How to simplify $y = \sin(\frac{\arcsin(x)}{n}), n≥1$? $y = \sin(\frac{\arcsin(x)}{n}), n≥1$
I know that:
$\lim \limits_{x \to 0} \frac{x}{y} = n$
But I can't figure out what the curve of $x/y$ practically represents. Is there an obvious simple solution?

Let $\theta = \arcsin x$ and we have to assume that $|x| \lt \dfrac{1}{\sqrt 2}$.
Then $\tan \theta = \dfrac{x}{\sqrt{1-x^2}}$ and $|\tan \theta| \lt 1$.
\begin{align}
    \cos \dfrac{\theta}{n} + i \sin \dfrac{\theta}{n}
    &= (\cos \theta + i \sin \theta)^{\frac 1n} \\
    &= (\cos \theta)^{\frac 1n}(1 + i \tan \theta)^{\frac 1n} \\
    &= x^{\frac 1n}\left(1 + i\dfrac{x}{\sqrt{1-x^2}}\right)^\frac 1n\\
    &= x^{\frac 1n}\sum_{k=0}^\infty
       \binom{1/n}{k} i^k\left(\dfrac{x}{\sqrt{1-x^2}}\right)^k \\
    \cos \dfrac{\theta}{n}
    &= x^{\frac 1n}\sum_{k=0}^\infty
       (-1)^k\binom{1/n}{2k}\left(\dfrac{x^2}{1-x^2}\right)^k \\
    \sin\dfrac{\theta}{n}
    &= x^{\frac 1n} \dfrac{x}{\sqrt{1-x^2}} \sum_{k=0}^\infty
       (-1)^k\binom{1/n}{2k+1}\left(\dfrac{x^2}{1-x^2}\right)^k \\
\end{align}
NOTES

We define $\binom zn$ where $z \in \mathbb R$ and $0 \le n \in \mathbb Z$ as follows
$(z)_n =
\begin{cases}
    1 & \text{If $n = 0$.}\\
    z(z-1)(z-2)\cdots(z-n+1) &\text{If $n \ge 1$.}
\end{cases}$
then $\binom zn = \dfrac{(z)_n}{n!}$
It can be shown that, if $|x| < 1$, then $(1 + x)^z = \sum_{k=0}^\infty \binom zk x^k$
