We have that the exterior derivative acts over a wedge product in the following manner. Let $\alpha,\beta$ be $p,q$ forms respectively. Then we have that \begin{equation} d(\alpha\wedge \beta) = (d\alpha)\wedge\beta+ (-1)^p \alpha\wedge(d\beta) \end{equation}
Is it then true, by the fact that the hodge codifferential $\delta$ is the adjoint operator to $d$, that it acts in the same way, considering it has hidden inside the exterior derivative itself? For example, \begin{equation} \delta(\alpha \wedge\beta)=(\delta\alpha)\wedge\beta + (-1)^p\alpha \wedge (\delta\beta) \end{equation} Should this be obvious? Thanks!
Edit: Attempt
I tried to derive the relation and this is as far as I got in a coordinate basis: Let $\alpha = \frac{1}{p!}\alpha_{a_1\ldots a_p}f^{a_1}\wedge\ldots\wedge f^{a_p}, \, \beta = \frac{1}{q!}\beta_{b_1\ldots b_q}f^{b_1}\wedge\ldots\wedge f^{b_q}$, then one can show that
$$\star d \star (\alpha \wedge \beta)=\frac{(-1)^{n(p+q-1)}s}{p!q!}\nabla_b \alpha^{[b a_2\ldots a_p}\beta^{a_{p+1}\ldots a_{p+q}]}f_{a_{2}}\wedge\ldots\wedge f_{a_{p+q}} $$ Where $s$ is the signature of the metric, $f^a \equiv dx^a$ are the coordinate $1$-forms, and $\nabla$ is the (at least Koszul, but could be Levi-Civita) connection on the manifold.
Is there a way someone can see to simplify this? Thanks again!
Edit: So, by staring at it, I believe you can convince yourself that we can apply Leibniz rule to the $\alpha\beta$ term, yielding
$$\star d \star (\alpha \wedge \beta)=\frac{(-1)^{n(p+q-1)}s}{p!q!} \left[ (\nabla_b \alpha^{[b a_2\ldots a_p})\beta^{a_{p+1}\ldots a_{p+q}]} + \alpha^{[b a_2\ldots a_p}(\nabla_b\beta^{a_{p+1}\ldots a_{p+q}]}) \right] f_{a_{2}}\wedge\ldots\wedge f_{a_{p+q}} $$
Then we have that, in coordinates, the codifferential acting on a $p$ form is $\delta = (-1)^{n(p-1)+1}s \star d \star$ and in coordinates, its action on $\alpha$ would be
$$\delta\alpha = \frac{1}{(p-1)!}\left[-\nabla^b \alpha_{b a_2 \ldots a_p}\right]f^{a_2}\wedge\ldots\wedge f^{a_p} $$
I think I'm onto something...