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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.

  1. 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).

  1. 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.

Isamu Isozaki
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2 Answers2

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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).

John Bentin
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  • Thanks for the answer! But isn't there a way to at least find a specific solution for a, b and c? – Isamu Isozaki Feb 15 '18 at 20:48
  • I think I can manage even if the solution is not simple. Any solution is very much appreciated. – Isamu Isozaki Feb 15 '18 at 20:50
  • From the way that you posed the problem, it looks as though you want to find $a$ in terms of $b$ and $c$. Well, trivially you could write it as $$a=\arcsin(\sin b+\sin c).$$There are two independent variables (any two of $a$, $b$, and$c$); so you cannot find the value of any one without specifying the value of the other two from further information. – John Bentin Feb 15 '18 at 21:29
  • Actually, I was fine with finding even one specific value where the above equation is valid. Is there a way to do that? – Isamu Isozaki Feb 16 '18 at 06:53
  • @IsamuIsozaki: Well, just choose any two angles at random (but not too big), say $b=\pi/5$ and $c=\pi/10$. Then $a=1.112488037$, by the formula in the above comment. – John Bentin Feb 16 '18 at 08:45
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If we plot a vector (or complex number) sketch in which $\sin \alpha = \sin \beta + \sin \gamma $ as follows

Rel_3_sin_1

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.

G Cab
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  • Thanks for the answer! So, basically, there are no ways to compute even specific(values) solutions unless through brute force? – Isamu Isozaki Feb 16 '18 at 06:55
  • Thanks for the diagram! I will try to think about it on my side quite a bit more. – Isamu Isozaki Feb 16 '18 at 06:56
  • @IsamuIsozaki: given two of the values, you can always compute the third quite easily as $\alpha=\arcsin{(\sin \beta+\sin \gamma)}$ and similar. I was answering to the question of whether it was possible to give a simple geometric interpretation as angles in a certain quadrilateral (the sketch shows one), or other ... – G Cab Feb 16 '18 at 09:39