# 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)?

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.

• @MFV: By Vieta, yes. (The coefficient of $z^{n-1}$ is zero). – J. M. is a poor mathematician Oct 24 '10 at 23:48
• A tiny correction @MFV: $z^5-1=(z-1)(1+z+z^2+z^3+z^4)$. Here, the Vieta relations become even more evident. – J. M. is a poor mathematician Oct 25 '10 at 3:15
• Where exactly did the polynomial in $x$ come from? – Geano Oct 3 '14 at 7:34
• Since $$z \neq 0$$, we have $$z^{-2}+z^{-1}+1+z+z^2=0$$. Expand $$z^{-2}+z^{-1}+1+z+z^2 - 4x^2$$. – Plop Oct 3 '14 at 20:46
• @Plop Where did you get $4x^2$? – hhh Apr 3 '16 at 16:37 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}

• @MFV: Given the context of your question, it seemed that you'd want a proof that was more geometrically oriented. In my head, the triangle I used is the triangle-equivalent of the golden rectangle. Also, if you draw a regular pentagon and all its diagonals, you'll see the diagram I drew as a small part of the picture (AB is a side, C and D are inside the pentagon at intersections of diagonals). – Isaac Oct 24 '10 at 21:51
• This was the proof I always used to quickly remember the value of $\cos \frac{2 \pi}{5}$. – Beni Bogosel Apr 17 '12 at 17:38

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.

• Also look up Kastelyn's method on counting the number of tilings of a 2mx2n chessboard with 2x1 rectangles. Choose m=1 and n=2 for this particular problem. – Aryabhata Oct 24 '10 at 20:58

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 $\cos \dfrac{2\pi}{5} = x$. 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*} Another way, $\cos \dfrac{2\pi}{5} > 0$, then $\cos \dfrac{2\pi}{5} = \dfrac{-1 + \sqrt{5}}{4}$.

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.

• @MFV: cut-the-knot.org/pythagoras/cos36.shtml might give you some help if you get stuck. But I agree with the suggestion to work it out yourself; that's much more interesting than trying to decipher someone else's "AB=AE=PQ=etc" description. – Hans Lundmark Oct 24 '10 at 17:37
• One of the first things I did upon learning coordinate geometry was to prove to myself that the compass-straightedge construction of the regular pentagon works as advertised. Do it when you can, it's fun. :) – J. M. is a poor mathematician Oct 24 '10 at 23:51

Or you can go to Mathworld for $\pi/5$ and use the multiple angle formula

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 $$\sin\frac{\pi}{10}=\frac{-1+\sqrt{5}}{4}$$ $$\implies \cos\left(\frac{\pi}{2}-\frac{\pi}{10}\right)=\frac{-1+\sqrt{5}}{4}$$ $$\implies \color{blue}{\cos\frac{2\pi}{5}=\frac{-1+\sqrt{5}}{4}}$$

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 $\frac{4\pi}5$

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.

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$$\cos2\pi/5=2c^2-1=\tfrac14\surd5-\tfrac14.$$