I found this problem in a collection of contest problems of a Russian competition in 1995 and wasn't able to solve it.

Solve for real $x$: $$ \cos (\cos (\cos (\cos(x))))=\sin (\sin (\sin (\sin (x)))) $$

My guess is that there is no solution, but how do I prove it? I tried to estimate

$LHS\ge \cos (1) \ge \cos(\pi/3)=1/2 $

and RHS similarly but the ranges overlap..

Do you have a better idea?

  • $\begingroup$ Of course there is no solution because $ cos (cos (cos (x)))\ge 1/2 $ as explained. What do you mean? $\endgroup$
    – user129798
    Feb 18, 2014 at 21:44
  • $\begingroup$ Plotting both functions with Wolfram Alpha does indeed indicate that the two functions don't overlap: wolframalpha.com/input/… $\endgroup$
    – Arno
    Feb 18, 2014 at 21:46
  • $\begingroup$ Ooops! I deleted the comment because I misunderstood your question. Sorry! $\endgroup$ Feb 18, 2014 at 21:47
  • $\begingroup$ Related: math.stackexchange.com/questions/589663/… $\endgroup$ Feb 25, 2014 at 22:08
  • $\begingroup$ Using MATLAB we can determine the approximate complex solution $x^\star = 0.756888895567887 -0.610156387039054 \cdot i$. If we define $f(x) = |\cos \cos \cos \cos x - \sin \sin \sin \sin x|$ and $f(x^\star) = 9.289 \cdot 10^{-9}$. $\endgroup$ Aug 12, 2022 at 21:38

3 Answers 3


Rephrase the problem as:

$$ \sin( \pi/2 + \sin(\pi/2 + \sin(\pi/2 + \sin(\pi/2 + x) ) ) ) = \sin(\sin(\sin(\sin(x))))$$ Our strategy will be to assume the expression above is true for some $x$, "invert" the sine on both sides, and show that the resulting expression cannot possibly have a solution. Doing this process once, we have

\begin{align} \pi/2 + \sin(\pi/2 + \sin(\pi/2 + \sin(\pi/2 + x) ) ) & = \sin(\sin(\sin(x))) + 2\pi n \\ \textrm{or} & \\ \pi/2 + \sin(\pi/2 + \sin(\pi/2 + \sin(\pi/2 + x) ) ) & = \pi - \sin(\sin(\sin(x))) + 2\pi n \end{align} for some integer $n$, which implies $$ \sin(\pi/2 + \sin(\pi/2 + \sin(\pi/2 + x) ) ) = \pm (\sin(\sin(\sin(x))) + 2 \pi (n - 1/4)),$$ where the $\pm$ covers the two cases above, and indicate that at least one of the two possibilities must hold, but not necessarily both. We must have $n = 0$ since both sine terms must have values in $[-1,1]$. So $$ \sin(\pi/2 + \sin(\pi/2 + \sin(\pi/2 + x) ) ) = \pm (\sin(\sin(\sin(x))) - \pi/2),$$ and so it must be that both sides of the above expression lie in either the interval $[-1, 1 - \pi/2]$ or the interval $[-1 + \pi/2, 1]$. Applying our strategy once more, we have $$\sin(\pi/2 + \sin(\pi/2 + x) ) = \pm (\arcsin( \sin(\sin(\sin(x))) - \pi/2) - \pi/2).$$ On the left side, clearly the expression must be in $[-1,1]$, while on the right side, the expression must be either in the interval $[-\pi, \arcsin(1 - \pi/2) - \pi/2]$ or in the interval $[\pi/2 - \arcsin(1 - \pi/2), \pi]$.

