In my groups course at university, we’ve spent a while on the orthogonal group, leading up the the conclusion that $$ \mathrm{O}_n = \mathrm{SO}_n \mathbin{\dot{\cup}} \begin{pmatrix} -1 & & & \\ & 1 & & \\ & & \ddots & \\ & & & 1 \end{pmatrix} \mathrm{SO}_n \,. $$ The proof given was just “because cosets partition”.
I just want to know: Is this true because of a specific choice of matrix or will any matrix in $\mathrm{O}_n \setminus \mathrm{SO}_n$ do? How do we know there arent any other cosets? (I.e., that this expression covers everything in $\mathrm{O}_n$).
The subgroup $\mathrm{SO}(n)$ of $\mathrm{O}(n)$ has index $2$. Indeed, $\mathrm{SO}(n)$ is the kernel of the surjective group homomorphism $$ \det \colon \mathrm{O}(n) \longrightarrow \{ 1, -1 \} \,, $$ so $$ [\mathrm{O}(n) : \mathrm{SO}(n)] = | \mathrm{O}(n) / \mathrm{SO}(n) | = | \{1, -1\} | = 2 \,. $$
Let more generally $G$ be a group and $H$ a subgroup of $G$ of index $2$. The group $G$ is the disjoint union of the left cosets with respect to $H$, and there are precisely $2$ such cosests. (Because the number of cosets is given by $[G : H] = 2$.) One of these cosets is $H$ itself, whence $$ G = H \mathbin{\dot{\cup}} gH $$ for every element $g$ of $G$ with $gH ≠ H$. We have $gH = H$ if and only if $g ∈ H$, so any element $g$ of $G$ with $g ∉ H$ does the trick.
To explicitly answer your questions:
Yes, any matrix of $\mathrm{O}(n)$ that is not contained in $\mathrm{SO}(n)$ will do the trick.
There are only two cosets because $\mathrm{SO}(n)$ has index $2$ in $\mathrm{O}(n)$. This in turn is true because $\mathrm{SO}(n)$ is the kernel of a surjective homomorphism from $\mathrm{O}(n)$ into a group of order $2$.