Mollweide's Formulas

Ching-Kuang Shene (冼鏡光), Professor Emeritus
Department of Computer Science
Michigan Technological University
Houghton, MI 49931
USA
Created December 9, 2025
Modified February 25, 2026

Given a triangle \( \bigtriangleup ABC \) with sides \( a=\overline{BC}\), \( b=\overline{CA}\) and \( c=\overline{AC}\) and the angle opposite to each side \( \alpha\), \( \beta\) and \( \gamma\) as shown below, Mollweide's formula can be stated as follows: $$ \begin{eqnarray*} \frac{a-b}{c} &=& \frac{\sin\left( \frac{\alpha-\beta}{2}\right)}{\cos\left( \frac{\gamma}{2}\right)} \\ \frac{a+b}{c} &=& \frac{\cos\left( \frac{\alpha-\beta}{2}\right)}{\sin\left( \frac{\gamma}{2}\right)}. \end{eqnarray*} $$ Given the three side lengths \( a, b \) and \( c \), there are six permutations: \(a-b-c, a-c-b, b-a-c, b-c-a, c-a-b \) and \(c-b-a\). Each permutation has its corresponging formula, and, hence, there are six pairs of Mollweide's formulas.

An Algebraic Proof

The following proof uses the Law of Sines, which is simple and straightforward. From the above diagram, the Law of Sines shows the following,: \[ \frac{a}{\sin(\alpha)} = \frac{b}{\sin(\beta)} = \frac{c}{\sin(\gamma)} = k \] As a result, the sides are: \[ a = k\sin(\alpha)\mbox{,\ \ \ \ \ } b = k\sin(\beta)\mbox{\ \ \ \ \ and\ \ \ \ \ } c = k\sin(\gamma). \] In this way, with the help of the Sum-to-Product identities, we have $$ \begin{eqnarray*} a - b &=& k\left[ \sin(\alpha) - \sin(\beta) \right] = 2k\left[ \sin\left(\frac{\alpha-\beta}{2}\right)\cos\left(\frac{\alpha+\beta}{2}\right)\right] \\ a + b &=& k\left[ \sin(\alpha) + \sin(\beta) \right] = 2k\left[ \sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)\right] \end{eqnarray*} $$ Applying the double angle identity to \( \sin(\gamma)\) gives $$ \begin{eqnarray*} \frac{a-b}{c} &=& \frac{\sin\left(\frac{\alpha-\beta}{2}\right)\cos\left(\frac{\alpha+\beta}{2}\right)}{\sin\left( \frac{\gamma}{2}\right)\cos\left( \frac{\gamma}{2}\right)} \\ \frac{a+b}{c} &=& \frac{\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)}{\sin\left( \frac{\gamma}{2}\right)\cos\left( \frac{\gamma}{2}\right)} \end{eqnarray*} $$ The terms with \( \frac{\alpha+\beta}{2}\) can be cancelled using \( \frac{\gamma}{2}=\frac{180^{\circ}-(\alpha+\beta)}{2} = 90^{\circ} - \frac{\alpha+\beta}{2}\), \( \sin\left(\frac{\gamma}{2}\right) = \cos\left(\frac{\alpha+\beta}{2}\right)\) and \( \cos\left(\frac{\gamma}{2}\right) = \sin\left(\frac{\alpha+\beta}{2}\right)\). $$ \begin{eqnarray*} \frac{a-b}{c} &=& \frac{\sin\left(\frac{\alpha-\beta}{2}\right)}{\cos\left( \frac{\gamma}{2}\right)} \\ \frac{a+b}{c} &=& \frac{\cos\left(\frac{\alpha-\beta}{2}\right)}{\sin\left( \frac{\gamma}{2}\right)} \end{eqnarray*} $$

A Geometric Proof

Given a triangle \( \bigtriangleup ABC\), let \( \alpha=\angle BAC\), \( \beta=\angle ABC\), \( \gamma=\angle BAC\), \( a=\overline{BC}\), \( b=\overline{CA}\) and \( c=\overline{AB}\). We have two cases to consider: \( a> b \) and \( a < b \).

