Below is an attempt at proving that if a curve satisfies an equation $ f(X,Y)= 0$ , its image under a counterclockwise rotation by $\theta$ radians is defined by the equation :
$$f( x cos \space \theta + y sin \space\theta , y cos \space\theta - x sin \space\theta) = 0 $$
My question is : a) concerning step (2) how to prove in a more formal way ( less relying on intuition) the fact that the matrix to be used is the matrix defining a clockwise rotation ( though the curve is defined as a rotated counterclockwise one)? b) also, concerning step (5), is it admissible to turn the column vector into an ordered pair?
I think this question is somewhat related to the fact that vectors are contravariant. Is it the case, and how to make this statement more precise, if true?
(1) Let $C$ be a curve defined by the equation :
$$ f(X,Y)= 0 , $$
for example $Y- cos(X) = 0 \space \space ( \iff Y= cos(X) ) $ , or $ (X^2 / 5² + Y^2 /3^2) -1 = 0 . $
(2) Let $C'$ be the image of curve $C$ under a rotation of $\theta$ radians counterclockwise.
One can define $C'$ as the set of all points $P=(x,y)$ such that the image of $P$ under a clockwise rotation of $ \theta$ radians satisfies the equation $f(x,y)=0$.
(3) The matrix of a clockwise $\theta$ radians rotation can be found by asking what are the images of the basis vectors $e_1 = \left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right)$ and $e_2 = \left( \begin{array}{c} 0 \\ 1 \\ \end{array} \right)$ under a clockwise $\theta$ rotation . Vector $e_1$ is mapped to $\left( \begin{array}{c} cos \theta \\ - sin\theta \\ \end{array} \right)$ and vector $e_2$ is mapped to $\left( \begin{array}{c} sin \theta \\ cos\theta \\ \end{array} \right)$.
Consequently, the matrix of a clockwise rotation by $\theta$ radians is : $\begin{bmatrix}cos \theta & sin \theta\\- sin\theta & \cos \theta\end{bmatrix}$ ( that is, the columns of this matrix are the images of $e_1$ and $e_2$ under a $ - \space\theta $ rotation).
(4) Now, we can restate the definition of curve $C'$ as follows. $C'$ is the set of all points $(x,y)$ such that :
$$f \Bigg{(} \begin{bmatrix}cos \theta & sin \theta\\- sin\theta & \cos \theta\end{bmatrix} \left( \begin{array}{c} x\\ y\\ \end{array} \right) \Bigg {)}=0$$
that is
$$f \Bigg {(} \left( \begin{array}{c} x cos \theta + y sin \theta\\ x ( - sin \theta) + y cos \theta\\ \end{array} \right) \Bigg {)} =0$$.
(5) If we finally allow ourselves to turn the column vector (defining the image of point $P$ under the clockwise rotation) into an ordered pair, we therefore can write that the equation of $C'$ is :
$$ f( x cos \space \theta + y sin \space\theta , y cos \space\theta - x sin \space\theta) = 0 $$.
For example, the image of the cosine curve will be defined by :
$$[y cos \space\theta - x sin \space\theta] - [cos( x cos \space \theta + y sin \space\theta)] =0$$.