Demostrar que $\tan20^°\tan40^°\tan60^°\tan80^°=3$

Question

Demostrar que $\tan20^°\tan40^°\tan60^°\tan80^°=3$

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\begin{align} \tan20^°\tan40^°\tan60^°\tan80^°&=\frac{\sin20^°\sin40^°\sin60^°\sin80^°}{\cos20^°\cos40^°\cos60^°\cos80^°} \\ &=\frac{2^4(\sin20^°\sin40^°\sin60^°\sin80^°)^2}{\sin 160^°} \end{align}

Estoy atrapado aquí.

Answer 1

Utilice $$\tan{(3x)}=\tan{x}\tan{(60^{\circ}+x)}\tan{(60^{\circ}-x)}$$ y que $x=20^\circ$ .

Answer 2

Es el caso $m=4$ de la fórmula general:

$$ \prod_{k=1}^m\tan\left(\frac{k\pi}{2m+1} \right)=\sqrt{2m+1} $$

Ver aquí

Answer 3

Entonces $$4\cos \alpha \cos (60^{\circ}-\alpha)\cos (60^{\circ}+\alpha)=\cos 3 \alpha$$ De la misma manera, $$4\sin \alpha \sin (60^{\circ}-\alpha)\sin (60^{\circ}+\alpha)=\sin 3 \alpha$$

Entonces $$\tan \alpha \tan (60^{\circ}-\alpha)\tan (60^{\circ}+\alpha)=\tan 3 \alpha$$