proof of Mollweide’s equations

We transform the equation

$$(a+b)\sin \frac{\gamma }{2}=c\cos \left(\frac{\alpha -\beta }{2}\right)$$

to

$$a\cos \left(\frac{\alpha }{2}+\frac{\beta }{2}\right)+b\cos \left(\frac{\alpha }{2}+\frac{\beta }{2}\right)=c\cos \frac{\alpha }{2}\cos \frac{\beta }{2}+c\sin \frac{\alpha }{2}\sin \frac{\beta }{2},$$

using the fact that $\gamma =\pi -\alpha -\beta$. The left hand side can be further expanded, so that we get:

$$a\left(\cos \frac{\alpha }{2}\cos \frac{\beta }{2}-\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\right)+b\left(\cos \frac{\alpha }{2}\cos \frac{\beta }{2}-\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\right)=c\cos \frac{\alpha }{2}\cos \frac{\beta }{2}+c\sin \frac{\alpha }{2}\sin \frac{\beta }{2}.$$

Collecting terms we get:

$$(a+b-c)\cos \frac{\alpha }{2}\cos \frac{\beta }{2}-(a+b+c)\sin \frac{\alpha }{2}\sin \frac{\beta }{2}=0.$$

Using $s:=\frac{a+b+c}{2}$ and using the equations

$\sin {\displaystyle \frac{\alpha }{2}}$$=$$\sqrt{{\displaystyle \frac{(s-b)(s-c)}{bc}}}$

$\cos {\displaystyle \frac{\beta }{2}}$$=$$\sqrt{{\displaystyle \frac{s(s-a)}{bc}}}$

we get:

$$2\frac{s(s-c)}{c}\sqrt{\frac{(s-a)(s-b)}{ab}}-2\frac{s(s-c)}{c}\sqrt{\frac{(s-a)(s-b))}{ab}}=0,$$

which is obviously true. So we can prove the first equation by going backwards. The second equation can be proved in quite the same way.