Can we apply the formula below to find the new coordinates of a given point $A':=(x',y')$ after applying a rotation with an angle $\theta$ on this point (I emphasise on the point, not on the coordinates system)?
$$ x' = x\cos(\theta) + y\sin(\theta)$$
$$y' = -x\sin(\theta) + y\cos(\theta)$$
whete $(x,y)$ are the coordinates of $A$ before applying the rotation .
This is almost correct. You may consider the Rotation mapping defined by $$f_\theta:\begin{cases} \mathbb{R}^2 \to \mathbb{R}^2\\ \begin{pmatrix} x\\y\end{pmatrix} \mapsto \begin{pmatrix} \cos\theta&-\sin\theta\\ \sin\theta & \cos \theta\end{pmatrix} \begin{pmatrix} x\\y\end{pmatrix}\end{cases}$$
How do we find this mapping? We consider $\mathbb{R}^2$ equipped with the standard basis $$e_1 := \begin{pmatrix} 1\\0\end{pmatrix} \qquad \text{and} \qquad e_2 :=\begin{pmatrix} 0\\1\end{pmatrix}$$
Then the rotation of $e_1$ about an angle $\theta$ yields $$f_\theta(e_1) = \cos\theta e_1 + \sin\theta e_2$$
as one can easily seen using pythagoras or the unit circle. Furthermore $$f_\theta(e_2) = -\sin\theta e_1 + \cos\theta e_2$$ And since the coordinates of the images of the basis vectors are the columns of the matrix, we get the above result.