I found this topic in a book ‘Metric Affine Geometry’ by Ernst Snapper and Robert J. Troyer.

I call a field $k$ trigonometric iff there is a quadratic form $q$ over $k^2$ such that every two lines through the origin in $k^2$ is isometric with respect to $q$. This condition is sufficient to introduce trigonometric functions over $mathbf{SO}(k^2,q)$ in a geometric fashion. Hence, a name.

Obviously, $mathbb{R}$ is trigonometric. I know, that to be trigonometric the field $k$ must Pythagorean, that is for every finite sequence of values $(alpha_i)^n_{i=1} in k^n$ there is a $gamma in k$ such that

$$

sum^n_{i=1} alpha_i^2 = gamma^2,

$$

namely every sum of squares is a square. Secondly it must be a formally real field, which means that $-1$ is not a sum of squares. Hence, sadly $mathbb{Q},mathbb{C},mathbb{R}(x),mathbb{Q}_p,mathbb{F}_p$ are all not trigonometric. Probably some extension of $mathbb{R}(x)$ which allows square roots of formally positive functions may work. But I still doubt that it can be totally-ordered, and probably there are some clews in differential Galois theory. Maybe $hat{mathbb{Q}} cap mathbb{R}$, where $hat{mathbb{Q}}$ are algebraic numbers will work, or just adjoining enough real algebraic square roots to mathbb{Q} (call it a Pythagorean closure $overline{mathbb{Q}}$). At least it is Pythagorean and formally real. But I don’t think it is interesting enough.

But I’m very a curious about finding an interesting example of trigonometric field different from $mathbb{R}$. Trigonometric field $k$ different from $mathbb{R}$ may mean formally that $k$ is not between $overline{mathbb{Q}}$ and $mathbb{R}$. I would be very grateful if you could suggest one.

If the result are negative, this would mean that class of all trigonometric fields has certain lower and upper bounds.