# Proving $f(x) := \sin(ax) - \sin(bx)$ has absolute value greater or equal to 1 for some $x>0$

I am trying to prove that for any choice of reals $a$ and $b$ such that $a \neq b$ the function $f(x) := \sin(ax) - \sin(bx)$ has $|f(x)| \geq 1$ for some $x>0$

(This is not an exercise question but something I am trying to prove for myself. Plots with Wolfram Alpha seem to indicate it might be true, and so counterexamples are very much welcome)

Here is my attempt. For this I wrote $f$ of 3 variables $a,b,x$. We find that value of $x_0$ for which $f$ attains a maximum and minimum and then show that $|f(x_0)|$ greater than 1.

$g(a,b,x)=f(x)$ and then zeroing the gradient. (pending the 2nd derivative test)

$f_a = x \cos(ax)$

$f_b = -x \cos(bx)$

$f_x = a\cos(ax) - b\cos(bx)$

Since the point of minima/maxima are conjectured $x>0$ , this means $\cos(ax)$ and $\cos(bx)$ are both zero and the third equation is redundant.

If $a$ and $b$ are rational then obviously we can find such an $x$. But when one of them is irrational, I don't know how I would prove this.

Another attempt was using the $\sin(u)-\sin(v)$ product formula, but that led nowhere too.

Any pointers?

• Simple counterexample: $a=b=0$. – celtschk May 26 '13 at 17:02
• Ah yes. I should have added non-zero condition :-D Please see edit. – smilingbuddha May 26 '13 at 17:03
• In general, just $a = b$ is a counterexample. They have to be distinct, since we would have $f(x) = \sin ax -\sin ax \equiv 0$. – Stahl May 26 '13 at 17:03
• Good point. Please see edit. – smilingbuddha May 26 '13 at 17:05
• You still need $\neq 0$ even with the condition $a\neq b$, because with $a=0$ you'll get just $\cos bx$ which obviously never goes above $1$. – celtschk May 26 '13 at 17:07

If $a=b$, then $f(x)=0$. If $a=0$ or $b=0$ then we have just a single sine function and hence $|f(x)|\le 1$ for all $x$. Therefore these cases must be excluded.
If $a=-b$, we have $f(x)=2\sin (ax)$, hence $f(\frac{\pi}{2a})=2$ and we are done. We may therefore assume that $$\tag1a,b,a+b,a-b\text{ are nonzero.}$$ Note that $$\tag2\sin (ax)-\sin (bx)=2\cos{\left(\frac{a+b}2x\right)}\sin\left(\frac{a-b}2x\right).$$ The cosine is $\pm1$ at $x\in \frac{2k\pi}{a+b}\mathbb Z$. Thus it is sufficient to find $k\in\mathbb Z$ with $$\tag3 \left|\sin\frac{(a-b)k\pi}{a+b}\right|>\frac12.$$ For which $c\in\mathbb R$ can we conclude that there exists $k\in\mathbb Z$ with $|\sin( ck\pi)|>\frac12$? Since $|\sin(ck\pi)|=|\sin((c+1)k\pi)|$, we may restrict to the case $c\in[0,1)$. And since $|\sin((1-c)k\pi)|=|\sin(ck\pi)|$, we can restrict further to $c\in[0,\frac12]$. If $\frac16<c\le \frac12$, we succeed by choosing $k=1$. If $0<c\le\frac16$, let $k=\lceil\frac1{6c}\rceil$, i.e. we have $\frac16<kc\le\frac16+c\le \frac13$ and thus obtain $\sin(ck\pi)>\frac12$. Allowing arbitrary $c\in\mathbb R$ again, we therefore conclude that there exists $k\in\mathbb Z$ with $|\sin(ck\pi)|>\frac12$ unless $c$ is not an integer. Back to our original problem, we can thus show the existence of $x$ with $|f(x)|>1$ unless $c:=\frac{a-b}{a+b}$ is an integer. Note that $c\notin\{-1,0,1\}$ because of $(1)$. Therefore $\left|\frac{\pi}{2c}\right|\le\frac\pi 4 3$ and $$f\left(\frac{\pi}{a-b}\right)=2\cos \frac{\pi}{2c}\sin\frac\pi2\ge2\cos\frac\pi4=\sqrt2>1.$$
For $x_k=\pi/2a + 2*k\pi/a$, $\sin(ax_k) =1$ You "just" have to prove that for a certain $x_k$ $\sin(bx_k)$ is inferior than 0.
By this method, you could also prove a greater result : There exists $x$ where $f(x)>A$ for every $A<2$