$|\sin(\sin( \cdots \sin(x)\cdots))|$ ($N$ times) is always $\leq|\sin( \cdots \sin(1)\cdots)|$ ($N-1$ times)

Question

Is this inequality always true? $$ \bigl\lvert\,\underbrace{\sin(\sin(\cdots \sin}_{N\text{ times}}(x)\cdots))\bigr\rvert\le\bigl\lvert\,\underbrace{\sin(\sin( \cdots \sin}_{N-1\text{ times}}(1)\cdots))\bigr\rvert $$

I think so, for example for $N=2$ we have:

$|\sin(x)|\leq 1$ therefore $|\sin(\sin(x))|\leq |\sin(1)|$

Answer 1

Sorry, but I remember you that the $\sin$ function is strictly increasing only in $Z_1=\displaystyle\bigcup_{k\in\mathbb{Z}}\left[-\frac{\pi}{2}+2k\pi,\frac{\pi}{2}+2k\pi\right]$ and therefore: \begin{equation} \forall x\in Z_1,\,\sin(\sin x)\leq\sin(1),\,1>\sin 1>0; \end{equation} while in $Z_2=\displaystyle\bigcup_{k\in\mathbb{Z}}\left[\frac{\pi}{2}+2k\pi,\frac{3\pi}{2}+2k\pi\right]=Z_1+\pi$ the $\sin$ function is strictly decreasing and therefore: \begin{gather} \forall x\in Z_2,\exists y\in Z_1:x=y+\pi\Rightarrow\\ \Rightarrow\sin(\sin(x))=\sin(\sin(y+\pi))=\sin(-\sin(y))=-\sin(\sin(y))\geq-\sin(1). \end{gather} From all this: \begin{equation} \forall x\in\mathbb{R},\,|\sin(\sin(x))|\leq|\sin1|, \end{equation} and by induction you can prove the claim.

Is it all clear?

Answer 2

First fact to use:

For $x\ge0$, $\sin(x)\le x$

Second fact to use:

$1<\pi/2$

Third fact to use:

The sine function is increasing in the interval $(0,\pi/2)$

Also, it's not restrictive to assume $x\ge0$, because $\sin(-x)=-\sin(x)$ and the minus sign is “eaten” by the absolute value.

For $x\ge0$, we have $0\le \sin(x)\le 1$, which implies the simplest case of your statement, namely that $$ \sin(\sin(x))\le\sin(1) $$ because we are in the interval $(0,\pi/2)$ where the sine is increasing.

For the general case, denote by $\sin^{[n]}$ the “$n$-fold iteration”: $$ \sin^{[1]}(x)=\sin(x),\qquad \sin^{[n+1]}(x)=\sin(\sin^{[n]}(x)) $$ so the statement above can be written $$ \sin^{[2]}(x)\le\sin^{[1]}(1) $$ which we can now use for a proof by induction of your general statement, namely that $$ \sin^{[n+1]}(x)\le\sin^{[n]}(1) $$ for $n\ge1$.

First a side step: $$ 0\le\sin^{[n]}(x)\le 1 $$ for $n\ge1$. Indeed, this is true for $n=1$. Suppose it is for some $n\ge1$; then $$ \sin^{[n+1]}(x)=\sin(\sin^{[n]}(x))\le \sin^{[n]}(x) $$ from the first fact. By the induction hypothesis, $\sin^{[n]}(x)\le 1$.

Now, suppose we know that, for some $n\ge1$, $$ \sin^{[n+1]}(x)\le\sin^{[n]}(1) $$ Then $$ \sin^{[n+2]}(x)=\sin(\sin^{[n+1]}(x)) \le \sin(\sin^{[n]}(1) = \sin^{[n+1]}(1) $$ where we use that $\sin^{[n]}(1)\le1$, as proved in the side step.