5
$\begingroup$

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!?

$\endgroup$
  • 2
    $\begingroup$ Also mention that you are working in radians. But it is not surprising as $$\arcsin(\sin(2))= \pi -2$$ $\endgroup$ – Chinny84 Feb 18 '15 at 9:18
  • $\begingroup$ @Chinny84 I would have put the degrees sign ($^{\circ}$) if I wasn't $\endgroup$ – AvZ Feb 18 '15 at 9:20
  • $\begingroup$ 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 :). $\endgroup$ – Chinny84 Feb 18 '15 at 9:22
10
$\begingroup$

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 $\pi-2$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Who is right here? Casio, WA or Ultra High Precision calculator? $\endgroup$ – AvZ Feb 18 '15 at 9:16
  • 3
    $\begingroup$ $\arcsin(\sin 2)+2-\pi$ is exactly zero. (That $10^{-600}$ is just a numerical fluke, since the precision is only "ultra high", not infinite!) $\endgroup$ – Hans Lundmark Feb 18 '15 at 9:18
  • $\begingroup$ Okay. Thanks (+1) $\endgroup$ – AvZ Feb 18 '15 at 9:19
5
$\begingroup$

$\arcsin(x)$ 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

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.