# How can I find the following product? $\tan 20^\circ \cdot \tan 40^\circ \cdot \tan 80^\circ.$

How can I find the following product using elementary trigonometry?

$$\tan 20^\circ \cdot \tan 40^\circ \cdot \tan 80^\circ.$$

I have tried using a substitution, but nothing has worked.

A simpler version of my answer to Evaluate $\tan ^{2}20^{\circ}+\tan ^{2}40^{\circ}+\tan ^{2}80^{\circ}$ works here. Basically: Write $\tan 3x$ as a polynomial in $\tan x$. The equation $\tan 3x = \sqrt{3}$ has roots $x = 20, 140$, and $260$ degrees. Thus the constant term of the polynomial in $\tan x$ you get with this equation gives $$-\tan(20^{\circ})\tan(140^{\circ})\tan(260^{\circ})$$ Since $\tan$ has period $180^{\circ}$ this is the same as $$-\tan(20^{\circ})\tan(-40^{\circ})\tan(80^{\circ})$$ $$= \tan(20^{\circ})\tan(40^{\circ})\tan(80^{\circ})$$

$$\tan x\tan(60^\circ-x)\tan(60^\circ+x)=\tan3x$$

Set $x=20^\circ$

tan 20 tan 40 tan 80 = (sin 20 sin 40 sin 80) / (cos 20 cos 40 cos 80)

Solving for numerator :

sin 20 sin 40 sin 80 = (sin 80 sin 20)(sin 60 sin 40) / (sin 60)
= (1/2)(cos 60 - cos 100)(1/2)(cos 20 - cos 100) / (sin 60) ... (product rule)
= (1/4)(1/2 + sin 10)(cos 20 + sin10) / (sin 60) ... (cos 60 = 1/2 & cos 100 = -sin 10)
= [(1/8)cos 20 + (1/8)sin 10 + (1/4)sin 10 cos 20 + (1/4) sin^2 10] / (sin 60)
= [(1/8)cos 20 + (1/8)sin 10 + (1/8)(sin -10 + sin 30) + (1/4)(1/2)(1 - cos 20)] / (sin 60)
= [(1/8(1/2) + 1/8] / (√3/2) ... (sin 30 = 1/2)
= √3/8

Solving for denominator :

sin 160
= 2sin 80 cos 80
= 4sin 40 cos 40 cos 80
= 8 sin 20 cos 20 cos 40 cos 80
But, sin 160 = sin 20
So, sin 20 = 8 sin 20 cos 20 cos 40 cos 80
Cancel sin 20. Therefore, cos 20 cos 40 cos 80 = 1/8

So, tan 20 tan 40 tan 80 = numerator/denominator = (√3/8) / (1/8) = √3 = tan 60
Thus, x = 60º. (This is just the principal value, being one of an infinite other values) enter preformatted text here