prove $\frac{|\arcsin(\sin(\pi x))|}{\pi} = |x - \operatorname{round}(x)|$ for all real x

Question

I have looked at this on my graphing calculator but I want a mathematical proof. can somebody please give me mathematical proof or a counterexample.

Answer 1

Let's study the problem of determining $\arcsin\sin t$. There are two cases ($k$ denotes an integer):

1. if $-\pi/2+2k\pi\le t\le \pi/2+2k\pi$, then $t-2k\pi\in[-\pi/2,\pi/2]$ and therefore $\arcsin\sin t=t-2k\pi$;

2. if $\pi/2+2k\pi\le t\le 3\pi/2+2k\pi$, then $-\pi/2+2k\pi\le t-\pi\le \pi/2+2k\pi$ and we're in the same case as before, so $\arcsin\sin t=t-\pi-2k\pi$.

For $t=\pi x$, the conditions read

1. if $-1/2+2k\le x\le 1/2+2k$, then $\dfrac{\arcsin\sin(\pi x)}{\pi}=x-2k$

2. if $1/2+2k\le x\le 3/2+2k$, then $\dfrac{\arcsin\sin(\pi x)}{\pi}=x-2k-1$

Set $y=x+1/2$. Then we're in case 1 when $2k\le y\le 2k+1$, in case 2 when $1+2k\le y\le 2k+2$. We can remove the upper extreme in both cases, because the final value would be the same for numbers belonging to both classes of intervals.

Therefore we can say that $$ \frac{\arcsin\sin(\pi x)}{\pi}=x-\left\lfloor x+\frac{1}{2}\right\rfloor $$ Now you should be able to conclude.

Answer 2

Solution of the injective part of the $sin(x)$

Notice that the $\arcsin(\sin(\pi(x))) = \pi x$ exactly when $\frac{-\pi}{2} < \pi x <\frac{\pi}{2} \implies -\frac{1}{2} < x < \frac{1}{2}$ . Maybe you would expect that $\arcsin(\sin(\pi(x))) = \pi x$ for all $x$, but since the $\sin$ function lacks the property of injectivity, the identity does not hold everyhere. Now, the problem simplies to

\begin{equation} \dfrac{|\pi x|}{\pi} = |x| = |x-\textrm{round}(x)| \quad \textrm{for} \quad -\frac{1}{2} < x < \frac{1}{2} \end{equation} since $|x| < \frac{1}{2}$, we have that round$(x) = 0$. Hence, the problem simplifies to \begin{equation} |x| = |x|, \quad \textrm{for} \quad -\frac{1}{2} < x < \frac{1}{2} \end{equation} Hence, $x$ is a solution as long as it satisfies $-\frac{1}{2} < x < \frac{1}{2}$. For all other $x$, I do not think an algebraic proof is easy to make.