For purposes of this post, let us restrict our attention to theories of arithmetic even though the question indeed makes sense for other theories. In what follows, assume that $PA$ is consistent.
A theory of arithmetic is said to be (arithmetically) sound if every statement it proves is true in the intended model $(\mathbb{N},+,\times,S,0)$. Clearly an inconsistent theory, being able to prove every statement, is unsound and is of no interest. On the other hand, by Gödel's incompleteness theorem, one can easily find unsound consistent theories extending $PA$, for example, $PA+\neg Con(PA)$.
My question is the following: Can an unsound consistent theory extending $PA$ be practically useful?
One may look at theories of sets (in particular, ZF) to see why this approach is not as meaningless as it seems at first sight. Suppose for the moment that in "the intended model of set theory", whatever that may be, ZF holds but the axiom of choice is false. There are theorems of ZF that can be proven in ZFC with an easy or short proof, for example, see this question. In this imaginary scenario, ZFC is unsound but is able to prove a true statement in an easy manner by circumventing obstacles that arise from absence of choice, for example, see the answer of Carl Mummert on Hindman's theorem. Proving theorems via forcing (i.e. forcing absolute statements to be true in an extension) may also be seen as a set-theoretic analogue of this approach with the exception that the meaning of the word "sound" is not clear in this context.
We know that there are arithmetical analogues of this speed-up phenomenon that occur for certain sound extensions of $PA$.
Back to my question... Has anyone investigated the possible use of an unsound consistent theory of arithmetic extending $PA$ to prove statements about natural numbers that can already be proven in $PA$ with possibly long or complicated proofs? Can an unsound consistent theory of arithmetic extending $PA$ be useful in other ways?
