Formula for rotating a vector in 2D

Let’s say we have a point $(x_1, y_1)$. The point also defines the vector $(x_1, y_1)$.

We rotate this vector anticlockwise around the origin by $\beta$ degrees.

The rotated vector has coordinates $(x_2, y_2)$.

Can we get the coordintes of $(x_2, y_2)$ given $(x_1, y_1)$ and $\beta$?

$L$ is the length of the vectors $(x_1, y_1)$ and $(x_2, y_2)$ : $L = \|(x_1, y_1)\| = \|(x_2, y_2)\|$.

$\alpha$ is the angle between the x axis and $(x_1, y_1)$.

We can see from the picture that:

$$\begin{align}\begin{aligned}x_2 = r - u\\y_2 = t + s\end{aligned}\end{align}$$

We are going to use some basic trigonometry to get the lengths of $r, u, t, s$.

Because the angles in a triangle sum to 180 degrees, $\phi$ on the picture is $90 - \alpha$ and therefore the angle between lines $q, t$ is also $\alpha$.

Remembering the definitions of $\cos$ and $\sin$:

$$\begin{align}\begin{aligned}\cos\theta = \frac{A}{H} \implies A = \cos \theta H\\\sin\theta = \frac{O}{H} \implies O = \sin \theta H\end{aligned}\end{align}$$

Thus:

$$\begin{align}\begin{aligned}x_1 = \cos \alpha L\\y_1 = \sin \alpha L\\p = \cos \beta L\\q = \sin \beta L\\r = \cos \alpha p = \cos \alpha \cos \beta L = \cos \beta x_1\\s = \sin \alpha p = \sin \alpha \cos \beta L = \cos \beta y_1\\t = \cos \alpha q = \cos \alpha \sin \beta L = \sin \beta x_1\\u = \sin \alpha q = \sin \alpha \sin \beta L = \sin \beta y_1\end{aligned}\end{align}$$

So:

$$\begin{align}\begin{aligned}x_2 = r - u = \cos \beta x_1 - \sin \beta y_1\\y_2 = t + s = \sin \beta x_1 + \cos \beta y_1\end{aligned}\end{align}$$

Luckily this is the same result as wikipedia on rotation matrices.