Deducing that $cos( \frac{2π}{5}) = \frac{-1+ \sqrt{5}}{4}$

Question

Consider $w=e^{ \frac{2πi}{5}}$ where $1+w+w^2+w^3+w^4=0$

what's the inspiration to get to this polynomial and where does one go from here?

Answer 1

The polynomial comes from setting $w=\cos(2\pi/5)+i\sin(2\pi/5)$; by De Moivre's formula, $w^5=1$; since $w\ne1$, we can say that $$ w^4+w^3+w^2+w+1=0 $$ by factoring $w-1$ out of $w^5-1$.

The usual trick now is to divide by $w^2$: $$ w^2+\frac{1}{w^2}+w+\frac{1}{w}+1=0 $$ and then notice that $w+w^{-1}=2\cos(2\pi/5)$ and $w^2+w^{-2}=(w+w^{-1})^2-2$. Thus we have $$ 4\cos^2(2\pi/5)+2\cos(2\pi/5)-1=0 $$ You need to choose the positive root, that is, $$ \cos(2\pi/5)=\frac{-1+\sqrt{5}}{4} $$

Answer 2

Put $x=cos(2π/5)$, $x=\frac{ω+ω^{-1}}{2}$. Then $ω^5-1=(ω+1)(ω^4+ω^3+ω^2+ω+1)$. So the quartic is a polynomial that sends ω to 0, and x satisfies $x^2+x/2-1/4=0$.

edit: more detail

$x^2+x/2-1/4= (\frac{ω+ω^{-1}}{2})^2+\frac{\omega+\omega^{-1}}4-\frac{1}{4}=\frac{\omega^2+\omega^{-2}+2+\omega+\omega^{-1}-1}{4}=\frac{\omega^{-1}+\omega^{-2}+\omega^2+\omega^{1}+1}{4}=\frac{\omega^{4}+\omega^{3}+\omega^2+\omega^{1}+1}{4}=0$

Answer 3

(1) Justify that $$1+w+w^2+w^3+w^4=0$$

(2) Deduce the value of $$ 1+\cos\big(\frac{2\pi}{5}\big)+\cos\big(\frac{4\pi}{5}\big)+\cos\big(\frac{6\pi}{5}\big)+\cos\big(\frac{8\pi}{5}\big)$$

(3) Show that: $\cos\big(\frac{6\pi}{5}\big)=\cos\big(\frac{4\pi}{5}\big)$ and $\cos\big(\frac{8\pi}{5}\big)=\cos\big(\frac{2\pi}{5}\big)$

(4) Using the relation $\cos\big(\frac{4\pi}{5}\big)=2\cos^2{\big(\frac{2\pi}{5}\big)}-1$ show that $\cos\big(\frac{2\pi}{5}\big)$ is solution of the quadratic equation $4x^2+2x-1=0$.

(5) Justify that: $\cos\big(\frac{2\pi}{5}\big)>0$. Find $\cos\big(\frac{2\pi}{5}\big)$.

Answer 4

Here's a possibly less "magic" proof. First note that for general $k$, $w^k$ is a point $\frac{2k\pi}{5}$ radians around the complex unit circle, so $1+w+w^2+w^3+w^4=0$ follows geometrically because this sum is just ($5$ times) the average of the vertices of a regular pentagon centered at $0$.

Now we can simplify this equation a bit by noting that $w^3$ is just the complex conjugate of $w^2$ and $w^4$ is the conjugate of $w$; this can also be seen geometrically, or you can use that $w^5=1$ and $|w|=1$, so for example $\overline{w}=w^{-1}=w^5\cdot w^{-1}=w^4$.

Since a number plus its conjugate is just twice its $x$ coordinate (i.e., its real part), this gives the equation $1+2\cos(2\pi/5)+2a=0$, where $a$ is the $x$ coordinate of $w^2$. Noting that $w = \cos(2\pi/5) + i\sin(2\pi/5)$, we can explicitly compute the real part of $w^2$ as $\cos^2(2\pi/5)-\sin^2(2\pi/5)=2\cos^2(2\pi/5)-1$, using the trig identity $\cos^2+\sin^2=1$.

Now we have the quadratic equation $1+2\cos(2\pi/5)+2(2\cos^2(2\pi/5)-1)=0$, which can be solved for $\cos(2\pi/5)$ to give the stated formula. Again, the fact that $\cos(2\pi/5)$ is the positive root can be seen geometrically by looking at the $x$ coordinate of the point $\frac{2\pi}{5}$ radians around the circle.