While trying to deduce a condition for three points to be co-normal on an ellipse with eccentric angles $a,b,c$ respectively, I have arrived at the two sums listed above separately.
The first by equating the determinant of the triangle formed by the three normals to zero.
And the second by writing a fourth degree equation in $z=\cos{\theta}+i\sin{\theta}$. $$(a^2-b^2)z^4-2(ah-ibk)z^3+2(ah+ibk)z-(a^2-b^2)=0$$ where $i$ is the square root of $-1$ and $(h,k)$ is a point that the normal at $\theta$ passes through. In the latter, the product of the roots (in the fourth degree) happens to be $-1$ which gives the summation condition that if four points are co-normal then their eccentric angles sum to zero. Also the quadratic coefficient comes out to be zero. If one uses the above summation condition, one can easily derive the second sum.
I am unable to prove their equivalence, but have checked, through computation means that the second is a subset and the "missing" solutions are the ones where two of $a,b,c$ differ by a multiple of $2\pi$.
I wish to prove $$\sum_{cyc}\sin(a-b)\cdot\sin(2c)=0 \implies \sum_{cyc}\sin(a+b)=0$$ and that the aforementioned are the only "missing" solutions.