Let $D^b(X)$ is the derived category of coherent sheaves on the smooth projective variety X and an object $\mathcal{E} \in D^b(X)$ is spherical, i.e., (i) $\mathcal{E} \otimes \omega_X \simeq \mathcal{E}$, (ii) $\operatorname{Hom}(\mathcal{E}, \mathcal{E}[i]) = k$ (if $i=0$ or dim X), $\operatorname{Hom}(\mathcal{E}, \mathcal{E}[i]) = 0$ (otherwise).
Then I want to prove that the derived dual of $\mathcal{E}$ i.e., $\mathcal{R}\operatorname{Hom}(\mathcal{E}, \mathcal{O}_X)$, $\mathcal{E}[i]$ for any $i \in \mathbb{Z}$, and $\mathcal{E} \otimes_X L$ for any $L \in \operatorname{Pic}(X)$ are again spherical.
It may not be difficult at all, but I am asking for help because I am unfamiliar with these arguments. Can anyone provide proofs or hint assertions or literature?
Let $\mathscr C$ be any object in the derived category. Consider: $$\operatorname{RHom}(\mathscr C,\operatorname{RHom}(\mathscr E,\mathscr O_X)\otimes \omega_X) = \operatorname{RHom}(\mathscr C \otimes \omega_X^\vee,\operatorname{RHom}(\mathscr E,\mathscr O_X))$$ $$ = \operatorname{RHom}(\mathscr C\otimes\mathscr E\otimes \omega_X^\vee,\mathscr O_X) = \operatorname{RHom}(\mathscr C\otimes\mathscr E,\mathscr O_X) = \operatorname{RHom}(\mathscr C,\operatorname{RHom}(\mathscr E,\mathscr O_X))$$
Since this is true for all $\mathscr C$ in a functorial way, we get $(i)$ by Yoneda. We have repeatedly used the (derived) tensor-hom adjuntion, $\mathscr E\otimes \omega^\vee \cong \mathscr E$ and that: $$\operatorname{RHom}(\mathscr C,\mathscr D) = \operatorname{RHom}(\mathscr C\otimes \mathscr L,\mathscr D\otimes \mathscr L)$$ for any line bundle $\mathscr L$ (since tensoring with a line bundle is an autoequivalence).