These are easy to solve: $$\cos(ax)=0,\qquad \cos(ax)+\cos(bx)=0$$ Any insight into $$\cos(ax)+\cos(bx)+\cos(cx)=0$$ Can't find any information on the web about this last one. I know solutions can be approximated via series but I'm interested in the exact cases or more information about triple exact solutions to identities like the last one.

  • 1
    $\begingroup$ If $a, b, c$ are integers, I believe each term can be expanded as a polynomial in $cos x$. At that point the problem reduces to figuring out if a given polynomial has roots in $[-1, 1]$. $\endgroup$ – Jack M May 18 '14 at 12:41
  • $\begingroup$ Since "exact solutions" is a little vague, I'd be interested in knowing if the roots are always in $\mathbb Q(a, b, c, \pi)$. $\endgroup$ – Jack M May 18 '14 at 13:41

This isn't a complete answer, but I think it gets most of the way there.

The first step would be to expand the cosines as follows:

$$ \cos(ax) = \cos([(a+c) - c]x)\\ \cos(bx) = \cos([(b+c) - c]x)\\ \cos(cx) = \cos(cx) $$

With liberal use of trig angle formulae, $\cos(ax) + \cos(bx) + \cos(cx)$ becomes $$ \cos((a+c)x)\cos(cx) + \sin((a+c)x)\sin(cx) + \\ \cos((b+c)x)\cos(cx) + \sin((b+c)x)\sin(cx) + \\ \cos(cx) $$ Factoring will turn this into $$ \cos(cx)[\cos((a+c)x) + \cos((b+c)x) + 1] + \sin(cx)[\sin((a+c)x) + \sin((b+c)x)] $$

To find out where this is zero, we can consider the case where the 2 halves are simultaneously zero. Finding out where both halves are nonzero but cancel each other out is a headache that I don't want to deal with.

So, we can consider the following 3 cases to get some of the solutions:

1) $\cos(cx) = 0$ and $\sin((a+c)x) + \sin((b+c)x) = 0$


2) $\sin(cx) = 0$ and $\cos((a+c)x) + \cos((b+c)x) = -1$


3) $\sin((a+c)x) + \sin((b+c)x) = 0$ and $\cos((a+c)x) + \cos((b+c)x) = 0$

At this point we're really just doing algebra. For (1), I get

$x = \frac{(\pi/2 + n\pi)}{c}$ and $\cos(ax) + \cos(bx) = 0$.


$$ \cos(\frac{a}{c}(\pi/2 + n\pi)) + \cos(\frac{b}{c}(\pi/2 + n\pi)) = 0 $$

This will only work for certain values of $a, b, c$. Which values are left as an exercise to the reader.

Similarly, for (2), I get $x = n\pi/c$ and $$ \cos(cx)[\cos(ax) + \cos(bx)] = -1 $$ Again, only certain values of $a,b,c$ will result in solutions

For (3), I'll skip the details but I get that $$ x = \frac{4\pi}{3(a+c)} = \frac{5\pi}{3(b+c)} $$ which also implies that an answer for this scenario exists only if $$ \frac{5}{4} = \frac{a+c}{b+c}. $$

Expanding about $c$ was an arbitrary choice. You can use the same method in either $a$ or $b$ to get the same solutions in terms of those variables.


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.