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This is an Exercise 3.2.5 from Beardon's Algebra and Geometry:

Show that $\cos\left(\frac{\pi}{5}\right)=\frac{\lambda}{2}$, where $\lambda$ = $\frac{1+\sqrt{5}}{2}$ (the Golden Ratio).

[Hint: As $\cos(5\theta) = 1$, where $\theta = \frac{2\pi}{5}$, we see from De Moivre's theorem that $P(\cos\theta) = 0$ for some polynomial $P$ of degree five. Now observe that $P(z)=(1-z)Q(z)^2$ for some quadratic polynomial $Q$.]

So, by De Moivre's theorem: $$\left(\cos\left(\frac{2\pi}{5}\right)+i\sin\left(\frac{2\pi}{5}\right)\right)^5=\cos(2\pi)+i\sin(2\pi)=\cos(2\pi)=1$$ And so:

$$\left(\cos\left(\frac{2\pi}{5}\right)+i\sin\left(\frac{2\pi}{5}\right)\right)^5-1=0$$

Thus, $\left(\cos\left(\frac{2\pi}{5}\right)+i\sin\left(\frac{2\pi}{5}\right)\right)^5-1$ is our polynomial $P$ of degree five. But how can I get from here to $Q$ and from $Q$ to $\cos\left(\frac{\pi}{5}\right)$?

Simon Marynissen
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fr_
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  • There's a really slick way to prove it with regular pentagons, without using complex numbers at all. – J.G. Aug 25 '20 at 16:20

2 Answers2

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Start with something simpler, an expression for $e^{i\frac{\pi}5}$: $$\left(\cos\frac{\pi}5+i\sin\frac{\pi}5\right)^5=-1$$ Write explicitly the terms of the expansion. We know that we can ignore therms in even powers of $i$. Using $z=\cos\frac{\pi}5$ and $\sin^2\frac{\pi}5=1-z^2$ one gets: $$z^5-10z^3(1-z^2)+5z(1-z^2)^2+1=0$$ Hopefully you can continue from here. Just note that your final polynomial first term in the expansion might be $1+z$, not $1-z$.

Andrei
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Here is another way to prove it, we'll be using De moivre's theorem but without having to get to the $ 5^{\mathrm{th}} $ degree.

We have : $$ \cos{\left(\pi-\theta\right)}=-\cos{\theta} $$

With $ \theta $ being equal to $ \frac{2\pi}{5} $, we get : $$ \cos{\left(\frac{3\pi}{5}\right)}=-\cos{\left(\frac{2\pi}{5}\right)} \ \ \ \ \ \ \left(*\right) $$

Now using De moivre, for any $ \theta\in\mathbb{R} $, we have : \begin{aligned} \cos{\left(3\theta\right)}=\mathcal{Re}\left(\left(\cos{\theta}+\mathrm{i}\sin{\theta}\right)^{3}\right)&=\mathcal{Re}\left(\cos^{3}{\theta}+3\mathrm{i}\cos^{2}{\theta}\sin{\theta}-3\cos{\theta}\sin^{2}{\theta}-\mathrm{i}\sin^{3}{\theta}\right)\\ &=\cos^{3}{\theta}-3\cos{\theta}\left(1-\cos^{2}{\theta}\right)\\ \cos{\left(3\theta\right)}&=4\cos^{3}{\theta}-3\cos{\theta} \end{aligned}

\begin{aligned} \cos{\left(2\theta\right)}=\mathcal{Re}\left(\left(\cos{\theta}+\mathrm{i}\sin{\theta}\right)^{2}\right)&=\mathcal{Re}\left(\cos^{2}{\theta}+2\mathrm{i}\cos{\theta}\sin{\theta}-\sin^{2}{\theta}\right)\\ &=\cos^{2}{\theta}-\left(1-\cos^{2}{\theta}\right)\\ \cos{\left(2\theta\right)}&=2\cos^{2}{\theta}-1 \end{aligned}

Applying those two formulas, and setting $ X=\cos{\left(\frac{\pi}{5}\right)} $, the equation $ \left(*\right) $ becomes : \begin{aligned} 4X^{3}+2X^{2}-3X-1&=0\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(X+1\right)\left(4X^{2}-2X-1\right)&=0\\ \iff 4\left(X+1\right)\left(X-\frac{1+\sqrt{5}}{4}\right)\left(X-\frac{1-\sqrt{5}}{4}\right)&=0\end{aligned}

Since $ X $ cannot be negative, we get : $$ X=\frac{1+\sqrt{5}}{4} $$

CHAMSI
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