How to prove $\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}$? I would like to find the apothem of a regular pentagon. It follows from 
$$\cos \dfrac{2\pi }{5}=\dfrac{-1+\sqrt{5}}{4}.$$
But how can this be proved (geometrically or trigonometrically)? 
 A: Since $x := \cos \frac{2 \pi}{5} = \frac{z + z^{-1}}{2}$ where $z:=e^{\frac{2 i \pi}{5}}$, and $1+z+z^2+z^3+z^4=0$ (for $z^5=1$ and $z \neq 1$), $x^2+\frac{x}{2}-\frac{1}{4}=0$, and voilà.
A: Notice that $$\sin 3\theta=3\sin \theta-4\sin^3 \theta$$ Now substitute $\theta=\frac{\pi}{10}$ in the above equation, we get 
$$\sin \frac{3\pi}{10}=3\sin \frac{\pi}{10}-4\sin^3 \frac{\pi}{10}$$ $$\implies \cos\left( \frac{\pi}{2}-\frac{3\pi}{10}\right)=3\sin \frac{\pi}{10}-4\sin^3 \frac{\pi}{10}$$ $$\implies \cos\frac{\pi}{5}=3\sin \frac{\pi}{10}-4\sin^3 \frac{\pi}{10}$$ $$\implies 1-2\sin^2\frac{\pi}{10}=3\sin \frac{\pi}{10}-4\sin^3 \frac{\pi}{10}$$ $$\implies 4\sin^3\frac{\pi}{10}-2\sin^2 \frac{\pi}{10}-3\sin \frac{\pi}{10}+1=0$$ It is obvious that the sum of all the coefficients of above cubic equation is $0$ hence  above cubic equation has one real root $1$. Now it can be easily factorized as follows $$\left(\sin \frac{\pi}{10}-1\right)\left(4\sin^2 \frac{\pi}{10}+2\sin \frac{\pi}{10}-1\right)=0$$ 
$$\color{red}{ \text{if}\quad \sin\frac{\pi}{10}-1=0 \implies \sin\frac{\pi}{10}=1}$$ but  $\frac{\pi}{10}<\frac{\pi}{2} \implies \sin\frac{\pi}{10}<1$ Hence, above value is unacceptable 
$$\color{blue}{ \text{if}\quad 4\sin^2\frac{\pi}{10}+2\sin\frac{\pi}{10}-1=0 \implies \sin\frac{\pi}{10}=\frac{-2\pm \sqrt{(-2)^2-4(4)(-1)}}{2(4)}}$$ $$\sin\frac{\pi}{10}=\frac{-1\pm \sqrt{5}}{4}$$ but  $0<\frac{\pi}{10}<\frac{\pi}{2}\implies 0<\sin\frac{\pi}{10}<1$  Thus we get 
$$\sin\frac{\pi}{10}=\frac{-1+\sqrt{5}}{4}$$
$$\implies \cos\left(\frac{\pi}{2}-\frac{\pi}{10}\right)=\frac{-1+\sqrt{5}}{4}$$
$$\implies \color{blue}{\cos\frac{2\pi}{5}=\frac{-1+\sqrt{5}}{4}}$$
A: 
Consider a $\triangle ABC$ with $AB=1$, $\mathrm{m}\angle A=\frac{\pi}{5}$ and $\mathrm{m}\angle B=\mathrm{m}\angle C=\frac{2\pi}{5}$, and point $D$ on $\overline{AC}$ such that $\overline{BD}$ bisects $\angle ABC$.  Now, $\mathrm{m}\angle CBD=\frac{\pi}{5}$ and $\mathrm{m}\angle BDC=\frac{2\pi}{5}$, so $\triangle ABC\sim\triangle BCD$.  Also note that $\triangle ABD$ is isosceles so that $BC=BD=AD$.
Let $x=BC=BD=AD$.  From the similar triangles, $\frac{AB}{BC}=\frac{BC}{CD}$ or $\frac{1}{x}=\frac{x}{1-x}$, so $1-x=x^2$ and $x=\frac{\sqrt{5}-1}{2}$ (the other solution is negative and lengths cannot be negative).
Now, apply the Law of Cosines to $\triangle ABC$:
$$\begin{align}
\cos\frac{2\pi}{5}=\cos C&=\frac{a^2+b^2-c^2}{2ab}
\\\\
&=\frac{\left(\frac{\sqrt{5}-1}{2}\right)^2+1^2-1^2}{2\cdot\frac{\sqrt{5}-1}{2}\cdot 1}
\\\\
&=\frac{\frac{\sqrt{5}-1}{2} \cdot \frac{\sqrt{5}-1}{2}}{2\cdot\frac{\sqrt{5}-1}{2}}
\\\\
&=\frac{\sqrt{5}-1}{4}.
\end{align}$$
A: Or you can go to Mathworld for $\pi/5$ and use the multiple angle formula
A: for any $\theta \in \mathbb{R}$ the transformation $\psi: x \rightarrow 2x^2 -1$ sends $cos \theta$ to $cos 2\theta$ . hence $\psi^2 $ sends $cos \theta$ to $cos4\theta$
if $\alpha = \frac{2\pi}5$ then $cos 4\alpha = cos \alpha$ so that $cos \alpha$ is a fixed point for $\psi$ and if $c=cos \alpha$ we have
$$
\psi^2(c) = c
$$
i.e.
$$
2(2c^2-1)^2-1= c
$$
or 
$$
(c-1)(8c^3-1) = 0
$$
i.e.
$$
(c-1)(2c-1)(4c^2+2c-1) = 0
$$
if we disregard the roots corresponding to angles with rational cosines we have:
$$
4c^2+2c-1 =0,
$$
giving
$$
c = \frac14\left(-1 \pm \sqrt{5}\right)
$$
the negative root corresponds to the angle $\frac{4\pi}5$
A: How about combinatorially?  This follows from the following two facts.


