This comes from physics, we know the configutation space of rigi body's motion is lie group $SO(3)$, and we know the phase space of the cinfiguration space is its cotangent bundle. So this tells us the phase space of rigid body's motion is $T^{*}(SO(3))= \bigcup T_gSO(3)$.
We first know $T_eG \cong \mathfrak g$. The lie algebra $\mathfrak g$ of $SO(3)$ is the 3-dimensional skey symmetric matrix = $\mathfrak{so}(n)=\left\{M \in \operatorname{Mat}(n \times n, \mathbb{R}) \mid M+M^{T}=0\right\}$.
Next we compute the $T_g (SO(3))$ for any $g\in SO(3)$. First take a vector $X\in T_e(SO(3))$, then we know $T_g(SO(3))= gX$ by differential left multiplication. Obviously, this can be transfered in matrix language and matrix multiplication.
In a word, $T_g(SO(3))$= (any skew symmetric matrix) $\cdot$ (element g in $SO(3))$.
Since all the matrix is inverse to each other, so I believe we've solved how cotangent bundle looks like.
In a word, I think it looks like $SO(3)\cdot$ {$3$-dimensional skew-symmetric matrix}
So is my proof or calculation correct ? And does this have a much prettier formula or a physical terminology ?