# Mollweide's formula

(Redirected from Mollweide's formulas)
Figure 1 – A triangle. The angles α, β, and γ are respectively opposite the sides a, b, and c.

In trigonometry, Mollweide's formula, sometimes referred to in older texts as Mollweide's equations,[1] named after Karl Mollweide, is a set of two relationships between sides and angles in a triangle.[2]

It can be used to check solutions of triangles.[3]

Let a, b, and c be the lengths of the three sides of a triangle. Let α, β, and γ be the measures of the angles opposite those three sides respectively. Mollweide's formula states that

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

and

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

Each of these identities uses all six parts of the triangle—the three angles and the lengths of the three sides.

## Derivation

### First Mollweide's formula

#### Method 1: Law of sines and angle bisector theorem

Figure 2 - Derivation of the first Mollweide's formula

The first Mollweide's formula can be derived from the law of sines and the angle bisector theorem.

In Figure 2, CD is the angle bisector of ${\displaystyle \gamma }$. From the angle bisector theorem, ${\displaystyle {\frac {b}{c_{1}}}={\frac {a}{c_{2}}}}$ but ${\displaystyle c_{2}=c-c_{1}}$. Rearranging, ${\displaystyle c_{1}={\frac {bc}{a+b}}}$ . ${\displaystyle \angle ADC=180^{\circ }-(\alpha +{\tfrac {\gamma }{2}})}$ and ${\displaystyle \gamma =180^{\circ }-(\alpha +\beta )}$. Applying the law of sines on ΔACD,

${\displaystyle {\frac {\sin {\frac {\gamma }{2}}}{c_{1}}}={\frac {\sin[180^{\circ }-(\alpha +{\frac {\gamma }{2}})]}{b}}}$. Substitute ${\displaystyle c_{1}={\frac {bc}{a+b}}}$ into the left denominator and using difference formula on the right numerator, after simplifying, we get ${\displaystyle {\frac {a+b}{c}}={\frac {\sin(\alpha +{\frac {\gamma }{2}})}{\sin {\frac {\gamma }{2}}}}}$. Substitute ${\displaystyle \gamma =180^{\circ }-(\alpha +\beta )}$ into the right numerator and upon simplifying, we get ${\displaystyle {\frac {a+b}{c}}={\frac {\sin[90^{\circ }+({\frac {\alpha -\beta }{2}})]}{\sin {\frac {\gamma }{2}}}}}$. Applying sum formula on the right numerator, we get the first Mollweide's formula ${\displaystyle {\frac {a+b}{c}}={\frac {\cos({\frac {\alpha -\beta }{2}})}{\sin {\frac {\gamma }{2}}}}}$.

#### Method 3: Law of sines and cosines

Applying the law of sines on Figure 1, ${\displaystyle a=b{\frac {\sin \alpha }{\sin \beta }}}$ and ${\displaystyle c=b{\frac {\sin \gamma }{\sin \beta }}}$. Applying the law of cosines, ${\displaystyle a^{2}=b^{2}+c^{2}-2bc\cos \alpha }$. Rearranging, ${\displaystyle {\frac {(a+b)(a-b)}{c}}=c-2b\cos \alpha }$. This is the beginning to get both the formulas. To get the first Mollweide's formula, rearranging further, ${\displaystyle {\frac {a+b}{c}}={\frac {c-2b\cos \alpha }{a-b}}}$. Substitute ${\displaystyle a=b{\frac {\sin \alpha }{\sin \beta }}}$ and ${\displaystyle c=b{\frac {\sin \gamma }{\sin \beta }}}$ in the right hand side of the equation and simplifying, ${\displaystyle {\frac {a+b}{c}}={\frac {\sin \gamma -2\cos \alpha \sin \beta }{\sin \alpha -\sin \beta }}}$. Substitute ${\displaystyle \gamma =180^{\circ }-(\alpha +\beta )}$ and use the sum and difference formulas and simplifying the right numerator, ${\displaystyle {\frac {a+b}{c}}={\frac {\sin \alpha \cos \beta -\sin \beta \cos \alpha }{\sin \alpha -\sin \beta }}}$. Using difference formula for sine on the right numerator, the equation becomes ${\displaystyle {\frac {a+b}{c}}={\frac {\sin(\alpha -\beta )}{\sin \alpha -\sin \beta }}}$. Using double angle formula for the right numerator and sum-to-product formula for the right denominator, the equation becomes ${\displaystyle {\frac {a+b}{c}}={\frac {2\sin {\frac {\alpha -\beta }{2}}\cos {\frac {\alpha -\beta }{2}}}{2\cos {\frac {\alpha +\beta }{2}}\sin {\frac {\alpha -\beta }{2}}}}}$ . Simplifying, ${\displaystyle {\frac {a+b}{c}}={\frac {\cos {\frac {\alpha -\beta }{2}}}{\cos {\frac {\alpha +\beta }{2}}}}}$. Substitute ${\displaystyle \alpha +\beta =180^{\circ }-\gamma }$, it becomes ${\displaystyle {\frac {a+b}{c}}={\frac {\cos {\frac {\alpha -\beta }{2}}}{\cos {\frac {180^{\circ }-\gamma }{2}}}}}$. Using the complementary angle theorem on the right denominator, we arrive at the first Mollweide's formula ${\displaystyle {\frac {a+b}{c}}={\frac {\cos {\frac {\alpha -\beta }{2}}}{\sin {\frac {\gamma }{2}}}}}$.

### Second Mollweide's formula

#### Method 2: Law of sines and cosines

The method is similar to the one for the first Mollweide's formula. We rearrange ${\displaystyle {\frac {(a+b)(a-b)}{c}}=c-2b\cos \alpha }$ to get ${\displaystyle {\frac {a-b}{c}}={\frac {c-2b\cos \alpha }{a+b}}}$. The right numerator is the same as for the first formula, that is ${\displaystyle 2\sin {\frac {\alpha -\beta }{2}}\cos {\frac {\alpha -\beta }{2}}}$. However, the right denominator is an addition instead of a subtraction, that is ${\displaystyle \sin \alpha +\sin \beta }$. Using the sum-to-product formula, it changes to ${\displaystyle 2\sin {\frac {\alpha +\beta }{2}}\cos {\frac {\alpha -\beta }{2}}}$. Thus, ${\displaystyle {\frac {a-b}{c}}={\frac {2\sin {\frac {\alpha -\beta }{2}}\cos {\frac {\alpha -\beta }{2}}}{2\sin {\frac {\alpha +\beta }{2}}\cos {\frac {\alpha -\beta }{2}}}}}$. Simplifying and substitute ${\displaystyle \alpha +\beta =180^{\circ }-\gamma }$ in the right denominator, we get ${\displaystyle {\frac {a-b}{c}}={\frac {\sin {\frac {\alpha -\beta }{2}}}{\sin {\frac {180^{\circ }-\gamma }{2}}}}}$. Using the complementary angle theorem on the right denominator, we arrive at the second Mollweide's formula ${\displaystyle {\frac {a-b}{c}}={\frac {\sin {\frac {\alpha -\beta }{2}}}{\cos {\frac {\gamma }{2}}}}}$.