Cosine theorem on the sphere

The cosine theorem on a sphere is an interesting relation that holds for a spherical triangle. One of it’s main applications is the computation of the air distance between two geographical points given by their latitude – longitude coordinates (I’ll show how in the next post).

Consider the spherical triangle NA'B' (see figure). The cosine theorem express a relation between the angles a, b, c and u which I’ll prove with basic geometry knowledge.

First of all note that AN and BN have nothing to do with the problem, they are just an auxiliary construction used for the proof. So first look only at the sphere and ignore AA', AN, B'B, BN, AB.

Then let’s consider AN the tangent to the arc A'N in the point N, and BN, the tangent to the arc B'N in the point N. Thus we’ll have AN||A''O and BN||B''O.
To prove the spherical cosine theorem the length of the segment AB can be written in two different ways: using cosine theorem in the triangle NAB and using cosine theorem in the triangle OAB.

Note that OAN is a right triangle(N being the right angle) because AN is a tangent to the sphere. Thus AN = R \cdot tg(a)  and OA = \frac{R}{ cos(a)}. Note that ON=R. Take it slowly to understand it, and make sure you visualise correctly the 3D figure :).  (*)

Same for triangle OBN we get: BN = R \cdot tg(b) and OB = \frac{R}{cos(b)}. (**)

From the cosine theorem in triangle OAB we get:
AB^2 = OA^2 + OB^2 - 2 \cdot OA \cdot OB \cdot cos(c)

Note that the angle ANB is u. From generalised cosine theorem in NAB we get:
AB^2 = NA^2 + NB^2 -2 \cdot NA \cdot NB \cdot cos(u)

It follows that: OA^2 + OB^2 - 2 \cdot OA \cdot OB \cdot cos(c) = NA^2 + NB^2 -2 \cdot NA \cdot NB \cdot cos(u)

Replacing with the values computed in (*) and (**) leads to:

\left(\frac{R}{cos(a)}\right)^2+\left(\frac{R}{cos(b)}\right)^2-2 \cdot \frac{R^2 \cdot cos(c)}{cos(a) \cdot cos(b)} =
\left(\frac{R \cdot sin(a)}{cos(a)}\right)^2+\left(\frac{R \cdot sin(b)}{cos(b)}\right)^2-2 \cdot \frac{R^2 \cdot sin(a) \cdot sin(b) \cdot cos(u)}{cos(a) \cdot cos(b)}

Now add the fractions on both sides of the equality by making them have the same denominator, and divide both sides by R^2:
cos(b)^2 + cos(a)^2 - 2 \cdot cos(a) \cdot cos(b) \cdot cos(c)=sin(a)^2 \cdot cos(b)^2 + sin(b)^2 \cdot cos(a)^2 - 2 \cdot sin(a) \cdot sin(b) \cdot cos(a) \cdot cos(b) \cdot cos(u) (***)

Because cos(x)^2+sin(x)^2=1 for every x:

cos(b)^2 - sin(a)^2 \cdot cos(b)^2 = cos(a)^2 \cdot cos(b)^2 and cos(a)^2 - sin(b)^2 \cdot cos(a)^2 = cos(a)^2 \cdot cos(b)^2

Then (***) become:

cos(a)^2 \cdot cos(b)^2 + cos(a)^2 \cdot cos(b)^2 - 2 \cdot cos(a) \cdot cos(b) \cdot cos(c)=-2 \cdot sin(a) \cdot sin(b) \cdot cos(a) \cdot cos(b) \cdot cos(u)

By dividing with 2 \cdot cos(a) \cdot cos (b) we get:

cos(a) \cdot cos(b) - cos(c) = -sin(a) \cdot sin(b) \cdot cos(u)  that can be written in a final form as:

cos(a)cos(b) + sin(a)sin(b)cos(u) = cos(c)

A remarkable thing is that when u is a right angle, cos(u)=0, and we get the Pithagora theorem on the sphere:

cos(a)cos(b) = cos(c)

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2 Responses to Cosine theorem on the sphere

  1. Pingback: Dirichlet’s Theorem and Liouville’s Extension of Dirichlet’s Theorem | Gaurav Happy Tiwari

  2. Pingback: Air distance | AlfaNotes

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