# Ask Uncle Colin: An Arcsine Inversion

Dear Uncle Colin,

I’m trying to invert $y = \arcsin\br{x^2 + 4x - 5}$, but I get stuck at $x^2 + 4x = \sin(y) + 5$. Any pointers?

- All Routine Calculations Should Invert Naturally

Hi, ARCSIN, and thanks for your message!

Just before I dive into this, I have an itch I need to deal with: the domain of this (original) function is a bit strange, and if we want to invert it, we’ll have to limit it further.

### Finding the domain

So where is $\arcsin \br{x^2 + 4x - 5}$ defined? Well, $\arcsin$ only takes values between -1 and 1, so we need to solve $-1 \le x^2 + 4x - 5 \le 1$.

I’d complete the square and rewrite that as $8 \le (x+2)^2 \le 10$, so we have two distinct intervals: either $-2 - \sqrt{10} \le x \le -2 - 2\sqrt{2}$ or $-2 + 2\sqrt{2} \le x \le -2 + \sqrt{10}$.

The trouble is, those two intervals are each responsible for outputs everywhere from -1 to 1 – which means the function is not one-to-one and doesn’t permit an inverse as it stands. Let’s restrict it instead to the second of those domains, and say that our function is $f(x) = \arcsin \br{ x^2 + 4x - 5}$ for $-2+2\sqrt{2} \le x \le -2 + \sqrt{10}$.

### Finding the inverse

This has a lot in common, it turns out.

If we complete the square again, we get $(x+2)^2 - 9 = \sin(y)$, which we can rearrange to get $x = -2 \pm \sqrt{9+\sin(y)}$ - and the positive square root is the one we want.

Hope that helps!

- Uncle Colin