If $\frac{\sin(b+\theta)\sin a}{\sin b}=\frac{\sin(c+\theta)\sin b}{\sin c}=\frac{\sin(a+\theta)\sin c}{\sin a}$, for arbitrary $\theta$, then $a=b=c$ 
If $$\frac{\sin(b+\theta)\sin a}{\sin b}= \frac{\sin(c+\theta)\sin b}{\sin c}= \frac{\sin(a+\theta)\sin c}{\sin a}$$ for some arbitrary constant $\theta$, then prove $a=b=c$.

I’ve tried everything, from product to sum manipulation, setting them to some $k$, to brute force expansion. But nothing seems to work.
 A: First, let's combine the two equations into one and eliminate the $\theta$. Take the first equation, expand $\sin(b+\theta)$ and $\sin(c+\theta)$.
$$(\sin \theta \cot b + \cos \theta)\sin a = (\sin \theta \cot c + \cos \theta)\sin b$$
$$\sin \theta (\cot b \sin a - \cot c \sin b) = -\cos \theta (\sin a - \sin b)$$
Do the same with the second equation:
$$\sin \theta (\cot c \sin b - \cot a \sin c) = -\cos \theta (\sin b - \sin c)$$
If $\sin \theta \neq 0$ and $\cos \theta \neq 0$, flip the second equation, multiply them together and cancel the $-\sin \theta \cos \theta$:
$$(\cot b \sin a - \cot c \sin b)(\sin b - \sin c) = (\cot c \sin b - \cot a \sin c)(\sin a - \sin b)$$
If $\sin \theta = 0$, then the terms without the cotangents are both zero, and if $\cos \theta = 0$, then the terms with the cotangents are both zero. So this equation is true no matter what $\theta$ is.
Expand and cancel the two like terms. The result has symmetry when the variables are cycled.
$$\sum_{a,b,c} \cos a \sin c = \sum_{a,b,c} \cot a \sin b \sin c $$
Now use the fact that $a+b+c=\pi$ to simplify this. Replace the $\sin b$ with an expression in terms of $a$ and $c$.
$$\sum_{a,b,c} \cos c \sin b = \sum_{a,b,c} \frac{\cos a \sin b \sin c}{\sin a} $$
$$\sum_{a,b,c} \cos c (\sin a \cos c + \cos a \sin c) = \sum_{a,b,c} \frac{\cos a (\sin a \cos c + \cos a \sin c) \sin c}{\sin a} $$
$$\sum_{a,b,c} (\sin a \cos^2 c + \cos a \sin c \cos c) = \sum_{a,b,c} \frac{(\sin a \cos a \sin c \cos c + \cos^2 a \sin^2 c) }{\sin a} $$
$$\sum_{a,b,c} \sin a \cos^2 c = \sum_{a,b,c} \frac{\cos^2 a \sin^2 c}{\sin a} $$
$$\sum_{a,b,c} \sin a (1 - \sin^2 c) = \sum_{a,b,c} \frac{(1 - \sin^2 a) \sin^2 c}{\sin a} $$
$$\sum_{a,b,c} \frac{\sin^2 a}{\sin a} = \sum_{a,b,c} \frac{\sin^2 c}{\sin a} $$
Let's use the Rearrangement inequality. Let the $x_i$'s be $\sin^2 a$, etc., and let the $y_i$ be $-1/\sin a$, etc. If you arrange the $x_i$'s so they increase, the corresponding $y_i$'s will increase too (because the sines are all positive). So we have equality if and only if $\sin a = \sin b = \sin c$. Because $a$, $b$ and $c$ are nonnegative and sum to $\pi$, this forces $a=b=c$.
Edit: Here’s some more detail about the Rearrangement inequality. If we have equality, then there must be at least one matching element among either the $x_i$’s or the $y_i$’s. In either case, there two variables with the same sine value. Assume that these are $b$ and $c$. So replace $c$ with $b$ in the equation, and one of the terms on each side will cancel:
$$ \frac{\sin^2 a}{\sin a} + \frac{\sin^2 b}{\sin b}
= \frac{\sin^2 b}{\sin a} + \frac{\sin^2 a}{\sin b} $$
Now apply the Rearrangement inequality again for this case, and we have $\sin a = \sin b$.
