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One way to do it is just to plot the function, draw its projection about $y=x$ and find the intersections. But I've been trying to solve it more algebraically:

$$ \begin{align} f(x) & = f^{-1}(x) \\ f(f(x)) & = x \\ \sin(\sin(x)-x)-\sin(x) + x & = x \\ \sin(\sin(x)-x) & = \sin(x) \end{align} $$ Now, it is possible to just look at the equation here and see that the solution would be $x=n\pi$ where $n$ can be any integer, but that's not really rigorous, and doesn't prove that there is no other solution.

How should I proceed here?

kipf
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brainfreeze
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2 Answers2

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Not only looking, but solving it can also give the result. $$\sin(\sin x)\cos x-\sin x\cos(\sin x)=\sin x$$ $$\implies \sin(\sin x)\cot x-\cos(\sin x)=1$$ Let $\sin x=t$. $$\cos x=\frac{1+\cos t}{\sin t}=\frac{2\cos^2\frac{t}{2}}{2\sin\frac{t}{2}\cos\frac{t}{2}}$$ $$\implies \cot x=\cot \frac{t}{2}$$ $$\implies x=nπ+\frac{t}{2}=nπ+\frac{\sin x}{2}$$

This can only be true $\forall x=nπ,n\in \mathbb{Z}$

Gwen
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  • Even simpler, $\sin(\sin(x)-x) = \sin(x)$ gives $\sin(x)-x=2k\pi +x$ or $\sin(x)-x=(2k+1)\pi -x$. – Mostafa Ayaz Apr 25 '24 at 21:16
  • What? It's like that? Ohh I'm sorry I though the general formula of sin had some (-1)^n somthing something. I didn't remember it well hence I did all this :") – Gwen Apr 26 '24 at 02:23
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Looking at the graph it seems that $\sin(\sin(x)-x)>0$ and $\sin(x)<0$ if $x\in((2k-1)\pi,2k\pi)$, $k\in\mathbb{Z}$, and $\sin(\sin(x)-x)<0$ and $\sin(x)>0$ if $x\in (2k\pi,(2k+1)\pi)$, $k\in\mathbb{Z}$, with equality if and only if $x=k\pi$, $k\in\mathbb{Z}$.

And indeed, if $x\in((2k-1)\pi,2k\pi)$, $k\in\mathbb{Z}$, then $\sin(x)<0$ and $\sin(x)-x$ is between $-(2k-1)\pi$ and $-2k\pi$ (which follows from the fact that $\sin(x)-x$ is monotonically decreasing), which implies that $\sin(\sin(x)-x)>0$.

Analogously for the other case.

mathemagician99
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