Take an odd prime $p$ and put  $x_0:=\sum\limits_{j=0}^{p-1}\left(a_{j}\sqrt{p}\cos\dfrac{j\pi}p+b_{j}\sin\dfrac{j\pi}p +c_{j}\tan\dfrac{j\pi}p\right)$, where the $a_{ij}$ are integers. If $f$ denotes the minimal polynomial of $x_0$, can we **prove that $p$ divides all coefficients of $f$ except the leading one**? 

 
I have quite a bit of numerical evidence for this. Note that it obviously doesn't hold without the $\sqrt p$ factor, but more interestingly, it is also false if $\sqrt p$ goes with the other terms instead of the $\cos$ term. Moreover, it seems that in those cases, *none* of the non-zero coefficients is divisible by $p$.

(More generally, I think those coefficients are divisible by $p$ if we replace $\dfrac{j\pi}p$ by $\dfrac{j\pi}{p^r},\ r\in\mathbb N$ and do the sum over $j=0,...,p^r-1$.)

 
 
If all but one of the $a_j,b_j,c_j$ are $0$, the claim is quite easy to prove (and not new). For instance, for $x_0=\sin\dfrac{j\pi}p$ with any fixed $j$, we have explicitly $$f(x)=\sum\limits_{i=0}^k(-1)^k\dbinom p{2i+1}(1-x^2)^{k-i}x^{2i},$$ where $p=2k+1$. So the claim is obvious here.

**Added:** *It should be clear from Galois theory that in general, the conjugates of $x$ are the sums obtained by replacing all the $j$'s by $kj$ for a fixed $k=2,...,p-1$.*

**Literature:** 


Beslin, S., de Angelis, V., 2004. The minimal polynomials of sin(2π/p) and cos(2π/p). Mathematical Magazine 77, 146–149.

Heierman, William E., [Minimal  polynomials  for  trig  functions of angles  rationally  commensurate  with  π][4]

Lang, Wolfdieter, [Minimal Polynomials of sin (2π/n)][2]

Surowski, David, and McCombs, Paul,  [Homogenous polynomials  and the minimal polynomial of cos(2π/n)][1]

W. Watkins and J. Zeitlin, [The Minimal Polynomial of cos(2Pi/n)][3], Am. Math. Monthly 100,5 (1993) 471-4.



  [1]: http://www.math.ksu.edu/~dbski/preprints/cosine-mo-jo.pdf
  [2]: http://www-itp.particle.uni-karlsruhe.de/~wl/EISpub/A181872.pdf
  [3]: http://www.jstor.org/stable/2324301
  [4]: http://www.corunduminium.com/Trigpolys.html