How does one approximate $\cos(58^\circ)$ to four decimal places accuracy using Taylor's theorem? When one needs to compute say $\cos (58^\circ)$ with an error of at most $10^{-4}$, how does one go about it?
What is an appropriate centre of the Taylor expansion, and how does one determine the required degree of the Taylor polynomial?
 A: If you wish to calculate, then you can solve $input^x/x! = accuracy$ and round the result up to the nearest multiple of $4$ (so the final term is an addition).  For your case, the solution is $8$, as you should always choose the largest positive solution.  Thence, $8$ is the largest degree of the Taylor polynomial you ask for.
To ad hoc the number of terms you will need in the Taylor expansion for cosine, convert the value being given to cosine into radians, and multiply that number by $e$, rounding up the result to the nearest multiple of $4$.  That number is the degree of the expansion for which every greater degree yields a more accurate approximation of cosine for the given input.  This is because $n!$ dominates $x^n$ if $x$ is less than $n/e$.  If you are solving one term at a time, continue until the term's result is less than your error margin, then do one or two more terms*.  This works because every further term will not only retreat further from the margin, it will alternate addition and subtraction.
As for the question of "the centre", if I'm right about what you're asking, then the answer is $2 \pi e$, or $20$, due to the equivalence of inputs for cosine and the domination mentioned above.
*Mind, a proof requires far more detail than I am prepared to provide here.
