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(I'm new here, so I hope this question hasn't come up before)

A bit of motivation for the problem:
It is well-known that the equations $\cos(x) = \sin(x)$, $\cos(\cos(x)) = \sin(\sin(x))$, and $\cos(\cos(\cos(x))) = \sin(\sin(\sin(x)))$ all have infinitely many solutions in $\mathbb{C}$ (the first and third have infinitely many solutions in $\mathbb{R}$ as well). The proofs in these cases are elementary, but break down when applying it to further (lacking a better word) iterations. Using the fourth to illustrate:

Consider the two functions $\cos(\cos(\cos(\cos(z))))$ and $\sin(\sin(\sin(\sin(z))))$, and for convenience, let
$H(z) = \cos(\cos(\cos(\cos(z)))) - \sin(\sin(\sin(\sin(z))))$.
Also, let $V = \{z \in \mathbb{C} \, | \, H(z) = 0\}$.

The original question (while not phrased in this manner) was:
Find $V$.

It is not difficult to verify that $\not\exists z\in V$ such that $\Im(z) = 0$. To prove this, locate local extrema of the function $H(x)$ (where in an abuse of notation, I use $H(x)$ to denote the restriction of $H$ to the real numbers), and one will find that all (relative) maximum and minimum values of the function $H(x)$ are strictly positive. In fact, one can prove that $H(x) \geq \frac{1}{10} > 0$, $\forall x \in \mathbb{R}$. There is a sharper estimate, but this is sufficient for our purposes, and proves that there are no real solutions to the equation $H(x) = 0$.

Having proved that there exist no real solutions, the question now becomes:
Is it possible to analytically (i.e. without numerical methods) prove that $V$ is non-empty?

My first instinct was to try to use the Argument Principle, as applied either to a ball of radius $n\in\mathbb{N}$ centered at $0$, or a rectangle of side-length $n$ centered at $0$, but I'm not sure if those integrals can be computed explicitly (even as contour integrals).

Note: It would be elementary to write an algorithm based on Newton's Method or some improvement thereof and attempt to find roots. But stability is an issue if you are far from a root.

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Sorry...To clarify, it should have said "infinitely many solutions", instead of "infinitely many real solutions", as Colin pointed out. It is actually an isolated instance that in the first and third cases, there are infinitely many real solutions, as for any other iterations, there are no real solutions. –  Nicholas Stull Apr 13 '12 at 16:43
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Since your function $H$ is entire, this is essentially just showing that its 'omitted value' isn't $0$ (or equivalently, that it's not of the form $e^{f(z)}$ for some $f$), but unfortunately my analysis background isn't strong enough to go much past Picard's theorem for this... –  Steven Stadnicki Apr 13 '12 at 17:23
    
@Steven That was something I had thought of. $H$ is indeed analytic and entire, and has an essential singularity at the point at infinity. If we express this as $H(z) = e^{f(z)}$ for some $f$, I believe we would need to ensure that $f$ is analytic. Thanks for the suggestion. I'll update the question if I make any progress. –  Nicholas Stull Apr 13 '12 at 23:06
    
One can sharpen the condition (without analytical methods). Consider the vanishing of $H^2$ too to see, that its necessary for $cos(cos(cos(cos(z))))$ and $sin(sin(sin(sin(z))))$ to coincide, to have the same value $\pm i\sqrt{1/2}$, i.e. $V = \{z|cos(cos(cos(cos(z)))) = \pm i\sqrt{1/2} = sin(sin(sin(sin(z))))\}$. But I don't know how to go further. –  Ben A. Apr 14 '12 at 11:42
    
It appears that there are infinitely many roots on the line $\operatorname{Re} z=\pi/2$. In this case the function $H(z)=\cos^4 z −\sin^4 z$ becomes $H(y)=\cos^3(\sinh y)−\sin^3(\cosh y)$, where the exponent denotes repeated application of the function. –  Antonio Vargas Apr 14 '12 at 16:06
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1 Answer

Huh? $\cos(\cos x) = \sin(\sin x)$ has no real solutions:

$ \cos(\cos x) = \sin(\sin x) \Rightarrow \cos x = (4n+1)\frac{\pi}{2} \pm \sin x \Rightarrow |\cos x \pm \sin x| = \frac{\pi}{2}|4n+1|$. But $ |\cos x \pm \sin x| \le \sqrt{2} < \frac{\pi}{2}$ for real $x$.

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I apologize for that mistake. I meant to write "infinitely many solutions", instead of "infinitely many real solutions". I have edited my question to correct that error. Thanks for the heads-up on that. –  Nicholas Stull Apr 13 '12 at 16:29
    
Nice solution....... –  juantheron Apr 16 at 5:24
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