Symmetrical and skew-symmetrical part of rotation matrix Every matrix can be decomposed to symmetrical and skew-symmetrical part with the formula:
$ A=\dfrac{1}{2}(A+A^T)+\dfrac{1}{2}(A-A^T)$.  
However if it is known only symmetrical part (we assume here that the whole matrix is unknown) it's impossible without additional information to reconstruct exactly skew-symmetrical part and vice versa.
In the case of 3D rotation matrix we have additional constraints and probably such reconstruction is possible.
Let's look at Rodrigues formula and two (symmetrical and skew-symetrical) parts  of rotation matrix:
$R(v,\theta)= \{I+(1-cos(\theta))S^2(v)\} + sin(\theta)S(v)$
where
$S(v)=\begin{bmatrix}
v\times{i} &  v\times{j}&  v\times{k} 
\end{bmatrix}^T$, skew-symetric matrix itself  ($3$ DOF set by components of an axis $v$) 
One can notice that having skew-symmetrical part of rotation matrix it is  relatively easy to reconstruct symmetrical part.   
Indeed 
$skew(R)=sin(\theta)S(v)$ 
and the whole expression $skew(R)$ can be decomposed to the product $kK$ in such a way that the sum of squares of matrix $K$ entries
i.e. $ \begin{bmatrix}
1 &  1 & 1 
\end{bmatrix}  (K\circ{K}) \begin{bmatrix}
1  &  1  & 1 
\end{bmatrix}^T =2 $
...   
then $k=sin(\theta)$ and $K=S(v)$
and we can calculate $ cos(\theta)$ and $ S(v)^2$
what makes possible reconstruction of symmetrical part.
Exactly we have two solutions because $sin(\theta)=sin(\pi-\theta)$.
In the second case
when we want to reconstruct skew-symmetrical part
the solution seems to be difficult to find (at least for me) so my question is:


*

*how to obtain skew-symmetrical part of rotation matrix $skew(R)$ knowing its
symmetrical part $sym(R)$?

*additionally: why do such asymmetry in difficulty of solutions exist at all ? (symmetry and skew symmetry in the first formula for the decomposition of any matrix $A$ seem not to differ too much)

