Prove $\sin(\cos x) \neq \cos(\sin x)$

Question

Prove that the trigonometric equation

sin(cosx) = cos(sinx)

has no solutions.

I graphed the functions and I can see they are never equal to eachother, but how do I do a formal proof for this?

Answer 1

Compare both functions with $\cos x$

Answer 2

$$\cos(\sin x)=\sin(\cos x)=\cos\left(\frac\pi2-\cos x\right)$$

$$\text{So}, \sin x=2n\pi\pm \left(\frac\pi2-\cos x\right)$$

$$\text{Taking the '+' sign,} \sin x=2n\pi+ \left(\frac\pi2-\cos x\right)\implies \sin x+\cos x=\frac{(4n+1)\pi}2$$ 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