Finding the value of $\tan 20^\circ$. 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? 
 A: $\sin(20^\circ)$ 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. 
A: We have that $\tan 20^{\circ} = \dfrac{\sin 20^{\circ}}{\cos 20^{\circ}} = \dfrac{\sin 20^{\circ}}{\sin 70^{\circ}}$.
This list here, gives you the exact value of the sine of every integer angle between $1$ and $90$. This allows you to compute $\tan 20^{\circ}$.

Whilst this section doesn't answer your question the way you want it, it does prove an alternative from brutally disgusting surds or horrible calculators.
You can construct a $20$-$70$-$90$ triangle yourself using a protractor and a ruler. Then take the ratio of the sides to get $\tan 20^{\circ}$. This won't be accurate, but you may find it amusing to do. :-) 
A: $\cos(3t) = 4 \cos(t)^3 - 3 \cos(t)$
$\sin(3t) = 3 \sin(t) - 4 \sin(t)^3$.
Substituting $t = 20^\circ$ gives cubics that can be solved in closed form if you allow the operation of taking cube roots.
A: $20°$ is a third of $60°$, for which the value of the tangent is well known to be $\sqrt3$. Let us denote $x:=\tan(20°)$.
Then, by the addition formula, 
$$\tan(40°)=\frac{2x}{1-x^2},$$
and 
$$\tan(60°)=\frac{x+\dfrac{2x}{1-x^2}}{1-x\dfrac{2x}{1-x^2}}=\frac{3x-x^3}{1-3x^2}=\sqrt3,$$
or
$$x^3-3\sqrt3x^2-3x+\sqrt3=0.$$
It turns out that this cubic equation cannot be solved by real radicals, so  you need to use numerical methods such as Newton's.
Starting from $x=\dfrac\pi9$ ($20°$ in radians), the iterates are
$0.349065850399\cdots\\
0.364116885850\cdots\\
0.363970248087\cdots\\
0.363970234266\cdots\\
0.363970234266\cdots\\
0.363970234266\cdots\\
\cdots$
A: I'm assuming you mean via multiple angle formulas, and I'm assuming you want an exact answer. The simple answer is, it would require a ridiculous amount of work. You can write:
$$ \tan \left(45-30+5\right)$$
You could perform the double angle formula twice, and reduce the problem to finding tan(5), which still isn't easily solvable. Heres a fun site to give you an idea of how bad these things can get :)
http://www.jdawiseman.com/papers/easymath/surds_tan.html
Note that if you are happy with an approximate solution, you should look into the CORDIC algorithm, which is how many hand-held calculators still do trigonometry :P
A: Since, obviously, numerical calculations would be required, I cannot resist the pleasure of reusing a 1400 years old approximation proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}$$ A similar one $$\cos(x)\simeq \frac{\pi^2-4x^2}{x^2+\pi^2}$$ Applied to $x=\frac \pi 9$, this leads to $$\tan(\frac \pi 9)\simeq \frac{10496}{28721} \approx 0.365447$$ Using this as a starting point $x_0$ for Newton method, as Yves Daoust proposed, the iterative scheme will converge in a couple of iterations $$x_1 \approx 0.363971632203$$ $$x_2\approx0.363970234267$$ $$x_3\approx0.363970234266$$.
Another solution would be to use Taylor expansion around $x=a$ $$\tan(x)=\tan (a)+(x-a) \left(\tan ^2(a)+1\right)+(x-a)^2 \left(\tan ^3(a)+\tan
   (a)\right)+$$ $$(x-a)^3 \left(\tan ^4(a)+\frac{4 \tan
   ^2(a)}{3}+\frac{1}{3}\right)+$$ $$(x-a)^4 \left(\tan ^5(a)+\frac{5 \tan
   ^3(a)}{3}+\frac{2 \tan (a)}{3}\right)+O\left((x-a)^5\right)$$ and to use $a=\frac \pi 8$ for which the tangent is $\sqrt 2 -1$ ( easily obtained using the double angle formula). Using one term, the result is $\approx 0.363094$; with two terms  $\approx 0.364018$; with three terms  $\approx 0.363969$; with four terms $\approx 0.363970$.
Another solution is based on Pade approximants. The simplest would be
$$\tan(x)\simeq \frac{(x-a)+\tan (a)}{1-(x-a) \tan (a)}$$ Using $a=\frac \pi 8$ would give $\approx 0.364002$.
The next Pade approximant would be $$\tan(x)\simeq \frac{(x-a)+\tan(a)-\frac{1}{3} (x-a)^2 \tan (a)}{1-(x-a) \tan
   (a)-\frac{1}{3} (x-a)^2}$$ which would give $\approx 0.363970238$.
