# Solving the following system of trigonometrical equations

Solve the following system of equations $$\begin{cases} \sin(a)+2\sin(a+b+c)=0 \\ \sin(b)+3\sin(a+b+c)=0 \\ \sin(c)+4\sin(a+b+c)=0 \end{cases}$$

I added all the three equations and got $$\frac{\sin(a)+\sin(b)+\sin(c)}{9}+\sin(a+b+c)=0.$$ Then, I have very vague view on how to continue is using something like Jensen's inequality... Any help is appreciated.

• @VIVID For using Jensen you need $\{a,b,c\}\subset[0,\pi]$, for example. Commented Apr 17, 2020 at 15:31
• @MichaelRozenberg yes I know. So that $\sin(x)$ would be concave. Could one apply this to the question? Commented Apr 17, 2020 at 15:33
• I think I can solve your problem for $\{a,b,c\}\subset\left[0,\frac{\pi}{2}\right]$ Commented Apr 17, 2020 at 15:35
• Did you try substituting in the last equation into the previous ones? What did you get from there? Commented Apr 17, 2020 at 15:39
• @VIVID I can prove that for $\{a,b,c\}\subset\left[0,\frac{\pi}{2}\right]$ and unique solution it's $(0,0,0)$. Commented Apr 17, 2020 at 15:50

I am using that $$\sin a = A \Rightarrow a = \sin^{-1} A$$, which is true if they are in $$[- \frac{\pi}{2}, \frac{\pi}{2}]$$.

Let $$\sin a = A, \sin b = B, \sin c = C$$.
You have shown that $$\sin (a+b+c) = -\frac{ A + B + C } { 9 }$$.

Substituting back in, we get
$$7A -2B - 2C = 0$$
$$-3A + 6 B - 3C = 0$$
$$-4 A - 4 B + 5 C = 0$$
Solutions to this system are of the form $$A:B:C = 2:3:4$$.
Let $$A = 2n, B = 3n, C = 4n$$ where $$-\frac{1}{4} \leq n \leq \frac{1}{4}$$.

We then need to check against the condition of $$-9\sin (a+b+c ) = A+B+C = 9n$$, which gives us
$$\sin^{-1} (2n) + \sin^{-1} (3n) + \sin^{-1} (4n) = - \sin^{-1}n$$ or $$\pi + \sin^{-1} (n)$$ or $$2\pi - \sin^{-1} (n)$$.

For $$n \in ( 0, \frac{1}{4})$$, we can show that the LHS is $$\in (0, 3 )$$, hence is never equal to the RHS. A similar statement holds for $$n \in ( - \frac{1}{4} , 0)$$.

So, the only solution is $$n = 0 \Rightarrow A=B=C = 0$$.

To extend outside of the range, we need to consider $$a = \pi - \sin^{-1} A$$, which makes it seem like solutions exist. To be continued (?) ...

Expand out $$\sin(a+b+c)$$, take $$\sin(a)=s_a$$, $$\cos(a)=c_a$$, etc., adjoin the relations $$s_a^2 + c_a^2 - 1 = 0$$ etc, and we can solve using Groebner basis methods. All solutions have $$\sin(a) = \sin(b) = \sin(c) = 0$$, i.e. $$a,b,c$$ are integer multiples of $$\pi$$.