Prove that $\tan20^°\tan40^°\tan60^°\tan80^°=3$

Question

\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}

I am stuck here.

Related http://math.stackexchange.com/questions/455070/proving-a-fact-tan6-circ-tan42-circ-tan12-circ-tan24-cir?lq=1 — lab bhattacharjee, Apr 28 '16 at 13:18

score 11 · Accepted Answer · edited Apr 28 '16 at 13:08

11

Use $$\tan{(3x)}=\tan{x}\tan{(60^{\circ}+x)}\tan{(60^{\circ}-x)}$$ and let $x=20^\circ$.

edited Apr 28 '16 at 13:08

Jack D'Aurizio

353,855

answered Apr 28 '16 at 12:58

math110

93,304

1

+1 Very neat. Do you know a slick derivation of your $\tan3x$ equation? – almagest Apr 28 '16 at 13:05
1

@almagest Probably insert it into $\frac{\sin(3x)}{\cos(3x)}$ and calculate. It seems the most straight-forward, although there might be more elegant solutions out there. – Arthur Apr 28 '16 at 13:19

score 4 · Answer 2 · edited May 09 '16 at 13:44

4

It is the case $m=4$ of the general formula:

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

See here

edited May 09 '16 at 13:44

Roby5

4,287

answered Apr 28 '16 at 13:10

Emilio Novati

62,675

score 2 · Answer 3 · answered Apr 28 '16 at 13:13

Then $$4\cos \alpha \cos (60^{\circ}-\alpha)\cos (60^{\circ}+\alpha)=\cos 3 \alpha$$ Similarly, $$4\sin \alpha \sin (60^{\circ}-\alpha)\sin (60^{\circ}+\alpha)=\sin 3 \alpha$$

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

Prove that $\tan20^°\tan40^°\tan60^°\tan80^°=3$

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