Question

Mollweide’s Formula Another form of Mollweide’s Formula is $\frac{a-b}{c}=\frac{\sin \left[{\displaystyle \frac{1}{2}}\left(A-B\right)\right]}{\cos \left({\displaystyle \frac{1}{2}}C\right)}$

Derive it.

Short answer

This is proved that $\frac{a-b}{c}=\frac{\sin \left[{\displaystyle \frac{1}{2}\left(A-B\right)}\right]}{\cos \left({\displaystyle \frac{1}{2}C}\right)}$$\frac{a-b}{c}=\frac{\sin \left[{\displaystyle \frac{1}{2}\left(A-B\right)}\right]}{\cos \left({\displaystyle \frac{1}{2}C}\right)}$$\frac{a-b}{c}=\frac{\sin \left[{\displaystyle \frac{1}{2}}\left(A-B\right)\right]}{\cos \left({\displaystyle \frac{1}{2}}C\right)}$

Step-by-step solution

Step 1. Given Information

Use $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}LawofSines$

Sum of angles in a triangle$A+B+C={180}^{\circ }$

Sum to Product formula$\sin A-\sin B=2\sin \frac{A-B}{2}\cos \frac{A+B}{2}$