Inverse of Bhāskara I's sine approximation

Question

Bhāskara I's approximation of $\sin x$, for $x$ in radians, is

$$\sin x = \frac{16x\left(\frac{\tau}{2}-x\right)}{5\left(\frac{\tau}{2}\right)^2-4x\left(\frac{\tau}{2}-x\right)}$$

where $\tau := 2\pi$.

I now need the inverse of this formula in order to approximate the inverse sine. (I'd figure out how to "crop" it once I can visualise it in Desmos.) Wolfram Alpha suggests

$$\arcsin x = \frac{x}{4}+\frac{\sqrt{\left(\tau^2+3\tau-4\right)\left(-x^2\right)}}{2\left(\tau+4\right)}$$

Yet Desmos fails to show any points (except possibly one at the origin). See https://www.desmos.com/calculator/w0us6zngaq.

Thanks in advance!

Answer 1

You (or WolframAlpha) appear to have swapped the $\tau$s and $x$s in the inverse. (Also, you really want the negative square root.) The inverse is actually $$\frac{\tau}{4}-\frac{\sqrt{-\tau^2 (x^2+3x-4)}}{2(x+4)}$$ which you can write as $$\frac{\tau}{4}\left(1-2\sqrt{\frac{1 - x}{4 + x}}\right)$$

I'll let you verify this in Desmos.