Building a rotational matrix transformation

[adrs]

I am trying to build a rotational transformation matrix both for counterclockwise and clockwise angles.

The first matrix in the picture is for counterclockwise angles and the second one for clockwise angles. The first matrix I built corresponds to the one given in my linear algebra book so it seems the building process's OK.
However, in my book there isn't one for clockwise angles and that's why I've built one. I've been searching on the Internet and it seems that the rotational matrix for clockwise angles is the same as the one for counterclokwise ones but with the sines with opposite signs.
Nevertheless, that's not the one I've obtained If I thry for example with a 24.78 clokwise angle.

So, where does my reasoning fail? Thanks!

 

[HallsofIvy]

You don't really need different matrices for "clockwise" and "counterclockwise" rotattions because rotation "clockwise" is just the same as rotation counterclockwise but with negative angle.

Rotation through angle $\theta$, counterclockwise, is given, as you say, by the matrix
$$\begin{bmatrix}cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \end{bmatrix}$$

so that rotation through the same angle, clockwise is the same thing with $\theta$ replaced by $-\theta$:
$$\begin{bmatrix}cos(-\theta) & -sin(-\theta) \\ sin(-\theta) & cos(-\theta) \end{bmatrix}$$

Perhaps what you are missing is that cosine is an even function and sine is an odd function: $cos(-\theta)= cos(\theta)$ and $sin(-\theta)= -sin(\theta)$ so that the previous matrix is just
$$\begin{bmatrix}cos(\theta) & sin(\theta) \\ -sin(\theta) & cos(\theta) \end{bmatrix}$$
the "transpose" of the counterclockwise rotation.

Of course, rotating cockwise by a given angle and then counterclockwise by the same angle, or vice-versa, gets you right back to where you were before so a good exercise is to do the two multiplications
$$\begin{bmatrix}cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \end{bmatrix}\begin{bmatrix}cos(\theta) & sin(\theta) \\ -sin(\theta) & cos(\theta)\end{bmatrix}$$
and
$$\begin{bmatrix}cos(\theta) & sin(\theta) \\ -sin(\theta) & cos(\theta) \end{bmatrix}\begin{bmatrix}cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta)\end{bmatrix}$$
and show that they both give the identity matrix.