Qual o valor de ?.... Pergunta de ideia deFelipeQueiroz

April 2020 1 281 Report

Qual o valor de $\mathrm{tg \ 20^o}.\mathrm{tg \ 40^o}.\mathrm{tg \ 80^o}$ ?

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Celio

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Olá, Felipe.

Fatorando a fórmula da tangente do arco triplo, temos:

$\tan 3\theta=\\\\=\frac{3 \tan\theta - \tan^3\theta}{1-3 \tan^2\theta}=\\\\=\frac{\tan\theta(3 - \tan^2\theta)}{(1 +\sqrt3 \tan\theta)(1 -\sqrt3 \tan\theta)}=\\\\=\frac{\tan\theta(\sqrt3 - \tan\theta)(\sqrt3 + \tan\theta)}{(1 +\sqrt3 \tan\theta)(1 -\sqrt3 \tan\theta)}=$

$=\tan\theta\cdot\frac{(\sqrt3 - \tan\theta)(\sqrt3 + \tan\theta)}{(1 +\sqrt3 \tan\theta)(1 -\sqrt3 \tan\theta)}=\\\\=\tan\theta\cdot\frac{(\sqrt3 - \tan\theta)}{(1 +\sqrt3 \tan\theta)}\cdot\frac{(\sqrt3 + \tan\theta)}{(1 -\sqrt3 \tan\theta)}$

$=\tan\theta\cdot\frac{(\tan60\º - \tan\theta)}{(1+\tan60\º\tan\theta)}\cdot\frac{(\tan60\º+ \tan\theta)}{(1-\tan60\º\tan\theta)}=\\\\=\tan\theta\cdot\tan(60\º-\theta)\cdot\tan(60\º+\theta)$

Portanto, vale a relação:

$\tan3\theta=\tan\theta\cdot\tan(60\º-\theta)\cdot\tan(60\º+\theta)$

Fazendo $\theta=20\º,$ temos:

$\tan3\cdot20\º=\tan20\º\cdot\tan(60\º-20\º)\cdot\tan(60\º+20\º)\Rightarrow\\\\
\tan60\º=\tan20\º\cdot\tan40\º\cdot\tan80\º\Rightarrow\\\\
\boxed{\tan20\º\cdot\tan40\º\cdot\tan80\º=\sqrt3}
$

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