Symmetrical and skew-symmetrical part of rotation matrix

"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.