Find the domain of $f(x)=\sqrt{\sin(\sin(\sin x))}$

Question

Find the domain of $f(x)=\sqrt{\sin(\sin(\sin x))}$

My Attempt:

$\sin(\sin(\sin x))\ge0$

$\sin(\sin x)\in[0,1]$

$\sin x\in[0,\arcsin1]$

$x\in[0,\arcsin(\arcsin1)]$

Is this correct?

Wolfram doesn't seem to agree.

Answer 1

We need

• $\sin(\sin(\sin x))\ge0$

which requires

• $2k\pi \le \sin(\sin x)\le \pi+2k\pi \iff 0 \le \sin(\sin x)\le 1$

which requires

• $2k\pi \le\sin x\le \pi+2k\pi \iff 0\le \sin x \le 1$

which requires

• $2k\pi \le x\le \pi+2k\pi$

Answer 2

Case 1: $$ \begin{align} \sin(x) \ge 0 &\implies 0 \le \sin(x) \le 1 \\ &\implies 0 \le \sin(\sin(x)) \le 1\\ &\implies 0 \le \sin(\sin(\sin(x))) \le 1 \end{align} $$ Case 2: $$ \begin{align} \sin(x) < 0 &\implies -1 \le \sin(x) < 0 \\ &\implies -1 \le \sin(\sin(x)) \le < 0 \\ &\implies -1\le \sin(\sin(\sin(x))) < 0 \end{align} $$

This shows that $\sin(\sin(\sin(x)))\ge 0$ if and only if $\sin(x) \ge 0$, that is if $2k \pi \le x \le 2(k+1)\pi$ for some integer $k$.