Express $y$ through $R$ and $x$.

Question

How can I express $y$ through $R$ and $x$? $$y=R \pm \sqrt{R^2-\bigg(R\arccos \left(\mp1 \pm \frac{y}{R} \right) -x \bigg)^2}$$

It can be transformed into this : $$y=\pm R \cos\left( \frac{\sqrt{2yR-y^2}+x}{R}\right)-R$$

Answer 1

You can't, because $R$ and $x$ don't determine $y$. For example, consider $R=1, x=0$, and if you plot the function $\cos(\sqrt{2y-y^2})-1-y$ you can see it has two solutions.

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Edit: the second equation doesn't follow from the first. The plot of the first is this and it has two real solutions: maybe adding some conditions can help finding a closed formula for $y$. The correct second equation is $y = R \pm R \cos(\frac{x \pm \sqrt{-y^2+2Ry}}{R})$, with the two $\pm$ independent from each other, the internal one coming from resolving a second degree equation.