Question

Calcular a tg20.tg40.tg80. Resposta: raiz de 3. Muito obrigada a quem ajudar.

Answer 1

Boa noite!

Relembrando as fórmulas para soma e diferença de arcos (tangente)
$\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}\\\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$

Então, vamos calcular tangente de 3x:
$\tan(3x)=\tan(x+2x)=\frac{\tan(x)+\tan(2x)}{1-\tan(x)\tan(2x)}\\\tan(3x)=\frac{\tan(x)+\frac{2\tan(x)}{1-\tan^2(x)}}{1-\tan(x)\frac{2\tan(x)}{1-\tan^2(x)}}$

Chamando de $y=\tan(x)$, temos:
$\tan(3x)=\frac{y+\frac{2y}{1-y^2}}{1-y\frac{2y}{1-y^2}}\\\tan(3x)=\frac{\frac{y-y^3+2y}{1-y^2}}{\frac{1-y^2-2y^2}{1-y^2}}\\\tan(3x)=\frac{3y-y^3}{1-3y^2}\\\tan(3x)=\frac{y(3-y^2)}{1-3y^2}\\ \tan(3x)=\frac{y(\sqrt{3}+y)(\sqrt{3}-y)}{(1+\sqrt{3}y)(1-\sqrt{3}y}\\\tan(3x)=\frac{\tan(x)(\tan(60)+\tan(x))(\tan(60)-\tan(x))}{(1-\tan(60)\tan(x))(1+\tan(60)\tan(x))}\\\tan(3x)=\tan(x)\tan(60+x)\tan(60-x)$

Agora, só utilizar x=20:
$\tan(3\cdot{20})=\tan(20)\tan(60+20)\tan(60-20)=\tan(20)\tan(40)\tan(80)\\\tan(60)=\tan(20)\tan(40)\tan(80)=\sqrt{3}$

Espero ter ajudado!