Tensoring with vector bundle is a dense endofunctor of $D^b(\text{coh }X) $? A functor $F:T\to R$ between triangulated categories is dense if every object of $R$ is isomorphic to a direct summand in the image of $F$.
Let $R=T=D^b(\text{coh }X)$ for a variety $X$ and consider the functor $-\otimes \mathcal{V}$, $\mathcal{V}$ a vector bundle.
I do not understand the following 
claim: "$-\otimes\mathcal{V}$ is a is a dense functor, as any object $P\in D^b(\text{coh }X)$ is a summand of $(P\otimes V^\vee)\otimes V$."
Can anyone help?

What part of the claim you don't understand?
For any vector bundle $V$ the bundle $V\otimes V^\vee$ contains trivial 1-dimensional vector bundle (spanned by the section "Id"$\in V\otimes V^\vee$; the map in the opposite direction is the evaluation map). So any object $P\in D^b$ is a summand of the image of the object $P\otimes V^\vee$.
