TITLE: how to compare $\sin(19^{2013}) $ and $\cos(19^{2013})$
QUESTION [14 upvotes]: how to compare $ \sin(19^{2013})$  and $\cos (19^{2013})$ or even find their value range with  normal calculator?
I can take $2\pi k= 19^{2013} \to \ln(k)= 2013 \ln(19)- \ln(2 \pi)=5925.32 \to k= 2.089 \times 10^{5925}$, but it useless.(I can get final answer with WolframAlpha but it is not allowed.) 
Any hint? thanks!

REPLY [8 votes]: To find the sine and cosine, you'll need to reduce the angle ($19^{2013} \approx 1.352  \times 10^{2574}$) modulo $2\pi$.  So, for reasonable accuracy, you'll need about 2600 digits of $\pi$.  Fortunately, the first 100 000 or even million digits are readily accessible online.
It happens that $19^{2013} \approx 1.2329141525482654$ modulo $2\pi$, so:

$\sin(19^{2013}) \approx \sin(1.2329141525482654) \approx 0.9434588183383549$
$\cos(19^{2013}) \approx \cos(1.2329141525482654) \approx 0.3314897556480367$

All you need is a programming language that supports arbitrary-precision rational arithmetic, and a way of obtaining lots of digits of $\pi$.  A normal TI-89 calculator will provide the former; the hard part is implementing an algorithm for the latter.  But that's a topic for another question.