<s>[INST] 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?
 [/INST] $$\cos(\sin x)=\sin(\cos x)=\cos\left(\frac\pi2-\cos x\right)$$
$$\text{So}, \sin x=2n\pi\pm \left(\frac\pi2-\cos x\right)$$
$$\text{Taking the '+' sign,} \sin x=2n\pi+ \left(\frac\pi2-\cos x\right)\implies \sin x+\cos x=\frac{(4n+1)\pi}2$$ which can be $\cdots,-\frac{3\pi}2,\frac{\pi}2,\frac{5\pi}2,\cdots$
$$\text{Taking the '-' sign,} \sin x=2n\pi- \left(\frac\pi2-\cos x\right)\implies \sin x-\cos x=\frac{(4n-1)\pi}2$$  which can be $\cdots,-\frac{5\pi}2,-\frac{\pi}2,\frac{3\pi}2,\cdots$
Now let, $1=r\cos\theta,1=r\sin\theta$ where $r>0$
So, $(r\cos\theta)^2+(r\sin\theta)^2=1+1=2\implies r^2=2\implies r=\sqrt2$
$\sin x\pm\cos x=r\cos\theta\sin x\pm r\sin\theta\cos x=\sqrt2\sin(x\pm \theta)$
So, $-\sqrt2\le \sin x\pm\cos x\le \sqrt 2 $
Now, $\sqrt 2<1.5<\frac\pi2$ as $3<\pi$
$\implies -\sqrt 2>-\frac\pi2$
$\implies -\frac\pi2<-\sqrt2\le \sin x\pm\cos x\le \sqrt 2<\frac\pi2 $
Hence, there is no real soultion
</s>