$x=\frac{1}{2}a(1-\cos(\alpha))$ substition in Bernoullie's brachistochrone

Question

When Bernoulli in brachistochrone calculus got to step $$y=a \sin^{-1} \left(\sqrt{\frac{x}{a}}\right)-\sqrt{ax-x^2}$$ he substituted $$x=\dfrac{1}{2}a(1-\cos(\alpha))$$ and got $$y=\frac{1}{2}a\alpha-\frac{1}{2}a \sin(\alpha)$$ Where did this substitution come from?

Answer 1

In order to get rid of the $sin^{-1}(\sqrt {...} )$ one may need $sin^2$ for their parameter changing.

We can say:

$\sin^2(\alpha/2)=\dfrac{1+cos(2\alpha)}{2}=x/a \to x=a\sin^2(\alpha/2)=a \cdot \dfrac{1+cos(2\alpha)}{2}$ $\to x=\dfrac{1}{2}a(1-cos(\alpha))$

In other words:

$x=\dfrac{1}{2}a(1-cos(\alpha))=a\sin^2(\alpha/2) \to y=a \cdot \alpha/2 -asin(\alpha)/2$