Let $\zeta$ be a primitive $n$-th root of unity. Consider the field extension $\mathbb{Q}(\zeta+\zeta^{-1}) \subset \mathbb{Q}(\zeta+\zeta^{-1},\zeta^t)$, for some positive integer $t$. What is its degree?
Edit: as mentioned in the comments, if $p$ and $t$ are coprime, the degree is just $2$. What about the other cases?
If $n \leq 2$ then $\zeta = \pm 1$, so the fields you ask about are both $\mathbf Q$.
Now we will assume $n \geq 3$, so $\zeta$ is not real.
Let $u = \zeta + \zeta^{-1}$, which is real. Then $u\zeta = \zeta^2 + 1$, so $\zeta^2 - u\zeta + 1 = 0$. Thus $\zeta$ has degree at most $2$ over $\mathbf Q(u)$, and since $\zeta \not\in \mathbf Q(u)$ due to $\zeta$ not being real, $\zeta$ has degree $2$ over $\mathbf Q(u)$.
Since $\mathbf Q(u) \subset \mathbf Q(u,\zeta^t) \subset \mathbf Q(\zeta)$ and $[\mathbf Q(\zeta):\mathbf Q(u)] = 2$, the degree $[\mathbf Q(u,\zeta^t):\mathbf Q(u)]$ is $1$ or $2$, and it is $1$ if and only if $\zeta^t \in \mathbf Q(u)$. The only real roots of unity are $\pm 1$, and $\pm 1$ are in $\mathbf Q(u)$, so $\zeta^t \in \mathbf Q(u)$ if and only if $\zeta^t = \pm 1$, which is equivalent to $\zeta^{2t} = 1$, which is equivalent to $n \mid 2t$. Thus when $n \geq 3$, the field degree $[\mathbf Q(u,\zeta^t):\mathbf Q(u)]$ is $1$ if $n \mid 2t$ and it is $2$ if $n \nmid 2t$.