Strange approximation of $\pi$?

I was playing with my calculator (Casio fx-991MS) the other day.
I input $$\arcsin(\sin(2))$$
The result came out as $$1.141592653\ldots$$
I immediately noticed that the digits seem to resemble $\pi$.
So, I decided to investigate further.
I input $$\arcsin(\sin(2))+2-\pi$$
I was expecting some small number to pop out but instead I got $0$.
I know my calculator has precision only upto $10$ digits.
So, naturally, I decided to put the number on another (albiet higher precision) calculator. This time,
I got a number in the range of $10^{-19}$.
I also put it through WA which surprisingly gave $0$ as answer.
Finally,I put the result on an online ultra high precision calculator.
It was even more crazy.
I got a mind bogglingly small number in the range $10^{-600}$
My question is, what the hell is happening here!?

• Also mention that you are working in radians. But it is not surprising as $$\arcsin(\sin(2))= \pi -2$$ – Chinny84 Feb 18 '15 at 9:18
• @Chinny84 I would have put the degrees sign ($^{\circ}$) if I wasn't – AvZ Feb 18 '15 at 9:20
• That maybe true, but i never underestimate if someone was lazy enough not to (ie me from time to time) but I like the question anyway :). – Chinny84 Feb 18 '15 at 9:22

The identity $$\sin(\pi-x)=\sin x$$ holds for all $x$, in particular for $x=2$. And 2 radians is an angle between $\pi/2$ and $\pi$, so when you take the arcsine you get back the angle between $0$ and $\pi/2$ which has the same sine, in other words $\pi-2$.

• Who is right here? Casio, WA or Ultra High Precision calculator? – AvZ Feb 18 '15 at 9:16
• $\arcsin(\sin 2)+2-\pi$ is exactly zero. (That $10^{-600}$ is just a numerical fluke, since the precision is only "ultra high", not infinite!) – Hans Lundmark Feb 18 '15 at 9:18
• Okay. Thanks (+1) – AvZ Feb 18 '15 at 9:19

$\arcsin(x)$ is an inverse of $\sin(x)$ on the range $[-\pi/2,\pi/2]$. To evaluate $\arcsin(\sin(2))$, we can use symmetries of $\sin$ to get it in this range:

$$\sin(x)=\sin(\pi-x)\implies\arcsin(\sin(2))=\arcsin(\sin(\pi-2))$$

And since $\pi-2$ is in this range (since $\pi>4/3$ and $\pi<4$), we get

$$\arcsin(\sin(2))=\arcsin(\sin(\pi-2))=\pi-2.$$