As the title asks, I have some matrix $M(a,b,c)$, a function of three parameters. For any choice of $a,b,c$ this matrix is orthogonal, and has determinant 1. This means that it is an element of $SO(3)$, or in other words the set of all possible matrices, $\{M(a,b,c)\}$ is a subset of $SO(3)$.
Now there is a particular choice of say $a$ and $b$ that will produce rotations about the $z$ axis by angle $c$. There are similarly other choices that would produce any rotation around the $x$ and $y$ axis. I wonder if this is enough to say that not only is any $M(a,b,c) \in SO(3)$, but also that $\{M(a,b,c)\}=SO(3)$? This would require that any element of $SO(3)$ has a corresponding $M(a,b,c)$. Under what circumstances is this true?
It is of course possible to find a matrix that does this, it would have as parameters the three angles of rotation about $x,y,z$ and the matrix itself would be the multiplication of the rotation matrices about each axis $R_xR_yR_z$. However, I do not know if my question holds for any matrix that satisfies the specified criteria.
It's perhaps worthwhile to first clear something up:
There are not “the” angles of rotation about $x$, $y$ and $z$. Because rotations about different axes don't commute, there's no canonical way to decompose a rotation into rotations about the coordinate axes. Different conventions exist for specifying three Euler angles that parametrize all rotations.
Your function $M(a,b,c)$ generates all rotations if and only if the set of rotations it produces is closed under multiplication.
You can generate rotations about the coordinate axes, so if multiplying them again yields a rotation you can generate, you can generate all rotations by composing them from rotations about the coordinate axes (see above).
Conversely, since the group of all rotations is closed under multiplication, your set can only be that group if it's closed under multiplication.