# Tough contest problem

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?

-
Of course there is no solution because $cos (cos (cos (x)))\ge 1/2$ as explained. What do you mean? –  user129798 Feb 18 '14 at 21:44
Plotting both functions with Wolfram Alpha does indeed indicate that the two functions don't overlap: wolframalpha.com/input/… –  Arno Feb 18 '14 at 21:46
Ooops! I deleted the comment because I misunderstood your question. Sorry! –  enoughsaid05 Feb 18 '14 at 21:47
–  Hans Lundmark Feb 25 '14 at 22:08

## 2 Answers

Rephrase the problem as:

$$\sin( \pi/2 + \sin(\pi/2 + \sin(\pi/2 + \sin(\pi/2 + x) ) ) ) = \sin(\sin(\sin(\sin(x))))$$ Then

$$\pi/2 + \sin(\pi/2 + \sin(\pi/2 + \sin(\pi/2 + x) ) ) = \sin(\sin(\sin(x))) + 2\pi n$$ which implies $$\sin(\pi/2 + \sin(\pi/2 + \sin(\pi/2 + x) ) ) = \sin(\sin(\sin(x))) + 2 \pi (n - 1/4).$$ It must be that $n = 0$ since both sine terms must have values in $[-1,1]$. For the remaining computations, we can take only the principal value of $\arcsin$ by this same argument. So $$\sin(\pi/2 + \sin(\pi/2 + \sin(\pi/2 + x) ) ) = \sin(\sin(\sin(x))) - \pi/2,$$ and so it must be that both sides of the above expression lie in the interval $[-1, 1 - \pi/2]$. But then, taking the $\arcsin$ again, we have $$\pi/2 + \sin(\pi/2 + \sin(\pi/2 + x) ) = \arcsin( \sin(\sin(\sin(x))) - \pi/2)$$ Since the expression was originally in the interval $[-1, 1 - \pi/2]$, then taking the $\arcsin$ produces an expression with values in the interval $[ - \pi/2, \arcsin(1 - \pi/2)]$ on both sides. Then $$\sin(\pi/2 + \sin(\pi/2 + x) ) = \arcsin( \sin(\sin(\sin(x))) - \pi/2) - \pi/2$$ and on the left side, clearly the expression must be in $[-1,1]$, while on the right side, the expression must be in the interval $[-\pi, \arcsin(1 - \pi/2) - \pi/2]$.

It now suffices to show that these two intervals do not intersect. In particular, we will show that $\arcsin(1 - \pi/2) - \pi/2 < -1$. Using the fact that $\pi > 3$, $$\arcsin(1 - \pi/2) - \pi/2 < \arcsin(1 - 3/2) - \pi/2 = -\pi/6 - \pi/2 = - 2\pi/3 < -1$$

-
+1 Wow, the only numerical approximation you use is $\pi>3$ –  Hagen von Eitzen Feb 18 '14 at 23:08
I know the question was for real x. I am curious if we allow complex x if there are solutions. –  Michael Smith Feb 27 '14 at 11:48
@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... –  Christopher A. Wong Feb 27 '14 at 20:36
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. –  Michael Smith Feb 28 '14 at 0:52
Alternative just looking at the imaginary axis putting ix for x and using cosh(x) = cos(ix) and i sinh(x) = sin(ix). I will play around with this a bit. –  Michael Smith Feb 28 '14 at 0:59

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).

-
As I said, I know that picture. But I'm looking for a formal argument. –  user129798 Feb 18 '14 at 21:49
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. –  Michael Kasa Feb 18 '14 at 22:01
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 ;-) –  Ant Feb 18 '14 at 23:23