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

Question

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?

Answer 1

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...)

Answer 2

Let $a = \cos x$ and $b = \sin x$, and so $a,b \in [-1,1]$.

We have to solve $\sin a = \cos b$.

We can assume that $0 \le b \le 1$, because, if $x$ is a root, so is $-x$.

Since $\sin a = \cos b$ and $b \ge 0$, we must have that $a \ge 0$ (remember, $a,b \in [-1,1]$)

Thus if the equation has roots, at least one of those is such that $x \in [0,\pi/2]$.

It is clear that $0, \pi/2$ are not roots. So we can assume $x \in (0, \pi/2)$.

Now $ \sin y \lt y$ for all $y \in (0, \pi/2)$.

Thus $\cos (\sin x) \gt \cos x$ (as $\sin x \lt x$ and $\cos $ is decreasing )

We also have $\cos x \gt \sin (\cos x)$ (using $\cos x = y \gt \sin y = \sin (\cos x)$).

Thus we have that $$\cos (\sin x) \gt \sin (\cos x), \quad x \in (0, \pi/2)$$

Contradiction, and the equation has no real roots.

Answer 3

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 $$\sin(\cos x)<\cos x<\cos(\sin x)\qquad\bigl(0<x<{\pi\over2}\bigr).$$

Answer 4

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

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

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