Tough contest problem

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 $\pi > 3$, $$ \arcsin(1 - \pi/2) - \pi/2 < \arcsin(1 - 3/2) - \pi/2 = -\pi/6 - \pi/2 = - 2\pi/3 < -1 $$

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).