Let $\theta=\frac{2\pi}{5}$. Show $2\cos(2\theta)+2\cos(\theta)+1=0$. I have been at this for a while. Any ideas?
 A: Hint: Consider the roots of unity of $z^5=1$.
So we have $1+z+z^2+z^3+z^4=0.$ Now using $z+z^{-1}=2\Re(z)$, see if you can get the answer.
A: Create the following diagram:

Prove that the angle $\angle BAB'= 36^\circ$. Then show that:
$$\cos 2\theta = \cos \pi/5 = \frac{1+\sqrt{5}}{4}$$ 
And by the double angle identity:
$$\cos \theta = \cos 2\pi/5 = 2\left(\frac{1+\sqrt{5}}{4}\right)^2-1$$ 
A: I think this is a simple way. By using De Moivre’s formula, we obtain the fifth time angle formula:
\begin{equation}
\cos(5\theta)=16\cos^5\theta-20\cos^3\theta+5\cos\theta
\end{equation}
Set $\theta=2\pi/5$, we have:
\begin{equation}
\cos(2\pi)=16\cos^5\theta-20\cos^3\theta+5\cos\theta\\
16\cos^5\theta-20\cos^3\theta+5\cos\theta-1=0
\end{equation}
By using factorization, we have:
\begin{equation}
(\cos\theta-1)(4\cos^2\theta+2\cos\theta-1)^2=0
\end{equation}
Obviously, $\cos\theta-1\neq0$, then the equality have been prove.
A: Do you see that your left-hand side is the same as
$$\cos(2\theta)+\cos(-2\theta)+\cos(\theta)+\cos(-\theta)+\cos(0)$$
Consider five unit rays emanating from the origin at these angles: $0, \theta, 2\theta, -2\theta, -\theta$ (with $\theta=\frac{2\pi}{5}$). Consider these five rays to be force vectors, pulling away from the origin like some complicated tug-of-war in five equally spaced directions around the origin. Does any one direction win the tug-of-war? No, because there is symmetry in the pulling, and if any one direction was the winner, then there would be at least 5 total directions that win. So these force vectors must sum to the zero vector.
Now the $x$-components of the five vectors are the five terms in the sum above. And the $x$-component of what they all add up to is the $x$-component of the zero vector. This equality is what you set out to prove.
A: Hint: Show that for every $\theta$ such that $\theta \not= 2k\pi$ $\left(k\in \mathbb{Z}\right)$ we have: $$1 + 2\cos (\theta) +  2\cos (2\theta) = \frac{\sin\left(\frac{5}{2}\theta \right)}{\sin\left(\frac{1}{2}\theta \right)}. $$
The trigonometry based proof is here. It hinges on 
$$ \frac{1}{2}\cos(x)\sin(y) = \sin(x+y) - \sin(x-y)$$
and telescoping sum.
Its generalization and proofs are here: Dirichlet's kernel.
A: $2×\cos(2×{{2\pi}\over 5})=-{{1+\sqrt 5}\over 2}$
$2×\cos{{2\pi}\over 5}={{-1+\sqrt 5}\over 2}$
$-{{1+\sqrt 5}\over 2}+{{-1+\sqrt 5}\over 2}+1=0$
A: If $\xi=e^{i2\pi/5}$, then $0=\xi^5-1=(\xi-1)(\xi^4+\xi^3+\xi^2+\xi+1)$. Therefore,
$$
\xi^4+\xi^3+\xi^2+\xi+1=0\tag{1}
$$
Since $2\cos(2\pi/5)=\xi+\xi^4$ and $2\cos(4\pi/5)=\xi^2+\xi^3$, $(1)$ says
$$
2\cos(4\pi/5)+2\cos(2\pi/5)+1=0\tag{2}
$$
A: If $\displaystyle5\theta=2\pi, 3\theta=2\pi-2\theta$
$\displaystyle\implies\cos3\theta=\cos(2\pi-2\theta)=\cos2\theta$
Using Triple and Double angle formulas,
$$4c^3-3c=2c^2-1\iff 4c^3-2c^2-3c+1=0\  \ \ \ (1)$$
where $ c=\cos\theta$
Again, $\displaystyle\cos3\theta=\cos2\theta$
$\displaystyle\implies3\theta=2n\pi\pm2\theta$ where $n$ is any integer
Taking the '+' sign, $\displaystyle\implies3\theta=2n\pi+2\theta\iff\theta=2n\pi\implies\cos\theta=\cos2n\pi=1$ 
Taking the '-' sign, $\displaystyle\implies3\theta=2n\pi-2\theta\iff\theta=\frac{2n\pi}5$ where $n\equiv0,1,2,3,4\pmod5$
Again as $\displaystyle\cos(2\pi-y)=\cos y,\cos\frac{2\pi}5=\cos\left(2\pi-\frac{2\pi}5\right)=\cos\frac{8\pi}5$ and $\displaystyle\cos\frac{4\pi}5=\cos\frac{6\pi}5$
Clearly, $c-1$ is a factor of $(1)$ and $\displaystyle\cos\frac{2\pi}5,\cos\frac{4\pi}5\ne1$
So, 
$\displaystyle\cos\frac{2\pi}5=\cos\frac{8\pi}5$ and $\displaystyle\cos\frac{4\pi}5=\cos\frac{6\pi}5$ are the roots of $$\frac{4c^3-2c^2-3c+1}{c-1}=0\iff 4c^2+2c-1=0$$ 
Again use Double formula $\displaystyle\cos2\theta=2\cos^2\theta-1\implies c^2=\cos^2\theta=\frac{1+\cos2\theta}2$
