I was studying conics and came around the topic of eliminating cross-product terms when rotating coordinates of a quadratic curve of the form $$A x^2 + B x y + C y^2 + D x + E y + F = 0$$ where $$\begin{align*} A x^2 &= A\left(\cos(\alpha) x' - \sin(\alpha) y'\right)^2\\ &= A\left(\cos^2(\alpha) x'^2 -2\cos(\alpha)\sin(\alpha) x' y' + \sin^2(\alpha) y'^2\right)\\ B x y &= B\left(\cos(\alpha) x' - \sin(\alpha) y'\right)(\sin(\alpha) x' + \cos(\alpha) y') \\ &= B\left(\sin(\alpha)\cos(\alpha)\left(x'^2 - y'^2\right) + \left(\cos^2(\alpha) - \sin^2(\alpha)\right) x' y'\right) \\ C y^2 &= C\left(\sin(\alpha) x' + \cos(\alpha) y'\right)^2 \\ &= C\left(\sin^2(\alpha) x'^2 + 2\sin(\alpha)\cos(\alpha) x' y' + \cos^2(\alpha) y'^2\right) \\ D x &= D\left(\cos(\alpha) x' - \sin(\alpha) y'\right) \\ E y &= E\left(\sin(\alpha) x' + \cos(\alpha) y'\right) \end{align*}$$
I am stuck when they converted from the above equation to $$A' x'^2 + B' x' y' + C' y'^2 + D' x' + E' y' + F' = 0$$
I want to know how they found the value of $ A', B', C'$ etc. as shown below: $$\begin{align*} A' &= A\cos^2\theta + B\cos\theta\sin\theta+ C\sin^2\theta \\ B' &= B\left(\cos^2\theta - \sin^2\theta\right) + 2\left(C - A\right)\sin\theta\cos\theta\\ C' &= A\sin^2\theta - B\sin\theta\cos\theta + C\cos^2\theta \\ D' &= D\cos\theta + E\sin\theta \\ E' &= -D\sin\theta + E\cos\theta \\ F' &= F \end{align*}$$
I tried using double angle formulas but wasn't able to derive the value in the image and I need your help. Any of $A', B', C'$ would do. I will do the rest myself. Just need direction.
I believe that I'm missing some information to provide so please let me know.
Just expand and collect the expressions right above. For example, focusing on the $\,\color{red}{{x'}^2}\,$ terms:
$$ \begin{align} A x^2 + B x y + C y^2 + D x + E y + F &= \color{red}{A}\left(\color{red}{\cos^2(\alpha) x'^2} -2\cos(\alpha)\sin(\alpha) x' y' + \sin^2(\alpha) y'^2\right) \\ &\;\; + \color{red}{B}\left(\color{red}{\sin(\alpha)\cos(\alpha)}\left(\color{red}{x'^2} - y'^2\right) + \left(\cos^2(\alpha) - \sin^2(\alpha)\right) x' y'\right) \\ &\;\; + \color{red}{C}\left(\color{red}{\sin^2(\alpha) x'^2} + 2\sin(\alpha)\cos(\alpha) x' y' + \cos^2(\alpha) y'^2\right) \\ &\;\;+ \ldots \\ &= \underbrace{\color{red}{\left(A \cos^2(\alpha) + B \sin(\alpha)\cos(\alpha) + C \sin^2(\alpha)\right)}}_{\color{red}{\Large{A'}}}\color{red}{{x'}^2} + \ldots \end{align} $$