For example, $\sqrt 2 = 2 \cos (\pi/4)$, $\sqrt 3 = 2 \cos(\pi/6)$, and $\sqrt 5 = 4 \cos(\pi/5) + 1$. Is it true that any integer's square root can be expressed as a (rational) linear combinations of the cosines of rational multiples of $\pi$?

Products of linear combinations of cosines of rational multiples of $\pi$ are themselves such linear combinations, so it only needs to be true of primes. But I do not know, for example, a representation of $\sqrt 7$ in this form.

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    $\begingroup$ Look up Gauss sums (and more generally the Kronecker-Weber theorem). $\endgroup$ – Lucia Dec 7 '17 at 17:40

Someone should actually record the formula. If $p$ is a prime $\equiv 1 \bmod 4$, then $$\sqrt{p} = \sum_{k=1}^{p-1} \left( \frac{k}{p} \right) \cos \frac{2 k \pi}{p}$$ where $\left( \tfrac{k}{p} \right)$ is the quadratic residue symbol. Note that $\left( \tfrac{k}{p} \right) = \left( \tfrac{p-k}{p} \right)$, so every term appears twice.

Similarly, if $p \equiv 3 \bmod 4$, then $$\sqrt{p} = \sum_{k=1}^{p-1} \left( \frac{k}{p} \right) \sin \frac{2 k \pi}{p}.$$ Again, $k$ and $p-k$ make the same contribution.

These are usually both written together as $$\sqrt{(-1)^{(p-1)/2} p} = \sum_{k=1}^{p-1} \left( \frac{k}{p} \right) \exp \frac{2 k \pi i}{p}.$$ This is a formula of Gauss.

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    $\begingroup$ Should it be $2k\pi i/p$? $\endgroup$ – Ira Gessel Dec 7 '17 at 20:57

Yes, that is true. The general case is the Kronecker-Weber theorem as Lucia mentioned in the comments. For square roots one can be more explicit and prove $\mathbb{Q}(\zeta_p) \cap \mathbb{R} = \mathbb{Q}[\sqrt{ (-1)^{(p-1)/2} p }]$ for odd prime numbers $p$ using properties of the Legendre symbol. Therefore you can write $\sqrt{p}$ either as $\mathfrak{Re}(\sqrt{p})$ or $\mathfrak{Im}(\sqrt{-p})$ depending on $p\mod 4$ so that you can write it as rational (in fact: integral) linear combination of real- or imaginary parts of powers of $\zeta_p$, i.e. cosine or sine values.


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