Finding relationships between angles, a, b and c when $\sin a - \sin b - \sin c = 0$

Question

while I was doing some calculations, I came across the condition where $$ \sin a - \sin b - \sin c = 0$$ where $0 < b, c < a < \dfrac{\pi}{2}$. Is there a way to find a relationship between the angles without using the trigonometric functions above? I have not been able to even find a specific solution hence any help is extremely appreciated. Yet, if possible, I will like to avoid approximations, such as by Taylor expanding the above.

My(unsuccessful) approaches

This is included for the sake of potential elaboration of the question and to just show my thought process.

1. Finding the area on a unit circle.

As $\sin 2\theta = 2\sin\theta\cos\theta$, I thought I could somehow correlate it to the area of a rectangle on the unit circle(which is given by $\sin \cos \theta$). Where, if $x = \dfrac{a}{2}, y = \dfrac{b}{2}, z = \dfrac{c}{2}$, and $\cos x = c_0, \sin x = s_0$ and so on, $$ c_0s_0 -c_1s_1 - c_2s_2 = 0$$ yet I could not figure out how to proceed from here.

2. Trying to associate it with the Euler's formula

I tried transforming all the sins to the complex identities. As may be evident, that did not turn out especially well due to the fact that adding exponential functions is not an advisable act(please correct me if I am wrong).

3. I tried to associate it to waves in physics.

In particular, I tried to associate it to the triple slit interference pattern where the waves destructively interfere. Yet as I could not find any formula for this that was not an approximation and as I could not think of my own, I dropped this idea.

Answer 1

If we plot a vector (or complex number) sketch in which $\sin \alpha = \sin \beta + \sin \gamma $ as follows

we can establish some interesting consequential relations.

One is given by the length of segment $AH_B$, which reads $$ \matrix{ \cos \beta - \cos \alpha = 2\sin \left( {\left( {\alpha - \beta } \right)/2} \right)\cos \left( {\pi /2 - \left( {\alpha + \beta } \right)/2} \right) = \hfill \cr = 2\sin \left( {\left( {\alpha - \beta } \right)/2} \right)\sin \left( {\left( {\alpha + \beta } \right)/2} \right) \hfill \cr} $$ but it is obvious.

Another relates to the length of segment $BH_B$ $$ \sin \gamma = 2\sin \left( {\left( {\alpha - \beta } \right)/2} \right)\cos \left( {\left( {\alpha + \beta } \right)/2} \right) $$

We can continue and derive other relations, and manipulate them in various ways.

However it does not look that we can arrive to anything simpler, or more concise, than the starting one.

Answer 2

As a general rule, relationships involving trigonometric functions can be rewritten only in a form that still involves trigonometric functions. Your relationship is already very simple; so it seems unlikely that any simpler, or nontrigonometric, equivalent exists.

At school, we are used to seeing trigonometric equations that have solutions. But these have been carefully constructed to that aim, for educational purposes. Generally, trigonometric equations have no simple solutions, and this applies all the more so, the greater the number of variables involved (3 in the present case).