how to compare $\sin(19^{2013}) $ and $\cos(19^{2013})$ 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!

If $19^{2013}$ were measured in degrees we could use a simple trick.
$19$ is one more than $18$ which is a multiple of $6$. This means that any power of $19$ is also $1$ more than a multiple of $6$, in particular $19^2 = 361$
$19^2$ being $1$ more than a multiple of $360$ tells us that any even power of $19$ will be 1 more than a multiple of $360$. In particular we can conclude that $19^{2012}$ is one more than a multiple of $360$.
What does that mean? It means that if we travel $19^{2012}$ degrees around the unit circle we will just end up at the $1$ degree tick mark. 
$19^{2013}$ is just $19^{2012}$ $19$ times. This means we would make the trip described above $19$ times each time ending up one tick further than the last time meaning that we would finish at $19$ degrees. In other words $19$ degrees is located at the same position of the unit circle as $19^{2013}$.
If you've never worked with modular arithmetic you may not understand why $19^2 = 360 + 1$ implies that $19^{2012} = k \cdot 360+1$.
To see this consider powers of $(360+1)$.
$$360+1$$
$$ (360+1)^2 = 360^2 + 2\cdot360 + 1$$
$$ (360+1)^3 = 360^3 + 3^360^2 + 3*360 + 1$$
$$ \vdots$$
$$ (360+1)^n = 360^n + n *360^{n-1} + \cdots + n * 360 + 1 $$
Notice the result is always one more than a multiple of $360$.
