Spherical twist and shift functor Answers to questions on this page (Spherical Twist Equivalence) include:

$$\mathcal{E}[1] \boxtimes (\mathcal{E}[1])^\vee \cong \mathcal{E}[1]
 \boxtimes \mathcal{E}^\vee[-1] \cong \mathcal{E} \boxtimes
 \mathcal{E}^\vee.$$

What is the reason for $(\mathcal{E}[1])^\vee \simeq \mathcal{E}^\vee[-1]$ here? Also, do they use the assumption of spherical?
In the argument, $\mathcal{E}$ is a spherical object in $D^b(X)$, bounded derived category consisted by coherent sheaves on a smooth projective variety X, i.e., (i) $\mathcal{E} \otimes \omega_X \simeq \mathcal{E}$, (ii) $\operatorname{Hom}(\mathcal{E}, \mathcal{E}[i]) = k$ (if $i = 0$ or $\operatorname{dim} X$), $\operatorname{Hom}(\mathcal{E}, \mathcal{E}[i]) = 0$ (otherwise). 
Thank you. 
EDIT : 
I understand like that
$$(\mathcal{E}[1])^\vee = R\mathcal{H}om(\mathcal{E}[1], \mathcal{O}_X), $$
$$\mathcal{E}^\vee[-1] = (R\mathcal{H}om(\mathcal{E}, \mathcal{O}_X))[-1]. $$
Honestly, I don't have the right intuition about function: "Apply a shift". I would appreciate it if you could tell me how you understand this.
 A: $\newcommand{\rhomm}{R \mathrm{Hom}} \newcommand{\rhom}{R \mathcal{H} om} \newcommand{\Homm}{\mathrm{Hom}} \newcommand{\Hom}{\mathcal{H} om}$I'll be using notation from this Stacks project article, plus this one. Note that in the Stacks project, the arguments of the derived hom are not complexes, but they define it to be $\Homm^\bullet$ which does take complexes as arguments, so I don't think it matters that I will put complexes as arguments of the derived hom. As per the discussion in p. 76 of Huybrechts' book, I think it's also OK for me to switch $R\mathrm{Hom}$ in the Stacks article to $\rhom$, and $\Homm$ to $\Hom$.

Write $\rhom_{D(R)}(L^\bullet, M^\bullet) = \Hom^\bullet(L^\bullet, M^\bullet)$. In turn, $\Hom^\bullet(L^\bullet, M^\bullet)$ is given as 
$$ \Hom^n(L^\bullet, M^\bullet) = \prod_{n=p+q} \Hom_R(L^{-q}, M^p).  $$
Then $\rhom_{D(R)}(L^\bullet[1], M^\bullet) = \Hom^\bullet(L^\bullet[1], M^\bullet) $ which is given by 
$$ \Hom^n(L^\bullet[1], M^\bullet) = \prod_{n=p+q} \Hom_R(L^{-q+1}, M^p).  $$
On the other hand, $\rhom_{D(R)}(L^\bullet, M^\bullet)[-1] = \Hom^\bullet(L^\bullet, M^\bullet)[-1] $ which is given by 
\begin{align} \Hom^n(L^\bullet, M^\bullet)[-1] &= \prod_{n-1=p+q} \Hom_R(L^{-q}, M^p) \\
&= \prod_{n=p+q+1} \Hom_R(L^{-q}, M^p) \\ &= \prod_{n=p+\tilde{q}} \Hom_R(L^{-q}, M^p) \\ &= \prod_{n=p+\tilde{q}} \Hom_R(L^{-\tilde{q}+1}, M^p) \end{align}
where we have used $\tilde{q} = q+1$, so $\rhom_{D(R)}(L^\bullet[1], M^\bullet) \cong \rhom_{D(R)}(L^\bullet, M^\bullet)[-1] $. Your case $\mathcal{E}[1]^\vee \cong \mathcal{E}^\vee [-1]$ is a special instance of this.
