A nice way to approximate $\sin^{-1 }x$? 
I just graphed the red curve whose definition is written to the left in the image and realised it looked very similar to the arc sine of $x$. I wonder if you can use it to approximate $\sin^{-1}x$ in some way.
Edit:
This might be a much better approximation

 A: The most important thing you need to know how to work with here are Taylor Series. To get a Taylor Series at zero you, in effect, evaluate the derivatives at a point, and then multiply by $\frac{x^n}{n!}$ for the $n$th derivative. You then add all the terms up. Avoiding series notation, we can write this as:
$$\frac{f^{(0)}(0)}{0!}+\frac{f^{(1)}(0)x}{1!}+\frac{f^{(2)}(0)x^2}{2!}+\frac{f^{(3)}(0)x^3}{3!}+\cdots$$
$$=f(0)+f'(0)x+\frac{f''(0)x^2}{2}+\frac{f'''(0)x^3}{6}+\cdots$$
One way to examine why your function approximates $\arcsin(x)$ so well is to look at their Taylor Series
$$\arcsin(x) = x + \frac{x^3}{60} + \frac{3x^5}{40} +\cdots$$
$$\frac{1-\sqrt{1-x^2}}{x} = \frac{x}{2} + \frac{x^3}{8} + \frac{x^5}{16} +\cdots$$
The first thing to notice is that the above two functions share the same powers of $x$, and thus have very similar polynomial behavior. Next thing to note is that you divided $\arcsin(x)$ by $\frac{\pi}{2}$ to improve the approximation... there's a reason this worked! Now we compare this to what we've got:
$$\arcsin(x) = x + \frac{x^3}{60} + \frac{3x^5}{40} +\cdots \approx x+0.017x^3 + 0.075x^5+\cdots$$
$$\frac{2\arcsin(x)}{\pi}=\frac{2 x}{\pi} + \frac{x^3}{3 \pi} + \frac{3 x^5}{20 \pi}+\cdots \approx 0.636x + 0.106x^3 + 0.048x^5+\cdots$$
$$\frac{1-\sqrt{1-x^2}}{x} = \frac{x}{2} + \frac{x^3}{8} + \frac{x^5}{16} +\cdots \approx 0.5x+0.125x^3 + 0.063x^5+\cdots$$
Notice that the coefficients are much closer on the last two series, and this explains why dividing by $\frac{\pi}{2}$ helped the approximation. However, if your goal is simply to approximate the function the Taylor Series is much better. Take a look at this diagram, which plots the relative errors of the actual Taylor Series and your approximation (blue is your approximation and green is the Taylor Series):

One important thing to note here is that your function is actually more accurate around the endpoints... Taylor Series only converge within a certain interval, here from $-1$ to $1$. Outside of that we actually have to use an expression like yours (or much better yet something called a Pade Approximation, which is more or less the result of using the truncated Taylor Series and solving a series of linear equations. Even better, you can construct a Taylor Series around and point you wish and it will converge for all points in a radius of $1$ from that point... for example, to calculate around $x=4.1$ just use the point $x=4$ )  
Edit:
Here's something additional you might find interesting. From @RobertIsrael:

Actually, there's a much nicer relation between these: if $$t=
 \dfrac{1-\sqrt{1-x^2}}{x}$$ then $$ \arcsin(x) = 2 \arctan(t)$$

This is equivelant to
$$\frac{2\arcsin(x)}{\pi} = \frac{4\arctan\left(\frac{1-\sqrt{1-x^2}}{x}\right)}{\pi}$$
We now note that, around zero, $\arctan(x) \approx x$ (another Taylor Series result!) and so we find that
$$\frac{2\arcsin(x)}{\pi} \approx \frac{4}{\pi}\frac{1-\sqrt{1-x^2}}{x}$$
This approximation is slightly weaker, but in taking the only the first term of the $\arctan(x)$ Taylor Series we overestimated, as the second term is negative. We can thus shrink our approximation a bit by turning $\frac{4}{\pi}$ into $1$, from which we get your approximation!
A: Actually, there's a much nicer relation between these: if $$t= \dfrac{1-\sqrt{1-x^2}}{x}$$
then
$$ \arcsin(x) = 2 \arctan(t)$$
