# Ask Uncle Colin: A Misbehaving Inverse

Dear Uncle Colin,

Why is $\arcsin\br{\sin\br{\frac {6}{7}\pi}}$ not $\frac{6}{7}\pi$?

- A Reasonable Conclusion Seems Incorrect Numerically

Hi, ARCSIN, and thanks for your message!

On the face of it, it does seem like a reasonable conclusion: surely feeding the output of $\sin(x)$ into its inverse function should get you back where you started?

There’s a wrinkle, though: $\arcsin(x)$ is not exactly the inverse function of $\sin(x)$.

### Defining a function

A function is typically defined in three parts: it has:

- A name - here, $\sin(x)$
- A definition - for example, $\lim_{n\to \infty} \sum_{i=0}^{n} \frac{(-1)^n}{(2n+1)!}x^{2n+1}$, or $\myim \br{e^{ix}}$, or the $y$-coordinate of the point of intersection between the unit circle and a line drawn at an angle $x$ anti-clockwise from the $x$-axis. (All of these are reasonable definitions for $\sin(x)$, incidentally).
- A domain - the values you can put into the function. Implicitly, with $\sin(x)$, this is “any real number”.

But there’s a problem.

Not every function has an inverse. In particular, if you try to take the inverse of any function that’s *many-to-one* - that is, it gives the same output for several different inputs - you suddenly get several possible outputs for some inputs. And that’s bad news: your “inverse function” isn’t really a function!

### So what is $\arcsin$?

Speaking loosely, $\arcsin(x)$ is the inverse function of $\sin(x)$. Speaking strictly, though, it’s the inverse function of $\sin(x)$ *defined only on the domain $-\piby 2 \le x \le \piby 2$*.

By restricting the domain of $\sin(x)$, we end up with a one-to-one function - no output corresponds to more than one input. $\arcsin(x)$ is the inverse of this *restricted* version of $\sin(x)$.

### Back to the question

So how does this all fit together? The trouble comes because $\frac{6}{7}\pi$ is not in the restricted domain of $\sin(x)$. If you draw the graph of $\sin(x)$, you’ll see that $\sin\br{\frac{6}{7}\pi}=\sin\br{\frac{1}{7}\pi}$, and $\frac{1}{7}\pi$ *is* in the restricted domain.

*That’s* what comes out of the $\arcsin$ function here: $\frac{1}{7}\pi$.

Hope that helps!

- Uncle Colin