Approximation of $\arcsin(\sqrt{1-x^2})$ by $2\arcsin(\sqrt{\frac{1-x}2})$

Question

I stumbled upon an use of $2\arcsin(\sqrt{\frac{1-x}2})$ to approximate $\arcsin(\sqrt{1-x^2})$. The latter is equal to $\frac\pi 2-\arcsin(x)$ when $0\le x\le 1$ and is useful if you have a good approximation of $\arcsin$ around $0$ and want to use it for values near $1$. On the flip side the starting approximation needs to cover $[-\frac1{\sqrt2},\frac1{\sqrt2}]$ in order to use the previous fact to approximate $\arcsin$ on the entire domain.

However if we use the fact that $\arcsin(\sqrt{1-x^2})\approx2\arcsin(\sqrt{\frac{1-x}2})$, our starting approximation only needs to cover $[-0.5, 0.5]$. Empirically this approximation is extremely good, but I'm wondering how would one come up with this kind of approximation and are there any insights to explain why the two are so close?

Answer 1

Check the following ilustration:

$$ \begin{align} \theta&=\sin^{-1}{\sqrt{1-x^{2}}}\\ \\ \frac{\theta}{2}&=\sin^{-1}{\frac{\sqrt{1-x^{2}}}{\sqrt{2+2x}}}\\ &=\sin^{-1}{\sqrt{\frac{1-x}{2}}} \end{align} $$

Answer 2

Let $\theta = \arcsin \sqrt{1-x^2}$. Since the argument of $\arcsin$ is non-negative, consider the range $0\le \theta\le \frac \pi2$.

$$\begin{align*} \sin\theta &= \sqrt{1 - x^2}\\ \sin^2\theta &= 1 - x^2\\ x &= \cos \theta\\ \end{align*}$$

Then using half-angle formula,

$$\begin{align*} 1-2\sin^2\frac\theta2&= \cos\theta\\ \sin^2\frac\theta2 &= \frac{1-\cos \theta}2\\ \sin\frac\theta 2&= \sqrt{\frac{1-\cos \theta}2}\\ &= \sqrt{\frac{1-x}2}\\ \frac\theta2 &= \arcsin \sqrt{\frac{1-x}2}\\ \theta &= 2\arcsin \sqrt{\frac{1-x}2} \end{align*}$$