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 :).