4

I was messing around on Desmos trying to create trigonometry problems when I came across the following:

For what positive integers $a,b,c$ is it true that all possible roots of $$\sin(ax)+\sin(bx)+\sin(cx)=0$$ are rational multiples of $\pi?$

By inspection I found some triples $(1,2,3), (1,3,4), (1,3,5), (2,3,4), (3,5,7)$ but I do not see any pattern. I looked into Chebyshev polynomials but that seems extremely ugly. How would I go about determining

  1. whether there are infinitely many triples $(a,b,c)$
  2. what is the "criteria" for such a triple?
Thomas Andrews
  • 177,126
x3yukari
  • 79
  • Do you mean real roots? – Robert Israel Nov 23 '21 at 23:42
  • 1
    There are easily infinitely many $a,b,c$ unless you add that $\gcd(a,b,c)=1.$ – Thomas Andrews Nov 23 '21 at 23:44
  • Have you found any examples where it is not true? – Thomas Andrews Nov 23 '21 at 23:45
  • In all of your examples, $a,b,c$ are in arithmetic progression. Without knowing cases $a,b,c$ where it is not true, it is hard to see if this is a pattern or an accident od what you have tried. – Thomas Andrews Nov 23 '21 at 23:53
  • For example, it's not true for $(a,b,c) = (1,2,4)$: if $w = e^{ix}$, $\sin(x) +\sin(2x) + \sin(4x) = (-i/(2 w^4))(w+1)(w-1)(w^6+2w^4+w^3+2w^2+1)$ and the last factor is irreducible over the rationals but not a cyclotomic polynomial, so its roots (of which two are on the unit circle) are not roots of unity. – Robert Israel Nov 24 '21 at 00:08
  • @RobertIsrael No, I mean roots that are rational multiples of $\pi.$ For example, $\sin(2x)+\sin(3x)+\sin(4x)=0$ has the roots $2\pi n, \frac{\pi}{3} + 2\pi n, \frac{2\pi}{3}+ 2\pi n , \pi + 2\pi n, \frac{4\pi}{3}+ 2 \pi n, \frac{5\pi}{3} + 2 \pi n$ – x3yukari Nov 24 '21 at 00:25
  • @x3yukari Robert’s argument is to rephrase it about the roots of polynomials. – Thomas Andrews Nov 24 '21 at 00:42

3 Answers3

4

Partial answer.

Assume that $a<b<c.$ If $a,b,c$ is an arithmetic progression, then $b=\frac{a+c}2,$ and $\frac{c-a}2=b-a,$ so $$\sin (ax)+\sin(cx)=2\sin(bx)\cos((b-a)x).$$

So in this case, the equation is equivalent to $$0=\sin(bx)(1+2\cos((b-a)x))$$ and you have that all roots are rational multiples of $\pi.$

Not sure if this is necessary, but it is sufficient.

Thomas Andrews
  • 177,126
3

More generally, let $f(x)$ be a linear combination over the rationals of sines of positive integer multiples of $x$. Taking $w = e^{ix}$, so $\sin(ax) = (w^a - w^{-a})/(2i)$, this can be written as $P(w)/(i w^n)$ where $P$ is a polynomial with integer coefficients. All (real or complex) zeros of your $f$ are rational multiples of $\pi$ if and only if $P$ is a constant times a product of cyclotomic polynomials; all real zeros are rational multiples of $\pi$ if and only if $P$ is a constant times a product of cyclotomic polynomials times a polynomial with no roots on the unit circle.

Robert Israel
  • 448,999
0

Let $z_t=\cos(t)+i \sin(t)$, so it means $z_{at}+z_{\pi-at}+z_{bt}+z_{\pi-bt}+z_{ct}+z_{\pi-ct}=0$.

The paper provides the condition that the sum of at most 12 roots of unary is 0. And we only need analysis 3 types of candidate solutions:
3 R2, which means ct=0 and at=-bt, which means sin(ct)=0, sin(at)=-sin(bt)
2 R3, which means {at,bt,ct} forms equilateral triangle
R5:R3, which means the shared and eliminated element of R5&R3 at is $\frac{\pi}2$ or $\frac{3\pi}2$. For example, the the shared element is $\frac{\pi}2$, at,bt,ct are $\frac{\pi}2+\frac{2\pi}5, \frac{\pi}2+\frac{4\pi}5, \frac{\pi}2+\frac{2\pi}3$ respectively

Zhaohui Du
  • 1,816