Derive Mollweid's Formula

[nycmath]

Show your work in step by step fashion, preferably using the provided LaTex.

[Guest]

Everything comes out of the law of sines, so let's start there.

Since [tex]\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R[/tex], we can write [tex]a=2R\sin A[/tex], [tex]b=2R\sin B[/tex] and [tex]c=2R\sin C[/tex]. Put those into the left side and the [tex]2R[/tex] just cancles:

[tex]\frac{a-b}{c}=\frac{\sin A-\sin B}{\sin C}[/tex]

Now the top is a difference of sines and the bottom is a double angle:

[tex]\sin A-\sin B=2\cos\left[\tfrac{1}{2}(A+B)\right]\sin\left[\tfrac{1}{2}(A-B)\right][/tex]

[tex]\sin C=2\sin\tfrac{1}{2}C\cos\tfrac{1}{2}C[/tex]

The angles of a triangle add up to [tex]180^\circ[/tex], so [tex]\tfrac{1}{2}(A+B)=90^\circ-\tfrac{1}{2}C[/tex], and that means [tex]\cos\left[\tfrac{1}{2}(A+B)\right]=\sin\tfrac{1}{2}C[/tex]. Substitue that in:

[tex]\frac{a-b}{c}=\frac{2\sin\tfrac{1}{2}C,\sin\left[\tfrac{1}{2}(A-B)\right]}{2\sin\tfrac{1}{2}C,\cos\tfrac{1}{2}C}=\frac{\sin\left[\tfrac{1}{2}(A-B)\right]}{\cos\tfrac{1}{2}C}[/tex]

[nycmath]

Everything comes out of the law of sines, so let's start there.

Since [tex]\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R[/tex], we can write [tex]a=2R\sin A[/tex], [tex]b=2R\sin B[/tex] and [tex]c=2R\sin C[/tex]. Put those into the left side and the [tex]2R[/tex] just cancles:

[tex]\frac{a-b}{c}=\frac{\sin A-\sin B}{\sin C}[/tex]

Now the top is a difference of sines and the bottom is a double angle:

[tex]\sin A-\sin B=2\cos\left[\tfrac{1}{2}(A+B)\right]\sin\left[\tfrac{1}{2}(A-B)\right][/tex]

[tex]\sin C=2\sin\tfrac{1}{2}C\cos\tfrac{1}{2}C[/tex]

The angles of a triangle add up to [tex]180^\circ[/tex], so [tex]\tfrac{1}{2}(A+B)=90^\circ-\tfrac{1}{2}C[/tex], and that means [tex]\cos\left[\tfrac{1}{2}(A+B)\right]=\sin\tfrac{1}{2}C[/tex]. Substitue that in:

[tex]\frac{a-b}{c}=\frac{2\sin\tfrac{1}{2}C,\sin\left[\tfrac{1}{2}(A-B)\right]}{2\sin\tfrac{1}{2}C,\cos\tfrac{1}{2}C}=\frac{\sin\left[\tfrac{1}{2}(A-B)\right]}{\cos\tfrac{1}{2}C}[/tex]

Wow! Impressive. Do you have a math degree? Are you a math teacher?