I'm supposed to find the value of the expression $$4x^2 - 4xy\cos\alpha + y^2$$
Given, $$\cos^{-1}(x) - \cos^{-1}(\frac{y}{2}) = \alpha$$
Using some trigonometric manipulation I've arrived at the equation $$4x^2(y\sqrt{1-x^2}) - 4xy\cos\alpha + y^2(\sqrt{4-2y^2}) = 0$$ which looks oddly similar to the the expression which I need to find. Is there any method to go from the equation to that expression?
Adding to this, is there any method to go from $Ax^{n}\alpha + By^{m}\beta + Cxy\gamma + O(D,F,..,\delta,\varepsilon,...) = 0$ to $Ax^n + By^n + Cxy + O(D,F,..) = k$
I'm not sure how you got your equation, but here's a way to solve the problem using trigonometric manipulation. Let
$$\begin{equation}\begin{aligned} & a = \cos^{-1}(x) \;\;\to\;\; x = \cos(a), \; \sin(a) = \pm\sqrt{1-x^2} \\ & b = \cos^{-1}\left(\frac{y}{2}\right) \;\;\to\;\; \frac{y}{2} = \cos(b), \; \sin(b) = \pm\sqrt{1-\frac{y^2}{4}} \end{aligned}\end{equation}$$
Next, from the list of angle sum and difference identities, using
$$\cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b)$$
we get from the given equation by taking $\cos()$ of both sides that
$$\begin{equation}\begin{aligned} \cos(\alpha) & = x\left(\frac{y}{2}\right) \pm\sqrt{1-x^2}\sqrt{1-\frac{y^2}{4}} \\ 2\cos(\alpha) - xy & = \pm\sqrt{1-x^2}\sqrt{4-y^2} \\ (2\cos(\alpha) - xy)^2 & = (1 - x^2)(4 - y^2) \\ 4\cos^2(\alpha) - 4xy\cos(\alpha) + x^2y^2 & = 4 - y^2 - 4x^2 + x^2y^2 \\ 4x^2 - 4xy\cos(\alpha) + y^2 & = 4 - 4\cos^2(\alpha) \\ 4x^2 - 4xy\cos(\alpha) + y^2 & = 4\sin^2(\alpha) \end{aligned}\end{equation}$$