Mollweide’s formula, sometimes also referred to as Mollweide’s equation, is a set of two relationships between sides and angles in a triangle. This equation is particularly useful in checking one’s result after solving an oblique triangle since all six components of the triangle are involved.
$$\frac{a+b}{c}=\frac{\cos{\left(\frac{\alpha-\beta}{2}\right)}}{\sin{\left(\frac{\gamma}{2}\right)}}\tag{1}$$
and
$$\frac{a-b}{c}=\frac{\sin{\left(\frac{\alpha-\beta}{2}\right)}}{\cos{\left(\frac{\gamma}{2}\right)}}\tag{2}$$
The equations adopt their name from a German mathematician and astronomer Karl Brandan Mollweide. Nonetheless, this pair of equations was discovered earlier by Isaac Newton and fully developed by Thomas Simpson. An excellent overview of the history of Mollweide’s formula is given by Wu (2007).
For proofs of the Mollweide's formula we invite the readers to see Karjanto's article on the subject (see here).
Generalization. Let $a$, $b$, $c$ and $d$ be the sides of a cyclic convex quadrilateral. Let $\angle{DAB}=\alpha$, $\angle{ABC}=\beta$, $\angle{BCD}=\gamma$ and $\angle{CDA}=\delta$. Denote $\theta$ the angle between the diagonals (see figure below), then the following identity holds
$$\frac{\sin{\left(\frac{\alpha+\beta}{2}\right)}}{\cos{\left(\frac{\gamma-\delta}{2}\right)}}=\frac{a+c}{b+d}\cot{\left(\frac{\theta}{2}\right)}\tag{3}$$
Proof. We take advantage of the cyclic nature of the half-angle formulas (see $(1)$ here) in combination with the formulas of compound angles.
$$\frac{\sin{\left(\frac{\alpha+\beta}{2}\right)}}{\cos{\left(\frac{\gamma-\delta}{2}\right)}}=\frac{\sin{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}+\cos{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}}{\cos{\frac{\gamma}{2}}\cos{\frac{\delta}{2}}+\sin{\frac{\gamma}{2}}\sin{\frac{\delta}{2}}}$$
Substituting from the half-angle formulas
$$\frac{\sin{\left(\frac{\alpha+\beta}{2}\right)}}{\cos{\left(\frac{\gamma-\delta}{2}\right)}}=\frac{\sqrt{\frac{(s-a)(s-d)}{ad+bc}}\sqrt{\frac{(s-c)(s-d)}{ab+cd}}+\sqrt{\frac{(s-b)(s-c)}{ad+bc}}\sqrt{\frac{(s-a)(s-b)}{ab+cd}}}{\sqrt{\frac{(s-a)(s-d)}{ad+bc}}\sqrt{\frac{(s-a)(s-b)}{ab+cd}}+\sqrt{\frac{(s-b)(s-c)}{ad+bc}}\sqrt{\frac{(s-c)(s-d)}{ab+cd}}}$$
Simplifying and factorizing
$$\frac{\sin{\left(\frac{\alpha+\beta}{2}\right)}}{\cos{\left(\frac{\gamma-\delta}{2}\right)}}=\sqrt{\frac{(s-a)(s-c)}{(s-b)(s-d)}}\cdot{\frac{(s-d)+(s-b)}{(s-a)+(s-c)}}$$
It is well-known that $\tan{\left(\frac{\theta}{2}\right)}=\sqrt{\frac{(s-b)(s-d)}{(s-a)(s-c)}}$ (see here), so the formula reduces to
$$\frac{\sin{\left(\frac{\alpha+\beta}{2}\right)}}{\cos{\left(\frac{\gamma-\delta}{2}\right)}}=\frac{a+c}{b+d}\cot{\left(\frac{\theta}{2}\right)}$$
$\square$
Note that as with Mollweide's formulas, this version for a cyclic quadrilateral not only relates the four sides to the four angles, but also includes the angle between the diagonals. I have to admit I was skeptical as to whether it was really a generalization of Mollweide's formula (it was not clear to me). I have published the result in a forum where someone whose nickname is Blue has shown that indeed the formula $(3)$ generalizes Mollweide's formula, more specifically, it generalizes Newton's version (see here).
You can download the entire article here. And here, you can download comments from reviewers of the Mathematical Association of America.
Update. I am pleased to announce that this generalization was published by the Romanian journal MATINF.
This is another example of what the half angle formulas can do for us. See also
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