Prove there does not exist three Pythagorean triples such that $\sin{2\theta_1}+\sin{2\theta_2}=\sin{2\theta_3}$

Question

This question stems from another one that I asked earlier here. However, I have found a way to reformulate it and I believe this reformulation is sufficient to warrant posting it as a separate question. My initial question was, does there exist three different primitive Pythagorean triples such that,

$$\frac{Area_1}{c_1^2}+\frac{Area_2}{c_2^2}=\frac{Area_3}{c_3^2}$$

Multiplying by two this can be rewritten as,

$$\frac{a_1b_1}{c_1^2}+\frac{a_2b_2}{c_2^2}=\frac{a_3b_3}{c_3^2}$$

Looking at an individual term and applying the context of right triangles we can now obtain,

$$\frac{ab}{c}=\frac{a}{c}*\frac{b}{c}=\sin{\theta}\cos{\theta}$$

Now revisiting the original question and doing some algebra,

$$\sin{\theta_1}\cos{\theta_1}+\sin{\theta_2}\cos{\theta_2}=\sin{\theta_3}\cos{\theta_3}$$

$$\sin{2\theta_1}+\sin{2\theta_2}=\sin{2\theta_3}$$

Where $\theta$ is the angle for the origin of the associated primitive Pythagorean triple. My belief (for whatever that's worth) is that for all primitive Pythagorean triples,

$$\sin{2\theta_1}+\sin{2\theta_2}\neq\sin{2\theta_3}$$

So far I have tested this for many triples but have not been able to find a counter example. As far as proving the latter I am pretty lost as far as finding an approach. For clarity I am asking, prove that there does not exist three primitive triples such that $$\sin{2\theta_1}+\sin{2\theta_2}=\sin{2\theta_3}$$ Or find a counter example. If you look at my initial question there was some helpful input and I would encourage you to look at it for some insight/context.

EDIT: I am amending this to be also including non primitive triples too. This makes sense because the solution to that should imply the other as we are dealing with angles and a normal triple should have the same angle as the scaled non primitive triple.

Edit2: Additionally, a,b,c can have no common factors.

Answer 1

Some possibilities:

The triangles can be normalized so that each $c_i = 1$.

You can then write each $a_i = m_i^2-n_i^2, b_i = 2m_in_i$ where the $m_i, n_i$ are rationals such that $m_i^2+n_i^2 = 1$.

Then $\sum_{i=1}^2m_in_i(m_i^2-n_i^2) =m_3n_3(m_3^2-n_3^2) $.

That's all that occurs to me right now.