We derive a contradiction by showing that neither of these two intervals intersect with $[-1,1]$. In particular, we show that $\arcsin(1 - \pi/2) - \pi/2 < -1$. Using the fact that $\pi > 3$, $$ \arcsin(1 - \pi/2) - \pi/2 < \arcsin(1 - 3/2) - \pi/2 = -\pi/6 - \pi/2 = - 2\pi/3 < -1.$$

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    $\begingroup$ +1 Wow, the only numerical approximation you use is $\pi>3$ $\endgroup$ Feb 18, 2014 at 23:08
  • $\begingroup$ I know the question was for real x. I am curious if we allow complex x if there are solutions. $\endgroup$ Feb 27, 2014 at 11:48
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    $\begingroup$ @MichaelSmith, It seems likely there is a complex solution, simply because then the equation is equivalent to solving $F(z) = 0$ for an entire function $F$, and Picard's theorem says that the range of such a function can only exclude at most one point in $\mathbb{C}$. It's not a proof, but that's my intuition so far... $\endgroup$ Feb 27, 2014 at 20:36
  • $\begingroup$ Cool use of Picard's Theorem! Makes sense to me. I was thinking of converting the sin() and cos() to expressions in exp()s but it gets pretty complex due to the nesting. $\endgroup$ Feb 28, 2014 at 0:52
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    $\begingroup$ In the second display line, how do you exclude the possibility that $$\pi/2 + \sin(\pi/2 + \sin(\pi/2 + \sin(\pi/2 + x) ) ) =\pi- \sin(\sin(\sin(x))) ?$$ $\endgroup$ Aug 12, 2022 at 21:25

This is not a solution, but here's a pretty compelling picture of the the fourth iterates of cosine and sine (in blue and red, respectively).

It suggests that you can't uniformly bound the two apart from one another. (The functions appear to have slightly overlapping ranges).

enter image description here

  • 2
    $\begingroup$ As I said, I know that picture. But I'm looking for a formal argument. $\endgroup$
    – user129798
    Feb 18, 2014 at 21:49
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    $\begingroup$ You can turn the picture into a formal argument. 1. The functions are $2\pi$-periodic, so it suffices to check on $[-\pi,\pi]$. 2. Clearly one is negative on $[-\pi,0]$ while the other is positive, so it suffices to check on $[0,\pi]$. 3. $\mathrm{cos}(\mathrm{cos}(\mathrm{cos}(\mathrm{cos}(x))))$ is injective on this domain, so it's easy to compute exactly when it dips below $\mathrm{sin}(\mathrm{sin}(\mathrm{sin}(1)))$, which is the maximum of the other function. Then you make sure they don't intersect on this domain. $\endgroup$ Feb 18, 2014 at 22:01
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    $\begingroup$ Probably he wanted to know how this problems was supposed to be solved, since in this kind of contest you don't get pictures or wolframalpha :-) Usually you can't even bring a calculator ;-) $\endgroup$
    – Ant
    Feb 18, 2014 at 23:23

Let $f(x)=\cos(\cos(\cos(\cos x)))-\sin(\sin(\sin(\sin x)))$

For $x\in[\pi,2\pi], f(x)>0$ As cos term positive and sin term negative !

Also $f(x)=f(\pi-x)\implies \text{it is sufficient to solve it for}\,\,x\in[0,\pi/2]$

$\implies \cos x+ \sin x\le \sqrt 2<\frac{\pi}{2}\,\,\,\text{ or }\cos x<\frac{\pi}{2}–\sin x$

$\implies \cos(\cos x)>\sin(\sin x)\implies \cos(\cos(\cos x))<\cos(\sin(\sin x))$

$\implies \cos(\cos(\cos x))+\sin(\sin(\sin x))<\cos(\sin(\sin x)) +\sin(\sin(\sin x)) \le \sqrt2<\frac{\pi}{2}$

$\implies \cos(\cos(\cos x))<\frac{\pi}{2}-\sin(\sin(\sin x))$

$\implies\cos(\cos(\cos(\cos x)))>\sin(\sin(\sin(\sin x)))$

Hence no solution in real numbers !

  • $\begingroup$ This should be the accepted solution. $\endgroup$ Aug 12, 2022 at 21:27

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