Can the angle 2π/5 be trisected using ruler and compasses? I'm trying to figure out how to do this problem. I know that it can be trisected using ruler and compasses based on the theorem that 

For a positive integer N, an angle of measure ​$\frac{2\pi}{N}$ is
  trisectible if and only if 3 does not divide N

but I tried to understand why that was true and couldn't, so I didn't want to use this theorem since I couldn't find/do/understand a proof of it.
I've tried to see if $f(t) = 4t^3 - 3t + cos(\frac{2\pi}{5})$ is reducible over $\mathbb{Q}(cos(\frac{2\pi}{5}))$ as I know this would imply that it is constructible (and I have been able to show this), but wasn't sure how to tell if $f$ was reducible over $\mathbb{Q}(cos(\frac{2\pi}{5}))$
So I was looking for some assistance on how to either prove the statement in yellow or how to show $f$ is reducible over $\mathbb{Q}(cos(\frac{2\pi}{5}))$
Thank you!
 A: Both the equilateral triangle and the pentagon can be constructed with straightedge and compass, and $\frac{2\pi}{3}-\frac{2\pi}{5}=\frac{4\pi}{15}$ can be bisected:

In algebraic terms, the minimal polynomial of $\cos\frac{2\pi}{15}$ over $\mathbb{Q}$ is given by $1+8 x-16 x^2-8 x^3+16 x^4$ and 
$$ \cos\frac{2\pi}{15} = \frac{1}{8} \left(1+\sqrt{5}+\sqrt{6 \left(5-\sqrt{5}\right)}\right). $$
A: You want to be able to construct $e^{2\pi i/15}$, which is a solution of $x^8 - x^7 + x^5 - x^4 + x^3 - x + 1=0$.
Observe that this polynomial is symmetric $x^8P(1/x)=P(x)$. Therefore We can write it as $x^4Q(x+1/x)$ with $Q$ a polynomial of degree $4$.
$$\begin{align}Q(x)&=x^4-x^3-4x^2+4x+1\\&=-1/4 (-2 x^2 + (1 + \sqrt{5}) x - \sqrt{5} + 3) (2 x^2 + (\sqrt{5} - 1) x - \sqrt{5} - 3)\end{align}$$
A polynomial of degree $4$ that can be solved in quadratic radicals, and therefore so can the original one.
A: $\cos(2\pi/15)$ is contained in $\Bbb Q(\zeta_{15})$ where $\zeta_{15}=\exp(2\pi i/15)$. We have $[\Bbb Q(\zeta_{15}):\Bbb Q]=\phi(15)=8$, so $\Bbb Q(\zeta_{15})/\Bbb Q$ Galois extension with a $2$-group as a Galois group. For an element to be constructible over a field, it is sufficient that is contained in a Galois extension with a $2$-group as s Galois group. Thus $\cos(2\pi/15)$ and $\zeta_{15}$ are constructible even over $\Bbb Q$.
A: Call $$\phi=cos({2\pi\over5})$$
One root of $f$ is $T={1+\sqrt{5}\over4}$ which is somewhat conjugate to $\phi={-1+\sqrt{5}\over4 }$. By forming the product you find $T\phi={4\over 16}=\frac14$.
Therefore you expect $f$ to factor as $$f(t)=(t-\frac {1}{4\phi})(\text{some quadratic expression})$$
From there it is not hard to see that the quotient is itself with coefficients in $\mathbb{Q}(\phi)$.
A: GeoGebra
straightedge slip on the point B  - line i
divider  FIG , slides on straightedge , after slipping point F straighte line BC , point  G describes lokus1  
section lokus1 and line k point J , when changing the angle, the point J must be manually set to the intersection
construction of regular polygon is possible by means of the proportion of the angle
