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

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

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.

Title proof of Mollweide’s equations ProofOfMollweidesEquations 2013-03-22 12:50:10 2013-03-22 12:50:10 mathwizard (128) mathwizard (128) 5 mathwizard (128) Proof msc 51-00