# Could someone give a proof of $\sum_{i=1}^{12}\sin(2^{13i+1}\pi/169)=0$?

An alternative description is:

Suppose $\zeta$ is a primitive 169-th root. Prove that

$$t=\sum_{i=1}^{12}\zeta^{2^{13i}}\in\mathbb{R}$$

Besides, is it possible to find out the minimal polynomials of $t$ without computer assistance?

I encounter this problem when I am trying to find out the 13-degree subfield of the 169-th cyclotomic field. 2 is a primitive element in $\mathbb{R}_{169}$, therefore the Galois group is a cyclic group generated by $\sigma_2$ which sends $\zeta$ to $\zeta^2$. Therefore $(\sigma_2)^{13}$ generates a subgroup of the Galois group of order 12. Therefore I found out that $t$ is an invariant piece of the subgroup. I was surprised that $t\in R$ and moreover, the roots of the minimal polynomial of $t$ in $\mathbb{Z}[x]$ are real. Is this a coincidence, or is there some explanation?

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Cyclic extensions of odd degree over the rationals are always totally real. – franz lemmermeyer Apr 11 '12 at 9:17
$(\sigma_2^{13})^6$ is complex conjugation, so $t$ is fixed by complex conjugation. – David Speyer Apr 11 '12 at 16:52