Original question: Full question: Prove that the area of the triangle formed by three points of an ellipse, whose eccentric angles are $\theta , \phi$ and $\psi$ , is
$$\frac{1}{2}\sin\frac{\phi-\psi}{2}\sin\frac{\psi-\theta}{2}\sin\frac{\theta-\phi}{2}.$$
We first the area of triangle in a circle, then use linear transformations to boost the triangle into an ellipse
Let the coordinates be $ (\cos \alpha, \sin \alpha)$ , $ ( \cos \beta, \sin \beta)$, $ (\cos \gamma, \sin \gamma)$, then two of a triangle are given as: $d_1 = (\cos \alpha - \cos \beta, \sin \alpha - \sin \beta)$ and $d_2 = ( \cos \gamma - \cos \beta, \sin \gamma - \sin \beta)$, by trigonometric identities, we find:
$$ d_1 = (2 \sin \frac{\alpha + \beta}{2} \sin \frac{\alpha - \beta}{2}, 2 \cos \frac{\alpha + \beta}{2} \sin \frac{\alpha - \beta}{2} )$$
$$ d_2 = ( 2 \sin \frac{\gamma + \beta}{2} \sin \frac{\gamma - \beta}{2}, 2 \cos \frac{\gamma + \beta}{2} \sin \frac{\gamma-\beta}{2})$$
We find, $|d_1| =| 2 \sin \frac{\alpha- \beta}{2}|$ and $|d_2| = | 2 \sin \frac{\gamma - \beta}{2} |$, we may also find geometrically that the angle between $d_1$ and $d_2$ is $ |\frac{\alpha - \gamma}{2}|$ (draw a picture with the position vector as arrow). By the interpetation of cross product as area, we find:
$$ |d_1 \times d_2| =\frac{ |d_1| |d_2|}{2} |\sin \frac{\alpha - \gamma}{2}| = 2| \sin \frac{\alpha - \beta}{2} \sin \frac{\gamma-\beta}{2} \sin \frac{\alpha - \gamma}{2}|$$
I know an ellipse is formed by the map $x= \frac{X}{a}$ and $y= \frac{Y}{b}$, it is clear that the jacobian of this transform i.e: area factors by which squares in $x-y$ plane transform into $(X,Y)$ plane is $ ab$ and hence scale up the area I have by this factor but this still seems off from the required expression.
One immediate problem I see is, angle differences are not conserved when moving from the circular angle to the eccentric angle ( I think?), so I need to figure out a way to transform the circular angle differences into eccentric angle differences so that the equation works if there are no other mistakes.