I was wondering what are the simplest formulas (or iterative process) to compute $\cos(1°)$ using only methods which predate calculus.
To give a more precise meaning to my question, let me just recap how to compute small values of $\cos$ and $\sin$. Basically you
1- either start from a known value (like $\sin 30° = \dfrac{1}{2}$ and $\cos 30° = \dfrac{\sqrt{3}}{2}$). From this you can always compute the value for the half-angle (since $\cos 2 \alpha = 2 (\cos \alpha)^2 - 1$ one has that $\cos \tfrac{\beta}{2} = \sqrt{ \dfrac{\cos \beta +1}{2} }$.
2- or use two known values (in addition to the value for 15° one can obtain in the first step, on can use $\sin 18° = \tfrac{1}{4}(\sqrt{5}-1)$ and $\cos 18° = \sqrt{ \dfrac{ 5 + \sqrt{5}}{8}}$; one gets these values from the regular pentagon) and use the angle substraction formula $\cos 3° = \cos 15° \cos 18° + \sin 15° \sin 18°$.
These two tricks were historically used to get new values. But to get 1° one needs to "reverse" the formula for the triple angle $\cos 3 \alpha = 4 (\cos \alpha)^3 - 3 \cos \alpha$. Since the solution of the cubic by radicals predate calculus, one gets a value for $\cos 1°$ which involves only radicals.
However, the formula depends a lot on the order in which the tricks were used (subtracting, bisecting and trisecting). For example, I get $\cos 3° = \tfrac{1}{4} \sqrt{ 8 + \sqrt{15} + \sqrt{3} + \sqrt{ 10 - 2 \sqrt{5}} }$ or $\cos 3° = \sqrt{ \dfrac{(5 + \sqrt{5})(2+\sqrt{3})}{32}} + \tfrac{1}{8}(\sqrt{5} -1)(\sqrt{ 2 - \sqrt{3}})$. Although the formulas are "obviously" equal, they might not be equally hard to compute.
On the other there were other (iterative) methods known to compute $\cos 1°$ (a famous one is "Jost Bürgi's Kunstweg").
So what is the best formula or method to compute [numerically] the value of $\cos 1°$?