What amplification can I apply to $y=\sin x$ for it to be a perfect oscillating arc?

Question

A perfect arc is $y=\sqrt{|1-(x-1)^2|}$. A sin wave is $y=\sin({\pi x\over2})$

I am curious how I can amplify the sin wave so that it's a perfect alternating arc.

In the link below you can see graphed each of these two equations, as well as the delta between them. So in the sage code there's function $i(x) = $SOMETHING $\times f(x)$, where $f(x) = \sin({\pi x\over 2})$, but I want to replace the 'SOMETHING' with an expression that transforms the wave to $$y=\sqrt{|1-(x-1)^2|}+\sqrt{|1-(x-3)^2|}+\sqrt{|1-(x-5)^2|}... + \sqrt{|1-(x-n)^2|}$$

In other words, I want $g(x):g(x)\sin({\pi x\over 2}) = \sqrt{|(1-(x-1)^2|}$ where $0<x<2$

I have no particular goal, I am just curious how I would go about solving that as currently I'm a bit stumped...

sage math link to the various graphs

thank you kind internet maths wizards!

Answer 1

$\DeclareMathOperator{sgn}{sgn}$If $\sgn$ denotes the signum function $$ \sgn(x) = \begin{cases} \frac{x}{|x|} & x \neq 0, \\ 0 & x = 0, \end{cases} $$ the graph $$ y = \sgn(u)\sqrt{|u| \cdot \bigl(\pi - |u|\bigr)},\qquad u = \arcsin(\sin x) $$ (blue; $y = \sin x$ is black) seems to be what you're seeking:

Answer 2

$$y=\sqrt{\left({\pi\over a}\right)^2-\left({\pi\over a}-|x-{2\pi\over a}|\right)^2}$$ conincides with $\sin({ax\over 2})$ at its zeros and gives you recurring semi-circles for $0\le x\le{4\pi\over a}$. In case $a=\pi$ the graphs are