Case \( a > b \):

Let \( D\) be a point on \( \overline{BC}\) such that \( \overline{CD}=b\). Hence, \( \overline{BD}=a-b\). Let the perpendicular feet from \( C\) and \( B\) to \( \overleftrightarrow{AD}\) be \( E\) and \( F\), respectively, and let \( e = \overline{BF}\). In this way, \( \overleftrightarrow{CE}\) bisects \( \gamma = \angle ACD\) and hence \( \angle ACE = \angle DCE = \frac{\gamma}{2}\). Through \( A\) construct a line parallel to \( \overleftrightarrow{CE}\), meeting \( \overleftrightarrow{BC}\) at \( G\). Therefore, \( \bigtriangleup ACG\) ia an isosceles triangle with \( \overline{AC}=\overline{CG}=b\) and \( \angle CAG=\angle CGA = \frac{\gamma}{2}\). Finally, let the perpendicular foot from \( B\) to \( \overleftrightarrow{AG}\) be \( H\). Then, we have \( \overline{BG}=a+b\), \( \overline{BD}=a-b\). What we have to do is determing various angle with respect to \( \alpha\), \( \beta\) and \( \gamma\).

First, note that \( \bigtriangleup CAD\) is an isosceles triangle with \( \overline{CA}=\overline{CD}\). We have \( \angle CAD = \angle CDA\). Furthermore, we have the following: \[ \angle CAD = 90^{\circ} -\frac{\gamma}{2} = 90^{\circ}-\frac{180^{\circ}-(\alpha+\beta)}{2} = \frac{\alpha+\beta}{2}. \] Then, the \( \angle BAD\) is calculated as follows: \[ \angle BAD = \alpha - \angle CAE = \alpha - \frac{\alpha+\beta}{2} = \frac{\alpha-\beta}{2}. \] Of course, \( \angle BAH\) is \[ \angle BAH = 90^{\circ}-\frac{\alpha-\beta}{2}. \]

From \( \bigtriangleup BDF\), we have \[ e = \overline{BD}\cdot\sin(\angle BDF) = (a-b)\sin\left( \frac{\alpha+\beta}{2}\right). \] From \( \bigtriangleup BAF\), we have \[ e = \overline{BA}\cdot\sin(\angle BAF) = c\cdot\sin\left( \frac{\alpha-\beta}{2}\right). \] The above two equations give \[ (a-b)\sin\left( \frac{\alpha+\beta}{2}\right) = c\cdot\sin\left( \frac{\alpha-\beta}{2}\right). \] Because \( \frac{\alpha+\beta}{2}= 90^{\circ}-\frac{\gamma}{2}\) as shown earlier, we finally get one of the two desired results: \[ \frac{a-b}{c} = \frac{ \sin\left( \frac{\alpha-\beta}{2}\right)} { \sin\left( \frac{\alpha+\beta}{2}\right)} = \frac{ \sin\left( \frac{\alpha-\beta}{2}\right)}{\sin\left( 90^{\circ}-\frac{\gamma}{2}\right)} = \frac{ \sin\left( \frac{\alpha-\beta}{2}\right)}{\cos\left( \frac{\gamma}{2} \right)}. \]

The second formula can be proved in a similar way. From \( \bigtriangleup BHG\), we have \[ f = \overline{BG}\cdot\sin(\angle BGH) = (a+b)\sin\left( \frac{\gamma}{2}\right). \] From \( \bigtriangleup BAH\), we have \[ f = \overline{BA}\cdot\sin(\angle BAH) = c\cdot\sin\left( 90^{\circ}-\frac{\alpha-\beta}{2}\right) = c\cdot\cos\left( \frac{\alpha-\beta}{2}\right). \] The above two equations give \[ (a+b)\sin\left( \frac{\gamma}{2}\right) = c\cdot\cos\left( \frac{\alpha-\beta}{2}\right). \] Therefore, the secone equation is proved: \[ \frac{a+b}{c} = \frac{ \cos\left( \frac{\alpha-\beta}{2}\right)} { \sin\left( \frac{\gamma}{2}\right)}. \]

Case \( b < a \):

If an algebraic proof is preferred, the following is a simple one: \[ \frac{a-b}{c} = -\frac{b-a}{c} = -\frac{\sin\left(\frac{\beta-\alpha}{2} \right)}{\cos\left( \frac{\gamma}{2}\right)} = \frac{-\sin\left(\frac{\beta-\alpha}{2} \right)}{\cos\left( \frac{\gamma}{2}\right)} = \frac{\sin\left(\frac{-(\beta-\alpha)}{2} \right)}{\cos\left( \frac{\gamma}{2}\right)} = \frac{\sin\left(\frac{\alpha-\beta}{2} \right)}{\cos\left( \frac{\gamma}{2}\right)}. \] A geometric proof similar to the case of \( a > b \) is possible. In this case, \( D\) is outside the segment \( \overline{BC}\) and \( \angle CAD > \alpha \). As a result, \( \overline{BD}=b-a\). Please see the left diagram below.

