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I have been playing around with circles lately, and I have found an interesting limited relationship between prime factors and cosine. Have the form of:

$$\cos{\left(2\pi\frac{1}{p}\right)}$$

And that $p=f_1\cdot f_2\cdot f_3\cdot\ldots\cdot f_n$ where $f$ is a prime factor.

Say, for example, that $p=12$. Its prime factorization is $2\cdot 2\cdot 3$. I found that disregarding the first two $2$'s, and putting $3$ in a squareroot, thus $\sqrt{3}$ and halving it, thus $\frac{\sqrt{3}}{2}$, is equal to $\cos{\left(2\pi\frac{1}{12}\right)}$.

Another example, $p=8$. Its prime factorization is $2\cdot 2\cdot 2$. Applying the same "algorithm" as before, we yield $\frac{\sqrt{2}}{2}$. We find that $\cos{\left(2\pi\frac{1}{8}\right)}=\frac{\sqrt{2}}{2}$.

And as long as $f=2$ and that $f_n=2$ or $f_n=3$, then this applies. Take for example $p=768$. Its prime factorization is $2^8\cdot 3$. Yielding $\frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}}}}{2}$ find that it is equal to $\cos{\left(2\pi\frac{1}{8}\right)}$.

My question is why would it not apply to $f_n>3$ ? and why always $f=2$ ? Is there some way that all cosine results can be represented by nested roots of, not only two, but of any prime numbers?

I have been trying to resolve the constraint of $f_n>3$ by turning any prime factor larger than $3$ into the form of $2^n+1$. For example, $20=2^2\cdot 5$ can be represented by $2^2(2^2+1)=2^4+2^2$ but I cannot find a way to represent this in any other way, not even as nested roots.

I don't have a degree in mathematics, or a major in it so I'm kind of new to this whole thing.

John Clement Husain
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  • A related question: For which angles we know the sin value algebraically (exact)? – David K Jun 12 '24 at 13:25
  • $\cos\left(\frac{2\pi}{20}\right) = \frac14\sqrt{10+2\sqrt5}.$ You can make this into something like your formula for the case $p=24$: for $p=24$ you get $\frac{\sqrt{2 + \sqrt3}}{2}$ whereas for $p=20$ you get $\frac{\sqrt{10 + \sqrt{20}}}{4}$. – David K Jun 12 '24 at 17:05
  • The counter-question would be, why in our wildest dreams would we ever imagine that $\cos(2\pi/p)$ where $p = 2^{n+3}$ or $p=2^{n+2}\cdot 3$ could be written in radicals simply by throwing out the first two factors of $2$ and putting the rest of the prime factors into a sequence of nested radicals, then dividing by $2$? It seems like a completely unexpected coincidence that the formula should actually work. The fact that it doesn't generalize any further is the unsurprising part of this observation. – David K Jun 12 '24 at 17:16

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Let $r$ be a nonzero integer. $\cos(2\pi/r)$ is the real part of $e^{2\pi i/r}$, which is a root of the polynomial $x^r-1$. It is known that the Galois group of this polynomial is abelian, which implies that the group is solvable, which implies that $\cos(2\pi/r)$ can always be expressed in radicals. However, one usually needs radicals other than square roots to do it. The ones for which square roots suffice are the ones for which the regular $r$-sided polygon can be constructed with ruler and compass. These are the ones for which $r$ is a product of powers of two and distinct Fermat primes. Fermat primes are primes of the form $2^{2^m}+1$. The only known Fermat primes are $3,5,17,257$, and $65537$. There are good reasons for thinking there are only finitely many of them, and not-quite-as-good reasons for thinking that the five known examples are the only ones (but nothing has been proved).

In any event, $7$ is not a Fermat prime, so $\cos(2\pi/7)$ can't be expressed using only (nested) square roots; cube roots are necessary.

Gerry Myerson
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