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Forgive this seemingly basic question; I recently found out about cycloids and cannot find any answers on the web. My guess is that it’s not, due to some part of the definition of a cycloid, but I can’t really find something about cycloids which explicitly helps me answer my question/I am not rounded enough to decipher the necessary information.

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  • $\begingroup$ It is not. ${}{}{}{}{}$ $\endgroup$ – Seewoo Lee Feb 23 at 23:29
  • $\begingroup$ The derivative of the cycloid can take any real value, while the derivative of $| \sin x |$ is never outside the interval $[-1,1]$ when it exists. You will notice the difference in shape where $y=0$ $\endgroup$ – Henry Feb 23 at 23:39
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One possible explanation is using the parametrization of a cycloid. It is given by $$ x = r(t-\sin t)\\ y = r(1-\cos t) $$ for some $r>0$. If $y = |\sin x|$ is a cycloid, or more generally, homothetic to cycloid, then it will be possible to find some constant $a\in \mathbb{R}$ such that $$ |\sin(r(t-\sin t))| = ar(1-\cos t) $$ for all $t\in \mathbb{R}$. However, it is not true.

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