Finding the exact value of $\tan(\pi/5)$ 
Hi, I realise there has been a question already asked regarding this particular exact value, but this question requires for it to be done under different conditions, which is the part I require help in.
I have been having trouble with this for the past 30 minutes, I was able to do (i) and (ii) quite easily, but I am unsure how to actually get the exact value from the quadratic. 
Please note that I only stumbled upon this question during self-study of harder trigonometry, and is not homework.
Any help would be greatly appreciated, thank you.
 A: Consider an isosceles triangle $ABC$, having its vertex angle $\hat{C}$ equal to $\pi/5$, so the base angles $\hat{A}$ and $\hat{B}$ are $2\pi/5$. Draw the bisector of angle $\hat{B}$ meeting $AC$ in $D$.
Then the triangle $ADB$ is similar to $ABC$. Call $l$ the length of $AC$ and $x$ the length of $AB$. Then $BD$ and $DC$ also have length $x$ and $l\cos2\pi/5=x/2$.
By the similarity, we have
$$
\frac{l}{x}=\frac{x}{l-x}
$$
so
$$
l^2-lx=x^2
$$
and, easily,
$$
x=\frac{\sqrt{5}-1}{2}l
$$
(the negative root must be discarded, of course). Thus
$$
\cos\frac{2\pi}{5}=\frac{\sqrt{5}-1}{4}
$$
and it's easy to compute $\sin(\pi/5)$ and $\cos(\pi/5)$ from this.

One can use the outlined strategy, too. The key is using $\pi/10$ and not $x=\pi/5$. If $x=\pi/10$, then
$$
\frac{\pi}{2}-3x=\frac{5\pi}{10}-\frac{3\pi}{10}=\frac{2\pi}{10}=2x
$$
and therefore
$$
\tan3x=\tan\left(\frac{\pi}{2}-2x\right)=\cot2x=\frac{1}{\tan2x}
$$
Now we can apply the formulas and get
$$
\frac{3\tan x-\tan^3x}{1-3\tan^2x}=\frac{1-\tan^2x}{2\tan x}
$$
or, setting $t=\tan x$,
$$
6t^2-2t^4=1-3t^2-t^2+3t^4
$$
that reduces to
$$
5t^4-10t^2+1=0
$$
This gives
$$
t^2=\frac{5+2\sqrt{5}}{5}
$$
But $\tan(\pi/5)=\tan2x$ or
$$
\tan\frac{\pi}{5}=\frac{2t}{1-t^2}
$$
and just substituting will give the result.
A: Here is a different approach that more directly gives the correct polynomial.
Using De Moivre's Formula, we have
$$
\begin{align}
\color{#C00000}{\cos(5x)}+i\color{#5D60BD}{\sin(5x)}
&=(\color{#C00000}{\cos^5(x)-10\cos^3(x)\sin^2(x)+5\cos(x)\sin^4(x)})\\
&+i(\color{#5D60BD}{5\cos^4(x)\sin(x)-10\cos^2(x)\sin^3(x)+\sin^5(x)})\tag{1}
\end{align}
$$
Since the smallest positive angle whose sine is $0$ is $\pi$, $\sin(\pi/5)\ne0$. Since the smallest positive angle whose cosine is $0$ is $\pi/2$, $\cos(\pi/5)\ne0$. Therefore, we can divide the imaginary part of $(1)$ by $\cos^4(x)\sin(x)$, with $x=\pi/5$, to get
$$
0=\frac{\sin(\pi)}{\cos^4(\pi/5)\sin(\pi/5)}=\tan^4(\pi/5)-10\tan^2(\pi/5)+5\tag{2}
$$
$\tan(\pi/5)$ is the smallest positive root of $z^4-10z^2+5$, which is
$$
\tan(\pi/5)=\sqrt{5-2\sqrt5}\tag{3}
$$

De Moivre's Formula
Using the formulas for the sine and cosine of a sum, we get
$$
\begin{align}
&(\cos(x)+i\sin(x))(\cos(y)+i\sin(y))\\
&=(\cos(x)\cos(y)-\sin(x)\sin(y))+i(\sin(x)\cos(y)+\cos(x)\sin(y))\\
&=\cos(x+y)+i\sin(x+y)
\end{align}
$$
from which, induction gives De Moivre's Formula:
$$
(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)
$$
A: For first part
use 
$\tan(a+b)= \dfrac{\tan a+\tan b}{1-\tan a \tan b}$
Put $a=x$ , $b=2x$
Further evaluate $\tan 2x$ using addition property described above by substituting $a=x$ and $b=x$
For second part
use $\tan 3x$ and convert $\cot 2x$ into $\dfrac{1}{\tan 2x}$
Using the second result find $\tan 36$
by substituting square of $\tan x =p$
and solving quadratic equation
remember $\tan (0-45)$ is less than $1$ so reject $1$ of the solution of the quadratic equation
hope this helps.
