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?
 A: As mentioned in the comments, there is no way to simplify while maintaining the accurate relation. If approximately there could be convenient simplifications. First let me clarify the question a bit. Suppose $\arcsin(x) = \theta$, then $\sin(\theta) = x$, so what you try to achieve is if $\sin(\theta)$ is known, what's $\sin(\theta/n)$.
A straight forward attempt is do a Taylor expansion of $\arcsin(x)$, for large $n$ the sine function can be expanded around $0$ as well. If two terms are kept of the arcsine expansion:
$$
\arcsin(x) = x + \frac{x^3}{6}+O(x^5)
$$
Then
$$
y = \sin \left(\frac{1}{n} (x + \frac{x^3}{6}) \right) \approx \frac{1}{n} (x + \frac{x^3}{6})
$$
Below is a figure (red is approximation above, blue is the accurate value), the approximation works well up to $x = 0.8$ for $n \geq 2$.

A: 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 $(1 + x)^z = \sum_{k=0}^\infty \binom zk x^k$
A: THIS ANSWER IS ADDRESSED TO BEGINNERS.
In the figure below you have an approximate representation of $\sin (x)$ and  $\sin (\frac  xn)$. 
Your problem is to calculate the length of the segment $\overline {CD}$. Since you have $n( \frac xn)=x$, you can first find $\sin (nX)$ and then replace $X$ by $\frac xn$.
 The expression of $\sin (\frac  xn)$  for each value of $n$, given in function of $\sin( x)$ is more and more complicated as it grows $n$. For example, in
$$\sin (3x)=4\cos^2(x)\sin (x)-\sin (x)=3\sin (x)-4\sin^3(x)$$
replacing $x$ by $\frac x3$ you get $$\sin(x)=3\sin(\frac x3)-4\sin^3(\frac x3)$$ so you have the cubic equation to solve $$4t^3-3t+a=0$$ where $a=\sin (x)$.
Similarly, for $\sin(\frac x5)$, you have to solve the quintic
$$16t^5-20t^3+5t-a=0$$ deduced of the formula $\sin (5x)=16\cos^4(x)\sin(x)-12\cos^2(x)\sin (x)+\sin (x)$.
In general for $\sin (\frac xn)$ you have to solve an equation of degree $n$.

