# Roots of combination of trigonometric functions

Consider the following function of $\mathbb{R^+}^2$:

$$f(s_1,s_2)=r_1^2\sin\big(\tfrac{1}{2}(s_2-s_1)\omega_1\big)\sin\big(\tfrac{1}{2}(s_2+s_1)\omega_2\big) + r_2^2\sin\big(\tfrac{1}{2}(s_2+s_1)\omega_1\big)\sin\big(\tfrac{1}{2}(s_2-s_1)\omega_2\big)$$

where $r_1^2+r_2^2=1$. My goal is to find the solution curves of $f$ on $\mathbb{R^+}^2$. There is certainly no closed-form solution, so I'm doing it numerically, for some given values of $(r_1,r_2,\omega_1,\omega_2)$.

### Symmetries

Obviously, $f(s_1,s_2)=-f(s_2,s_1)$ so the solution curves of $f$ are symmetric w.r.t. to $s_2=s_1$: it suffices to focus on $s_1\geq s_2$. Note that also $f(-s_1,s_2)=-f(-s_2,s_1)$ so $s_2=-s_1$ is another axis of symmetry for the solutions of the extension of $f$ to $\mathbb{R}^2$, but it is a priori not intersting on $\mathbb{R^+}^2$.

$f$ is a scalar function of two variables so its solution curves are (generically) curves. That's what we can observe to some numeric values of $(r_1,r_2,\omega_1,\omega_2)$:

(Note the axial symmetry mentionned above.)

### Calculating some points on the solution curves

For some reason (reduction of computational cost in more complex cases), I'd like to get points on the solution curves of $f$ and then do numerical continuation to compute each curve. By curve, I mean a connex curve, and the solution curves is a family of such connex curves.

The idea is that when $\tfrac{1}{2}(s_2+s_1)\omega_i$ approaches $\pi\mathbb{N}$, only one term is the sum remains. For example, when $s_2+s_1=2\pi/\omega_1$,

$$f(s_1,2\pi/\omega_1-s_1)=r_1^2 \sin(\pi\tfrac{\omega_2}{\omega_1})\sin(s_1\omega_1)$$

and so the roots of $s_1\mapsto f(s_1,2\pi/\omega_1-s_1)$ are the $\pi/\omega_1\mathbb{N}$.

In the, we know that the set of points

$$\mathcal{S}=\Big\{ \frac{\pi}{\omega_i}(p,2p-q),\ p,q\in\mathbb{N}, i\in\{1,2\}\Big\} \cap {\mathbb{R}^+}^2$$

is a subset of the solution curves of $f$, as illustrated below:

or, for another set of parameters $(r_1,r_2,\omega_1,\omega_2)$:

Red and green points correspond to $i=1$ and $i=2$.

### Question

On both figures, all the curves pass through the set of points $\mathcal{S}$. Is this always true, or can there be some curves which do not pass through any green or red points?

I am also interested in references or keywords for mathematical tools which could be useful for the study of solution curves of functions of the "type" of $f$ (i.e. sum of products of sines), even though I don't think there are miraculous simplification.

• Nice curves!... A terminology issue: you use the word "roots" instead of (solution) curves. Usually, the term "roots" is reserved for discrete cases. – Jean Marie Nov 18 '16 at 17:27
• @JeanMarie Thank you I edited. So is correct to say e.g. that solutions of $g(x,y)=1$ are the solution curves of $g(x,y)-1$? I mean, does "solution curves" imply zeros? – anderstood Nov 18 '16 at 18:06
• Yes, one can say it in this way. – Jean Marie Nov 18 '16 at 18:07
• Have you considered defining $t_1 = \frac12(s_2-s_1), t_2 = \frac12(s_2+s_1)$? It would transform $f$ into a sum of two separable functions, which may simplify things a bit. – Rahul Nov 18 '16 at 18:15
• @Rahul It's make a slightly simpler indeed (see image) but does not help me find the idea why every curves should pass through the grid of points. That function "becomes": $r_1^2\sin(t_1\omega_1)\sin(t_2\omega_2)+r_2^2\sin(t_1\omega_2)\sin(t_2 \omega_2)$ with \$t. – anderstood Nov 18 '16 at 18:25