Suppose that K = $\mathbb{Q}(\zeta )$ where $\zeta $ is a primitive nth root of unity, then the matching totally real subfield is ${{\text{K}}^{+}}$ = $\mathbb{Q}(\zeta +\text{ }1/\zeta )$ . (These two have ‘matching’ rings of integers given by $\mathbb{Z}[\zeta ]$ and $\mathbb{Z}[\zeta +1/\zeta ]$ .)
So the ‘canonical’ generator of ${{\text{K}}^{+}}$ is $\zeta +1/\zeta ={{\lambda }_{n}}=2\operatorname{Cos}(2\pi /n)$ which has minimal degree $\varphi (n)/2$ for n > 2, but any Galois conjugate of ${{\lambda }_{n}}$ would suffice so $\mathbb{Q}(\zeta +1/\zeta )=\mathbb{Q}({{\zeta }^{k}}+1/{{\zeta }^{k}})$ for gcd(k,n) = 1. The question is how to characterize the possible generators of ${{\text{K}}^{+}}$ .
For example ${{\tan }^{2}}(\pi /n)$ will always be a generator of ${{\text{K}}^{+}}$ based on the ‘double-angle’ formula relating cos and tan. There are other possible candidates including any algebraic integer a in $\mathbb{Z}[\zeta + 1/\zeta ]$. Note that ${{\tan }^{2}}(\pi /n)$ is not an algebraic integer when n is twice-odd (a case that some authors omit). For example ${{\tan }^{2}}(\pi /6)$ is 1/3 but here ${{\text{K}}^{+}}$ is $\mathbb{Q}$ which is also $\mathbb{Q}(1/3)$ . It can be awkward doing calculations with non-integer generators, and there are often advantages to using unit generators. Even ${{\lambda }_{n}}$ sometimes fails here. For example ${{\lambda }_{12}}$ =$\sqrt{3}$ which is the ‘canonical’ choice of generator but it is not a unit and ${{\tan }^{2}}(\pi /12)$ = $7-4\sqrt{3}$ is a unit which can be used as an alternate generator of ${{\text{K}}^{+}}$ for n =12. (Geometrically this ‘scaling parameter’ has the advantage of properly describing how generations of polygons scale relative to the dodecagon as ‘parent’. For n = 6 above, this natural geometric scaling is the identity so in this case ${{\lambda }_{6}}$ = 1 is the correct scaling and ${{\tan }^{2}}(\pi /n)$ should only be used as an alternative generator when n is twice-even.) The real issue is understanding what choices are possible and any help along these lines will be appreciated.
For $p$ prime thus for $n$ square-free the conjugates of $\zeta_n$ are a normal basis so it is simply
$$\mathbb{Q}(\zeta_n) = \sum_{m \in (\mathbb{Z}/n\mathbb{Z})^\times}\zeta_n^m \mathbb{Q},\qquad \mathbb{Q}(\zeta_n+\zeta_n^{-1}) = \sum_{m \in (\mathbb{Z}/n\mathbb{Z})^\times/\{ \pm 1\}}(\zeta_n^m +\zeta_n^{-m}) \mathbb{Q}$$ $$\mathbb{Q}(\zeta_n+\zeta_n^{-1}) = \mathbb{Q}(\beta), \beta =\!\! \!\!\sum_{m \in (\mathbb{Z}/n\mathbb{Z})^\times/\{ \pm 1\}}\!\!\!\!(\zeta_n^m +\zeta_n^{-m}) b_m \iff \forall k \in (\mathbb{Z}/n\mathbb{Z})^\times/\{ \pm 1\}, \exists m, b_{mk} \ne b_m$$
Not sure if there is a way to enumerate the $\beta$ such that $\mathbb{Z}[\zeta_n+\zeta_n^{-1}]= \mathbb{Z}[\beta]$ other than enumerating the elements of $\mathbb{Z}[\zeta_n+\zeta_n^{-1}]$ and checking one by one