Finding 2D rotation matrices, and how are they derived? Is there a 2D matrix "formula" (for lack of a better known word) where I can substitute in the value of theta to find the matrix of a rotation about the origin through theta degrees clockwise and anticlockwise? 
How are these matrices derived?
Regards Tom
 A: If you are dealing with column vectors
$$
\begin{bmatrix}
\cos(\theta)&-\sin(\theta)\\
\sin(\theta)&\cos(\theta)
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix}
$$
rotates the vector by $\theta$ counterclockwise. A clockwise angle is the negative of a counterclockwise angle; therefore,
$$
\begin{bmatrix}
\cos(\theta)&\sin(\theta)\\
-\sin(\theta)&\cos(\theta)
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix}
$$
rotates the vector by $\theta$ clockwise.

If you are dealing with row vectors, the situation is reversed.
$$
\begin{bmatrix}
x&y
\end{bmatrix}
\begin{bmatrix}
\cos(\theta)&\sin(\theta)\\
-\sin(\theta)&\cos(\theta)
\end{bmatrix}
$$
rotates the vector by $\theta$ counterclockwise. Similarly,
$$
\begin{bmatrix}
x&y
\end{bmatrix}
\begin{bmatrix}
\cos(\theta)&-\sin(\theta)\\
\sin(\theta)&\cos(\theta)
\end{bmatrix}
$$
rotates the vector by $\theta$ clockwise.
A: Robjohn's answer addresses the main question. The second part of how the matrices are derived follows the usual principle. Find the images under rotation of the standard basis vectors $(1, 0)^t$ and $(0,1)^t$, and make them the first and second column of the matrix desired. Draw the right angle triangles with $\theta$ as one angle and use elementary trigonometry.
