Yo creo que usted puede ser capaz de calcular esto sin obsceno números de dígitos de $\pi$ si usted toma ventaja del hecho de que estos son factoriales. Para simplificar el álgebra, podemos calcular el $a_n=e^{i(n!)}$ (desea que la parte imaginaria). Entonces $$a_{n+1}=e^{i(n!)(n+1)}=a_n^{n+1},$$ and it's perfectly reasonable to calculate $a_{100000}$ recursively with a high-precision library.
The downside is that to start the recursion you need a very good approximation of $e^i$, and I don't know if the error dependence works out any differently than in the $\pmod{2\pi}$ approach.
But to answer your actual question, Mathematica doesn't even break a sweat with the mere million digits needed for this:
> Block[{$MaxExtraPrecision = 1000000}, N[Sin[100000!], 10]]
-0.9282669319
takes about 15 ms on my computer.
For calculating the sine or cosine of a large arbitrary precision real number $x$, the gains of this method (which are tuned for $\pecado, n$ for integer $n$) are mostly lost, so I would recommend your original idea of reducing the argument $\bmod 2\pi$. As has been noted, the main bottleneck is a high-precision estimation of $\pi$. Your answer will be useless unless you can at least calculate $\frac{x}{\pi}$ to within $1$ (otherwise you may as well answer "somewhere between $-1$ and $1$"), so you need at least $\log_2(x/\pi^2)$ bits of precision for $\pi$. With $x\approx100000!$, that's about $1516701$ bits or $456572$ digits. Add to this the number $a$ of bits of precision you want in the result, so about $1516734$ digits of $\pi$ to calculate $33$ bits ($\aproximadamente un 10$ digits) of $\sin x$ in the range $x\aprox 100000!$.
Once you have an integer $n$ such that $y=2\pi n$ is close to $x$ (ideally $|x-2\pi n|\le2\pi$, it doesn't have to be perfectly rounded), calculate $\pi$ to precision $+\log_2(n)$, so that $s$ is known to precision $$, and then $x-y$ is precision $$ and $\sen x=\sin (x-y)$ can be calculated to precision $$ así.
