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

The following problem was on a math competition that I participated in at my school about a month ago:

Prove that the equation $\cos(\sin x)=\sin(\cos x)$ has no real solutions.

I will outline my proof below. I think it has some holes. My approach to the problem was to say that the following equations must have real solution(s) if the above equation has solution(s):

$$ \cos^2(\sin x)=\sin^2(\cos x)\\ 1-\cos^2(\sin x)=1-\sin^2(\cos x)\\ \sin^2(\sin x)=\cos^2(\cos x)\\ \sin(\sin x)=\pm\cos(\cos x)\\ $$

I then proceeded to split into cases and use the identity $\cos t = \sin(\frac{\pi}{2} \pm t\pm y2\pi)$ to get

$$ \sin x=\frac{\pi}{2} \pm \cos x\pm y2\pi\\ $$

and the identity $-\cos t = \sin(-\frac{\pi}{2}\pm t\pm y2\pi) $ to get

$$ \sin x=- \frac{\pi}{2}\pm \cos x \pm y2\pi.\\ $$

where $y$ is any integer. I argued that $y=0$ was the only value of $y$ that made any sense (since the values of sine and cosine remain between $-1$ and $1$). Therefore, the above equations become

$$ \sin x=\frac{\pi}{2} \pm \cos x\implies \sin x \pm \cos x=\frac{\pi}{2}\\ $$

and

$$ \sin x=- \frac{\pi}{2}\pm \cos x\implies \cos x\pm\sin x= \frac{\pi}{2}.\\ $$

Then, by a short optimization argument, I showed that these last two equations have no real solutions.

First, does this proof make sense? Second, if my proof makes sense, then I feel that it was not very elegant nor simple. Is my approach the best, or is there a better (i.e., more elegant, shorter, simpler) proof?

Solution 1:

The function $$f(x):=\cos(\sin x)-\sin(\cos x)$$ is even and $2\pi$-periodic; therefore it suffices to consider $x\in[0,\pi]$. When $x=0$ or $x\in\bigl[{\pi\over2},\pi\bigr]$ then obviously $f(x)>0$. Finally, when $0<x<{\pi\over2}$ then $\cos x$ and $\sin x$ both lie in the interval $\ ]0,1[\ \subset\ ]0,{\pi\over2}[\ $. Therefore we also have $$\sin(\cos x)<\cos x<\cos(\sin x)\qquad\bigl(0<x<{\pi\over2}\bigr).$$

Solution 2:

A possibly-shorter way of getting there would be to write $$\cos (\sin x) - \sin(\cos x)=\cos (\sin x) - \cos(\pi/2-\cos x)$$ and then use a sum-to-product identity to turn this last expression into: \begin{eqnarray} &&−2 \sin \left(\frac{\sin x + \pi/2 - \cos x}{2}\right)\sin\left(\frac{\sin x - \pi/2 + \cos x}{2}\right)\\ &=& −2 \sin \left(\frac{\pi/2 + \sqrt{2}\sin (x-\pi/4)}{2}\right)\sin\left(\frac{- \pi/2 + \sqrt{2}\sin (x + \pi/4)}{2}\right) \, . \end{eqnarray} Since $\pi/2$ is not within $\sqrt{2}$ of any multiple of $2\pi$, this last expression never vanishes; thus your equation has no real solutions.

(Really, though, this is equivalent to your solution, except that we're outsourcing a lot of the case analysis to trig identities...)