International Mathematics Competition
for University Students
2026

IMC2026: Day 1, Problem 1

Problem 1. Show that the equation \(\displaystyle \cos(\cos x) = \sin(\sin x)\) has no real solutions.

Alexander Slávik, Charles University, Prague

Solution. Assume, for contradiction, that there exists \(\displaystyle x\in\mathbb{R}\) such that \(\displaystyle \cos(\cos x)=\sin(\sin x)\). Using \(\displaystyle \cos t=\sin\bigl(\frac{\pi}{2}-t\bigr)\), we may rewrite this as

\(\displaystyle \sin\left(\frac{\pi}{2}-\cos x\right)=\sin(\sin x). \)

Set

\(\displaystyle A=\frac{\pi}{2}-\cos x \qquad\text{and}\qquad B=\sin x. \)

Since \(\displaystyle \sin x,\cos x\in[-1,1]\), we have

\(\displaystyle A\in\left[\frac{\pi}{2}-1,\frac{\pi}{2}+1\right] \qquad\text{and}\qquad B\in[-1,1]. \)

The general solutions of \(\displaystyle \sin A=\sin B\) are

\(\displaystyle A-B=2k\pi \qquad\text{or}\qquad A+B=(2k+1)\pi, \qquad k\in\mathbb{Z}. \)

The above bounds on \(\displaystyle A\) and \(\displaystyle B\) force \(\displaystyle k=0\). Hence either

\(\displaystyle \frac{\pi}{2}-\cos x=\sin x \)

or

\(\displaystyle \frac{\pi}{2}-\cos x=\pi-\sin x. \)

After rearranging, these equations become, respectively,

\(\displaystyle \sin x+\cos x=\frac{\pi}{2} \)

and

\(\displaystyle \sin x-\cos x=\frac{\pi}{2}. \)

However,

\(\displaystyle \sin x\pm\cos x \leq \sqrt{2}, \)

as follows, for example, from

\(\displaystyle \sin x\pm\cos x=\sqrt{2}\,\sin\left(x\pm\frac{\pi}{4}\right). \)

Since

\(\displaystyle \sqrt{2}<\frac{\pi}{2}, \)

neither equality is possible. Hence the equation

\(\displaystyle \cos(\cos x)=\sin(\sin x) \)

has no real solutions.