22
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Your approach looks fine to me. $\endgroup$ – user7530 Apr 5 '13 at 6:21
22
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ @mavavilj: Mainly from $\sin u<u$ for all $u>0$. $\endgroup$ – Christian Blatter Aug 29 '15 at 18:40
7
$\begingroup$

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...)

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I'll delete my answer since any attempt to fix it will make it look like this or some of the other answers. $\endgroup$ – leo Apr 5 '13 at 22:02
2
$\begingroup$

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

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

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

$\endgroup$
-1
$\begingroup$

After spending many hours trying to find the real solutions of this equation, here are the ways of proving that it has no real solutions that I found:

For $\sin(\cos(x))=\cos(\sin(x))$ to be true, both $\cos(x)$ and $\sin(x)$ have to be equal to $\frac{\pi}{4}$ since $\cos(x)$ and $\sin(x)$ take same value in this number. (Note that I'm talking about the terms inside the sine on the left hand and the cosine on the right hand)

The first way to find such $x$ is by setting up the following system of equations: $$\cos(x)=\frac{\pi}{4}$$ $$\sin(x)=\frac{\pi}{4}$$

We sum both equations to get: $$\cos(x)+\sin(x)=\frac{\pi}{2}$$ Raising both sides of the equation by the exponent 2 we get: $$\cos^2(x)+2\cos(x)\sin(x)+\sin^2(x)=\frac{\pi^2}{4}$$ Which can be reduced to $$1+2\cos(x)\sin(x)=\frac{\pi^2}{4}$$ $$1+\sin(2x)=\frac{\pi^2}{4}$$ Solving for $x$ we get: $$x=\frac{1}{2}\sin^{-1}(\frac{\pi^2}{4}-1)$$ Which is impossible to perform since the value inside the inverse sine should be between $-1$ and $1$ Yet another way of proving it is by using the identity $\cos(x)=\sin(x+\frac{\pi}{2})$ so that we have: $$\sin(\cos(x))=\sin(\sin(x)+\frac{\pi}{2})$$ $$\cos^2(x)-2\cos(x)\sin(x)+\sin^2(x)=\frac{\pi^2}{4}$$ And in the end we have: $$1-\sin(2x)=\frac{\pi^2}{4}$$ Which will lead to the same conclusion as we did before.

And yet another way to prove it is by writting the equations as follows: $$\sqrt{1-\sin^2(x)}=\sin(x)+\frac{\pi}{2}$$ Which is the same as: $$\sin^2(x)+\frac{\pi}{2}\sin(x)+\frac{\pi^2-4}{8}=0$$ Performing a change of variable $u=\sin(x)$: $$u^2+\frac{\pi}{2}u+\frac{\pi^2-4}{8}=0$$ Both roots of the polynomial are complex so, again, there is no such real $x$ that can satisfy the above mentioned equation.

$\endgroup$
  • $\begingroup$ it's easier if you stop at the second line. add $+-2\pi n$ to the RHS to generalize it. write it as a single sine function Note that there are no solutions within the maximum/minimum range which is $-+\sqrt{2}<\pi/2$ $\endgroup$ – Dis-integrating Jun 21 '17 at 21:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.