*what is the situation for higher dimensions? (when we don't have a Rodrigues formula)
 A: "how to obtain skew-symmetrical part of rotation matrix skew(R), knowing its symmetrical part sym(R)?"
I will assume you are talking about rotations in $\mathbb{R}^3$. First note that a rotation about an axis of angle $\theta$ and a rotation about the same axis of angle $-\theta$ have the same symmetric parts. But apart from this ambiguity, it is possible to reconstruct the skew-symmetric part of a rotation matrix (up to a sign) by only knowing the symmetric part. Thus one can reconstruct the whole rotation matrix (apart from this ambiguity above) from the symmetric part.
Proof:
Case 1: $\operatorname{sym}(R)-I \neq 0$.
By Rodrigues formula, $\operatorname{sym}(R)-I = (1-\operatorname{cos}(\theta))S^2(v)$. For a square matrix $A$, you can define its norm by $||A||^2 = \frac{1}{2} Tr(A^TA)$. Then by normalizing $\operatorname{sym}(R)-I$, one gets rid of $1-\operatorname{cos}(\theta)$, and gets $S^2(v)$.
The kernel of $S^2(v)$ is the axis of rotation. It remains to recover $\operatorname{sin}(\theta)$, up to a sign. Well, in $\operatorname{sym}(R)-I$, the factor multiplying $S^2(v)$ is $1-\operatorname{cos}(\theta)$, so $\operatorname{cos}(\theta)$ is known, and from it we can get $\operatorname{sin}(\theta)$ up to a sign.
Case 2: $\operatorname{sym}(R)-I=0$
By Rodrigues formula, this corresponds to $\operatorname{cos}(\theta) = 1$, i.e. to $R = I$. This finishes the proof.
A: Let $A\in O(n)$. By an orthonormal change of basis, we can write $A$ in the form $A=diag(R_{\theta_1},\cdots,R_{\theta_p},I_q,-I_r)$ with $R_{\theta_i}=\begin{pmatrix}\cos(\theta_i)&-\sin(\theta_i)\\\sin(\theta_i)&\cos(\theta_i)\end{pmatrix}$, $\theta_i\in(0,2\pi)\setminus\{\pi\}$ and $2p+q+r=n$. Then the considered decomposition is $A=S+K$ with $S=diag(\cos(\theta_1)I_2,\cdots,\cos(\theta_p)I_2,I_q,-I_r)$.
Conversely, assume that we know $S$, the symmetric part of $A$; then by an orthonormal change of basis, we can write $S$ in the form $S=diag(\cos(\theta_1)I_2,\cdots,\cos(\theta_p)I_2,I_q,-I_r)$.
Case 1. The $(\cos(\theta_j)_j$ are distinct. Let $P_j$ be the eigenplane of $S$ associated to the double eigenvalue $\cos(\theta_j)$; note that $P_j$ is the invariant plane of $K$ s.t. the eigenvalues of $K_{|P_j}$ are $\pm i\sin(\theta_j)$ and that the $(\pm i\sin(\theta_j))_j$ are distinct; moreover, let $E$ be a subspace s.t. $S_{|E}=\pm I$, then $K_{|E}=0$. Thus $K$ is in the form $diag(U_1,\cdots,U_p,0_q,0_r)$ and, clearly, there are only two choices for each $U_j$: $\pm\begin{pmatrix}0&-\sin(\theta_j)\\\sin(\theta_j)&0\end{pmatrix}$. Finally, there are $2^p$ choices for the skew symmetric matrix $K$.
Case 2. The $(\cos(\theta_j)_j$ are not distinct. For example, assume that $n=4$ and $A$ is similar to $diag(R_{\theta},R_{\theta})$. Thus we know $S=\cos(\theta)I_4$. Note that if $P\in O(4)$, then $P^{-1}(\cos(\theta)I_4+K)P=\cos(\theta)I_4+P^{-1}KP$; consequently, the set of solutions in the skew matrix $K$ is composed with the $P^{-1}K_0P$ where $P\in O(4)$, $K_0=diag(\begin{pmatrix}0&-\sin(\theta)\\\sin(\theta)&0\end{pmatrix},\begin{pmatrix}0&-\sin(\theta)\\\sin(\theta)&0\end{pmatrix}),$. There is an infinity of such matrices $K$.
EDIT. Answer to @ Widawensen . Assume that $A$ is a real normal matrix ($AA^T=A^TA$). Then $SK=KS$ and there is a unitary matrix $P$ s.t. $S=P^{-1}diag(a_1 I_2,\cdots, a_p I_2,c_1,\cdots,c_q)P,K=P^{-1}diag(ib_1,-ib_1,\cdots,ib_p,-ib_p,0_q)P$ where $2p+q=n$, $(a_j,c_j)_j\in\mathbb{R}$ and $(b_j)_j\in\mathbb{R}^*$. More precisely, there is $P\in O(n)$ s.t. $S=P^{-1}diag(a_1 I_2,\cdots, a_p I_2,c_1,\cdots,c_q)P,K=P^{-1}diag(U_1,\cdots,U_p,0_q)P$ where $U_j=\begin{pmatrix}0&-b_j\\b_j&0\end{pmatrix}$.
Conversely, assume that we know $S$, that is $diag(a_1 I_2,\cdots, a_p I_2,c_1,\cdots,c_q)$. If $a_1,\cdots,a_p,c_1,\cdots,c_q$ are distinct, then $K$ is in the form $diag(U_1,\cdots,U_p,0_q)$; yet, unlike the case where $A\in O(n)$, we know nothing about the $(b_j)_j$ values.
