If you derive a right exact functor $F$ you get a functor normally denoted by $RF$ on the derived category. Similarly, if you start with a left exact functor $G$ you get a functor normally denoted by $LG$. These are simply two triangulated functors on a triangulated categories. Suppose they are both defined in the same category (for example, the derived category of bounded complexes in some abelian category). Is there any way to distinguish the rightness or leftness of $RF$ and $LG$ from inside the triangulated category? If there isn't, then the $R$ and $L$ don't seem to be a very good choice of notation...
I am not sure what your goal is with this question, but I think there is an inherent problem with your set up. $RF$ stands for a functor derived from $F$ and not just a functor on the derived category. If you consider the derived category only as a triangulated category via a forgetful functor then you are also forgetting $F$, so there is no reason to still denote your functor by $RF$. On the other hand if you remember the derived category structure or at least you still deal with complexes and so you can take cohomology of these complexes, then $RF$ induces $R^iF$ and you have the condition that $R^iF=0$ for $i<0$ and $R^0F=F$. In other words, the functor $RF$ is such that the leftmost nonzero cohomology of it is $F$ and all the interesting stuff is on the right. Homework: do the same for $LG$. It is important that $RF$ is the right derived functor of $F$. It can happen that $RF=LG$, but then $F$ and $G$ are different. Typically $G=R^nF$ such that $R^mF=0$ for $m>n$. 


$F=H^0(X, )$
and$\Phi=L(H^n(X, ))[n]$
for $X$ a smooth projective variety of dimension $n$. You probably had more in mind, but perhaps you should share it. :) 2) Yes, but I still think the OP is talking about triangulated categories and not derived categories. 3) Cheers! 4) Unfortunately, I accidentally deleted my first comment and I don't think I can reproduce it, especially not at that "slot". – Sándor Kovács Nov 3 '11 at 21:11