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I saw someone derive the closed form for $\cos(\frac{\pi}{5})=\frac{\varphi}{2}$, and got inspired to try to find a closed form expression for $\cos(\frac{\pi}{7})$ using the same method. In doing this, you get that $\cos(\frac{\pi}{7})$ is one of the solutions to the following polynomial. $$8x^3-4x^2-4x+1$$ If there is an accessible closed form for the roots of this polynomial, I cant seem to find it. I have tried factoring, guessing and also looked at wolfram alpha, which only gives massive closed forms involving imaginary numbers.

So if anyone can find a closed form or show that none is accessible, that would be great!

Riccardo.Alestra
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Simon
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2 Answers2

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I've got sad news: we cannot express the solution to this equation using sums, products, and square roots. The reasoning is strange to say the least: we can only find the solution to $\cos(\pi/p)$ in that form where $p$ is on odd prime when $p$ is a Fermat prime. This means that we can assign $p$ to be something like $3, 5, 17, ...$, but $7$ is not an option. That's why we can find the solution to $\cos(\pi/5)$ but not for $\pi/7$. As an aside, this means we can find a solution for $\cos(\pi/17)$, which is really weird.

Part of why this is not doable is that this cubic equation has three roots, and when a cubic has three roots we might not have an approach to finding them. Cardano's method finds a root if the cubic has only one root, but we can employ his method with complex numbers in an attempt to find the remaining roots. Once you employ complex numbers you will need to find the cosine of some strange angles, and after performing some simplifications, you will need the value of...you guessed it, $\cos(\pi/7)$. Thus solving this problem is circular and there is nothing we can do about it.

We can find the solution in terms of complex numbers if interested, but the solution is not particularly helpful.

Hope this helps!

Francisco Santos
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Josh B.
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  • This is linked to the Gauss rule for regular polygons that is possible to build with rule and compass. Heptagon in not one of them. – Raffaele Jan 21 '21 at 18:55
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    @Raffaele: Note that being constructible with ruler and compass corresponds to being expressible using a finite sequence of operations each of which is one of the four basic arithmetic operations or extraction of a square root (the earlier parts of this manuscript may be of interest -- click "File available"), but it doesn't prevent the possibility that an expression involving cube or higher roots is possible (without invoking the use of complex numbers anywhere). (continued) – Dave L. Renfro Jan 21 '21 at 21:02
  • However, in the case of roots of unity, ruler-and-compass constructible is equivalent to being expressible using radicals of any positive integer order (the order can vary within the expression for a number) -- see the 26 September 2005 sci.math post I linked to in another comment here. By the way, I didn't actually misspell constructible on the first page of that manuscript, as the italicized phrase that includes "Constuctable" was the actual spelling used in the original title of the blog post I mention. – Dave L. Renfro Jan 21 '21 at 21:13
  • This answer sort of mixes being "constructible" (i.e., expressible with rational operations plus square roots) and being "solvable" (expressible with rational operations plus roots of arbitrary order). All cubic and quartic polynomials (with integer coifs) are solvable, and this is one of them. – Francisco Santos Jun 28 '24 at 18:04
  • The connection between solvability byvreal radicals and constructibility is thus: you always get complex radicals, but the square roots of complex numbers can be resolved into real radicals and $i$, with the latter dropping out when we add the conjugates, by algebra alone. Those square roits are the constructible radicals. – Oscar Lanzi Jan 28 '25 at 16:01
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Maple says $$ \frac{1}{12}\,\sqrt [3]{-28+84\,i\sqrt {3}}+\frac{7}{3}\,{\frac {1}{\sqrt [3]{-28+84\, i\sqrt {3}}}}+\frac16 $$ despite that $i$ in there, the result is a real number. Of course complex cube roots may be done trigonometrically, but this whole question is to avoid the trigonometry in $\cos\frac{\pi}{7}$.

Cannot be expressed in radicals using real numbers only. For explanation, see "casus irreducibilis"

GEdgar
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  • Although nice, my wish is to find something that is expressed in only real numbers. It doesn’t matter if the expression is complicated, as long as it’s finite and only uses real numbers. – Simon Jan 21 '21 at 18:22
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    Cannot be expressed in radicals using real numbers only. For explanation, see "casus irreducibilis" https://en.wikipedia.org/wiki/Casus_irreducibilis . – GEdgar Jan 21 '21 at 19:50