# Mollweide's Formula Proof—Triangle Trigonometry

Mollweide's Formula (actually, a pair of formulas) is an equation that relates the three side lengths and three angles of a triangle. If a triangle has sides of length A, B, and C, and the opposite angles are respectively x, y, and z, then the first formula of Mollweide states that

(A+B)/C = cos[(x-y)/2] / sin(z/2)

The second formula is similar:

(A-B)/C = sin[(x-y)/2] / cos(z/2)

Both of these formulas can be used to check the measurements of a triangle or solve for a missing quantity. To prove these relations are true, we need to use several basic trigonometric identities and formulas for sums and products of sine and cosine. The proofs are given below with some examples of how to use Mollweide's Formulas.

## Derivation of (A+B)/C = cos[(x-y)/2] / sin(z/2)

We prove Mollweide's first formula by deriving it from first principles.

If x, y, and z are the angles of the a triangle, we start with the fact that x + y + z = 180, which implies that z = 180 - (x+y) and z/2 = 90 - (x+y)/2. We will also need the trigonometric identities sin(θ) = cos(90-θ) = sin(180-θ), for any angle θ between 0 and 180 degrees. Beginning from these basic trigonometric and geometric relations we get

**sin(z)** = sin(180 - z) = sin(180 - (180 - (x+y))) = **sin(x+y)**

Next, we need to invoke the double angle formula for sine, 2sin(θ)cos(θ) = sin(2θ). This can also be expressed as 2sin(θ/2)cos(θ/2) = sin(θ). This gives us

**sin(z)** = sin(x+y) = **2sin[(x+y)/2]cos[(x+y)/2] **

Now we do some algebraic rearrangement to get

**2sin[(x+y)/2] / sin(z) = 1 / cos[(x+y)/2]**

Using the fact that sin(θ) = cos(90-θ) = sin(180-θ), we can see that cos[(x+y)/2] is equal to cos[90 - z/2], which equals sin(z/2). This gives us a new expression for the right hand side:

**2sin[(x+y)/2] / sin(z) = 1 / sin(z/2)**

Multiplying both sides by cos[(x-y)/2] gives us

**2sin[(x+y)/2]cos[(x-y)/2] / sin(z) = cos[(x+y)/2] / sin(z/2)**

*Now that we have the right-hand side of Mollweide's first formula down, how do we convert the left hand-side into an expression involving the side lengths rather than the angles?*

First, the left hand side can be transformed using the sum-product identity for sines: sin(P) + sin(Q) = 2sin[(P+Q)/2]cos[(P-Q)/2]. This gives us

**[sin(x) + sin(y)] / sin(z) = cos[(x+y)/2] / sin(z/2)**

**sin(x)/sin(z) + sin(y)/sin(z) = cos[(x+y)/2] / sin(z/2)**

Finally, we use the **Law of Sines**, which states that if a triangle has sides A, B, and C with respective opposite angles x, y, and z, then sin(x)/A = sin(y)/B = sin(z)/C. These ratios can be reconfigured to produce A/C = sin(x)/sin(z) and B/C = sin(y)/sin(z). Making these substitutions on the left hand side gives us Mollweide's Formula:

**A/C + B/C = cos[(x+y)/2] / sin(z/2)**

**(A+B)/C = cos[(x+y)/2] / sin(z/2)**

and thus the proof and derivation is complete.

## Proof that (A-B)/C = sin[(x-y)/2] / cos(z/2)

The second version of Mollweide's equation for triangles can be proved similar to the method above. We start with the equality

**2sin(z/2)cos(z/2) = sin(z)**

Using the facts that z/2 = 90 - (x+y)/2 and that sin(θ) = cos(90-θ), we get

**2cos[(x+y)/2]cos(z/2) = sin(z)**

**2cos[(x+y)/2] / sin(z) = 1 / cos(z/2)**

**2cos[(x+y)/2]sin[(x-y)/2] / sin(z) = sin[(x-y)/2] / cos(z/2)**

Now we invoke the sine difference formula, which says that for any angles P and Q, sin(P) - sin(Q) = cos[(P+Q)/2]sin[(P-Q)/2]. Applying this trig identity gives us a new expression for the left hand side:

**[sin(x) - sin(y)] / sin(z) = sin[(x-y)/2] / cos(z/2)**

Finally, using the Law of Sines, sin(x)/A = sin(y)/B = sin(z)/C, we get

**sin(x)/sin(z) - sin(y)/sin(z) = sin[(x-y)/2] / cos(z/2)**

**A/C - B/C = sin[(x-y)/2] / cos(z/2)**

**(A-B)/C = sin[(x-y)/2] / cos(z/2)**

The next section shows how these formulas can be useful in solving geometric problems involving triangles, something that is often encountered in real life, for example, when constructing buildings or measuring irregularly shaped plots of land.

## Example 1: Missing Sides of Isosceles Triangle

An isosceles triangle has angles of 70 degrees, 55 degrees, and 55 degrees. The longest side of the triangle (opposite the 70-degree angle) has a length of 1089 cm. What are the lengths of the other two sides?

For this problem, we can set C = 1089, A = B, z = 70, and x = y = 55. The missing value is A. Using the first formula of Mollweide, we can write

(A+B)/C = cos[(x-y)/2] / sin(z/2)

2A/1089 = cos[(55-55)/2] / sin(70/2)

(2/1089)A = cos(0) / sin(35)

(2/1089)A = 1 / sin(35)

A = 1089 / [2sin(35)]

A ≈ 949.307

Therefore, the missing sides of the triangle both have a length of about 949.307 centimeters.

## Example 2: Existence of a Special Scalene Triangle

Does there exist a scalene triangle such that the following two conditions are met?

- The medium side is the average of the long and short sides.
- The medium angle is the average of the large and small angles.

Let the longest side be A , the shortest side be B, and the medium-length side be C = (A+B)/2. Let the largest angle be x, the smallest angle be y, and the medium angle be z = (x+y)/2. The first formula of Mollweide gives us

(A+B) / [(A+B)/2] = cos[(x-y)/2] / sin[(x+y)/4]

2 = cos[(x-y)/2] / sin[(x+y)/4]

Since x + y + z = 180, we get x + y = 120 and z = 60. This allows us to further simplify the equation:

2 = cos[(x-y)/2] / sin(30)

2 = cos[(x-y)/2] / (1/2)

1 = cos[(x-y)/2]

arccos(1) = 0 = (x-y)/2

0 = x - y

x = y

Since x = y and x + y = 120, we get x = y = z = 60. This is not a scalene triangle, but an equilateral one. The only triangle whose medium side is the average of the other two sides, and whose medium angle is the average of the other two angles is an equilateral triangle. Therefore, no scalene triangle exists that satisfies the given conditions.

## About Karl Mollweide

Karl Mollweide (1774 - 1825) was a German mathematician most famous for developing what is known as the Mollweide map projection, a 2-D map of Earth that corrects many of the distortions found in the Mercator projection. The other mathematical discovery that bears his name is the pair of trigonometric identities for triangles.

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