Mollweide's Formula Proof (Part 1)
Mollweide's Formula Proof (Part 2)

In this way, from \( \bigtriangleup BDF\) we have the following: \[ e =(b-a)\sin\left( \frac{\alpha+\beta}{2}\right). \] From \( \bigtriangleup BAF\), we have \[ \angle BAF = \angle DAC - \angle BAC = \frac{\alpha+\beta}{2}-\alpha = \frac{-(\alpha-\beta)}{2}. \] Hence, in \( \bigtriangleup BAF\) we have \[ e=c\cdot\sin\left( -\frac{\alpha-\beta}{2}\right) = -c\cdot\sin\left( \frac{\alpha-\beta}{2}\right). \] Consequently, the result is: \[ (b-a)\sin\left( \frac{\alpha+\beta}{2}\right) = -c\cdot\sin\left( \frac{\alpha-\beta}{2}\right). \] and hence \[ \frac{(a-b)}{c}= \frac{\sin\left( \frac{\alpha-\beta}{2}\right)}{\sin\left( \frac{\alpha+\beta}{2}\right)} = \frac{\sin\left( \frac{\alpha-\beta}{2}\right)}{\cos\left( \frac{\gamma}{2}\right)}\]

As for the second formula, please see the right diagram above in which only relevant portion is shown. From \( \bigtriangleup BAH\), we have \[ \angle BAH = \alpha + \frac{\gamma}{2} = \alpha + \frac{180^{\circ}-(\alpha+\beta)}{2} = 90^{\circ} + \frac{\alpha-\beta}{2} \] and hence \[ f = \overline{AB}\cdot\sin(\angle BAH) = c\cdot\sin\left( 90^{\circ}+\frac{\alpha-\beta}{2} \right) = c\cdot\cos\left( \frac{\alpha-\beta}{2}\right). \] From \( \bigtriangleup BHG\), we have \[ f = \overline{BG}\cdot\sin\left( \frac{\gamma}{2} \right) = (a+b)\sin\left( \frac{\gamma}{2} \right). \] Combining the above two equation gives \[ (a+b)\sin\left( \frac{\gamma}{2} \right) = c\cdot\cos\left( \frac{\alpha-\beta}{2}\right) \] and hence \[ \frac{a+b}{c} = \frac{\cos\left( \frac{\alpha-\beta}{2}\right)}{\sin\left( \frac{\gamma}{2} \right)}. \]

Therefore, we get the same results whether \( a > b\) or \( a < b \). The case of \( a =b\) is obvious and is not discussed.

Discussions

I personally like the algebraic proof more than the geometric one, because it does not have to deal with special cases.

There are other geometric proofs available, published and on the web, most of them only address the case of \( a > b\) and did not even mention the case of \( a < b\). This is particularly true for those PWW (i.e., proof-without-word) type proofs. Due to its diagrammatic nature, PWW proofs usually shows a single diagram, which is perhaps the most obvious and easiest case and the authors seem hesitate to explain how the lengths and angles are obtained. As a result, some PWW proofs are not so easy to understanding while some others lack of generality. The proof of Mollweide's Formula is a good example. This is the major reason for me to discuss the proof using the Law of Sines first. Comparing this algebraic proof with the geometric proof, which includes both cases, the algebraic one is not only easier but also straightforward and general.

Note that the seond Mollweide's formula is usually referred to as Newton's Formula. For more history regarding Mollweide's Formula, please refer to this short article by Rex H. Wu (2007): The Story of Mollweide and Some Trigonometric Identities. This article also includes a few proofs. You may also like Karjanto's Mollweide’s formula in teaching trigonometry. The proof presented here is a combination of the proofs of both formulas in one diagram and also considered the cases of \( a > b\) and \( a < b\). I just dislike to present anything that is half-baked.

From Mollweide Formulas, it is easy to proof the Pythagorean Theorem.

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