*

*The eigenvalues of the adjacency matrix of the path graph on $n$ vertices are $2 \cos \frac{k \pi}{n+1}, k = 1, 2, ... n$.  

*The number of closed walks from one end of the path graph on $4$ vertices to itself of length $2n$ is the Fibonacci number $F_{2n}$.
The first can be proven by direct computation (although it also somehow falls out of the theory of quantum groups) and the second is a nice combinatorial argument which I will leave as an exercise.  I discuss some of the surrounding issues in this blog post.
A: Note that $$2\cdot \dfrac{2\pi}{5} + 3\cdot \dfrac{2\pi}{5} = 2\pi,$$
therefore $$\cos\left(2\cdot \dfrac{2\pi}{5}\right) = \cos\left(3\cdot \dfrac{2\pi}{5}\right).$$
Put $\dfrac{2\pi}{5} = x$. Using the formulas
\begin{equation*}
\cos 2x = 2\cos^2 x - 1, \quad \cos 3x = 4\cos^3 x - 3\cos x,
\end{equation*}
we have
\begin{equation*}
4x^3 - 2x^2 -3x + 1 = 0 \Leftrightarrow (x - 1)(4x^2 + 2x - 1) = 0.
\end{equation*}
Because $\cos \dfrac{2\pi}{5}  \neq 1$, we get
\begin{equation*}
4x^2 + 2x - 1 = 0.
\end{equation*}
Solving the above quadratic equation for $x$ gives us $\cos \dfrac{2\pi}{5}  = \dfrac{-1 \pm \sqrt{5}}{4}$. Because $\cos \dfrac{2\pi}{5} > 0$, we take the positive sign, giving us $\cos \dfrac{2\pi}{5}  = \dfrac{-1 + \sqrt{5}}{4}$.
A: Look up the "construction of a regular pentagon" using the straightedge and compass. If you keep track of each step in this construction, you will find that the angle $72^\circ$ comes up in a few places, and this expression follows from it.
It's a fun exercise-- you should do it.
A: Let $\omega$ be a primitive fifth root of unity. The quadratic residues in $\mathbb{Z}/(5\mathbb{Z})^*$ are $1$ and $4$, hence by setting $q=\omega^1-\omega^2-\omega^3+\omega^4$ such Gauss sum fulfills $q^2=5$. On the other hand
$$\omega^1-\omega^2-\omega^3+\omega^4= 2\cos\frac{2\pi}{5}-2\cos\frac{4\pi}{5} \tag{A}$$
is real an positive, so $q=\sqrt{5}$, and 
$$\omega^1+\omega^2+\omega^3+\omega^4=2\cos\frac{2\pi}{5}+2\cos\frac{4\pi}{5}=-1,\tag{B}$$
so by summing $(A)$ and $(B)$ we get $4\cos\frac{2\pi}{5}=\sqrt{5}-1$ as wanted.
A: Let us write $c:=\cos\pi/5$ and $s:=\sin\pi/5$. Then $$2sc=\sin2\pi/5=\sin3\pi/5=3s-4s^3.$$ Dividing this through by (nonzero) $s$ gives $$2c=3-4s^2=4c^2-1.$$Thus $c$ is the positive solution of the quadratic equation $4c^2-2c-1=0$, namely  $\frac14+\frac14\surd5$. Then$$\cos2\pi/5=2c^2-1=\tfrac14\surd5-\tfrac14.$$
A: By my post, $$
\cos \frac{\pi}{5}-\cos \frac{2 \pi}{5}=\frac{1}{2}
$$
By $\cos (\pi-\dfrac{\pi}{5} )=-\cos \dfrac{\pi}{5}, $ we have
$$
-\cos \frac{4 \pi}{5}-\cos \frac{2 \pi}{5}=\frac{1}{2}
$$
By double-angle formula, $$
-\left(2 \cos ^{2} \frac{2 \pi}{5}-1\right)-\cos \frac{2 \pi}{5}=\frac{1}{2} \Leftrightarrow 
4 \cos ^{2} \frac{2 \pi}{5}+2 \cos \frac{2 \pi}{5}-1=0
$$
Using quadratic formula gives $$
\boxed{\cos \frac{2\pi}{5}=\frac{-1+\sqrt{5}}{4}}
$$
By the way, $$\cos \frac{\pi}{5}= \cos \frac{2\pi}{5}+\frac{1}{2}= \frac{1+\sqrt{5}}{4}
$$
