){sin[(α+β)+α]}/{[cos(α+β)](cosα)}
=(1/2){[sin(α+β)]cosα+[cos(α+β)]sinα}/{[cos(α+β)](cosα)}
=(1/2)[tan(α+β)+tanα]
∴tan(α+β)=3tanα
证毕