Since the day I met this cutie, I struggled with it.
I always knew that there are some cosines, sines, and one minus but never knew
how to place them to that bloody $2\times 2$ square matrix.

Today, I still don't remember how the rotation matrix looks like.
I don't want to remember.
I don't want to know.
So maybe the title is a bit misleading.
To be honest, this post is more about deriving the matrix than remembering it.
For most, I thought that to get the rotation matrix, I need to see the whole
thing with polar coordinates, use one (or more) goniometric identities (I need to derive
them because as with everything else I just don't remember it) and maybe after all this
tiresome work I obtain the result.

## Little brain method

Now, let's use a little bit of brain, and some knowledge of linear transformations (their matrices)
and get this thing sorted much more quickly.

What is the simplest rotation?
Of course, it's the $0\degree$ rotation, in other words, the identity.

How does the identity matrix look?
That's simple, right?
$\begin{bmatrix}
1& 0 \\
0& 1
\end{bmatrix}$

We know that the rotation matrix is composed of just sines, cosines, and $+/-$ signs and also that
$\sin 0\degree = 0,$
$\cos 0\degree = 1.$

Ha! Now, we know what's on the main diagonal — cosines.
And what remains for the other diagonal?
Sines, of course.
But we're not that sure about signs of our sines.

Our matrix now looks like this:
$\begin{bmatrix}
\cos \alpha& \pm \sin \alpha \\
\pm \sin \alpha & \cos \alpha
\end{bmatrix}$

Plus, or minus, that is the question.
To answer this question, let's think of $90\degree$-rotation and point with coordinates $[1, 1]$.
Because when talking about a rotation we usually mean the counterclockwise one, the resulting point
should lie on coordinates $[-1, 1]$.
It shouldn't be hard to see that one matrix that does that is
$\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}.$

If we put this together with the nearly-correct rotation matrix we can work out the signs and get
the neat result
$\begin{bmatrix}
\cos \alpha& - \sin \alpha \\
\sin \alpha & \cos \alpha
\end{bmatrix}.$

From here, we can also note another nice property.
To get the clockwise rotation, all we need is to swap signs accordingly :).