Just a small thought popped up in my mind; and now I'm stuck on it. Any idea on how to find the value of $\tan 20^\circ$? I tried doing it by using the multiple angle formulas, but I didn't get an answer… How do I proceed?
[Math] Finding the value of $\tan 20^\circ$.
trigonometry
Related Solutions
I considered the more general formula : $$T(m):=\prod_{k=1}^m \tan\left(\frac{k\pi}{4m}\right)$$ and noticed that the result for small values of $m$ was solution of a polynomial of degree $\le m$. For $m=45$ I found that the answer was solution of this irreducible polynomial of degree $24$ : $$1\\- 3256701697315828896312\,x^1 - 325994294876282580655116\,x^2 + 7220097128841103979624568\,x^3 + 112578453555034444841119842\,x^4 + 493898299320136273975435032\,x^5 + 649061666980531722406164708\,x^6 - 840700351973464244018822232\,x^7 -2457988129238279755530778353\,x^8 - 138286882106888055215208624\,x^9 +2474525072938192662606171624\,x^{10}+326024084648343835216068912\,x^{11} - 1088043811994145989051965476\,x^{12} + 5147738954805237173669808\,x^{13} +182273284200850360076819304\,x^{14} - 33045263177263307887100976\,x^{15} + 677463542076505961377071\,x^{16} +170537100491574073221480\,x^{17} - 6714674580553776884700\,x^{18} - 128584156182235814952\,x^{19} - 339010000890501150\,x^{20}\\ +776030507612856\,x^{21} - 397610115660\,x^{22}\\ + 37004040\,x^{23} + 6561\,x^{24}$$
This is an 'experimental' result (no proof) but rather satisfying from the 'not nice' point of view ! ;-)
Obviously, the best solution is to rationalize the numerator $$\frac{\sqrt{3}}{3}=\frac{\sqrt{3}}{3}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{3}{3\sqrt{3}}=\frac{1}{\sqrt{3}}$$
But lets say that completely went by us. One solution may be to try to build right triangle with an angle $\theta$ such that $$\tan\theta = \frac{\sqrt{3}}{3}$$
Setting $opposite = \sqrt{3}$, and $adjacent=3$, by Pythagoras we have $$hypotenuse = \sqrt{opposite ^2 + adjacent^2}=\sqrt{12}=2\sqrt{3}$$
Notice that the hypotenuse is exactly twice that of the of the opposite side.
By reflecting this triangle onto itself, we can build an equilateral triangle, giving us an angle of $\theta = \frac{\pi}{6}$
Best Answer
$\sin(20^\circ)$ and $\cos(20^\circ)$ can both be found using the triple-angle formulas. (I'm assuming your "20" was in degrees, not radians.)
Unfortunately, both of those will end up with you having to solve a cubic; you can use Cardano's formula to do that.
You can't solve it with just quadratics and algebra, for if you could, it'd be possible to trisect a 60-degree angle; but that is in fact exactly the example used generally to show that trisection is impossible, because $\cos(60^{\circ})$ is not a surd.