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