Spherical cosine law

This post is about the spherical cosine law which perhaps is well-known to many of you. However I am putting it here because I love and only know very simple geometry.

Let {\triangle ABC} be a triangle on the unit sphere {\mathbb{S}^2=\{(x, y, z)\in \mathbb{R}^3: x^2+y^2+z^2=1\}}, i.e. the 3 sides of {\triangle ABC} consists of geodesics segments on the sphere. Let {a, b, c} be the three sides opposing the three angles {A, B, C} respectively (we denote the vertices and the angle at these vertices by the same symbols).

Then we have the cosine law for spherical triangle

\displaystyle \boxed{ \cos c =\cos a \cos b + \cos C \sin a \sin b } \ \ \ \ \ (1)

When I first arrived at this formula, I did not realize that this is actually the spherical cosine law, which is closely related to the Euclidean cosine law for triangle:

\displaystyle \boxed{ c^2=a^2+b^2 -2ab \cos C } \ \ \ \ \ (2)

Actually the above cosine law can be viewed as the limiting case of the spherical version, and in turn the spherical cosine law is derived from Euclidean geometry. We will look at how these two are related after the proof.

The proof is quite simple. We can without loss of generality assume that {C} is the north pole and {A} lies on the {x-z} plane. Note that then {B} lies on the longitude whose plane containing it makes an angle {C} with the {x-z} plane (why?). Since the radius of the sphere is 1, the angle {OA} made with the z-axis is exactly the length of {CA} (arc-length{=r\theta}), which is {b} by definition, and similarly the angle {OB} made with the {z}-axis is {a}. So in spherical coordinates, the 3 points {A, B, C} are given by

\displaystyle A=(\sin b, 0, \cos b), B=(\sin a \cos C, \sin a \sin C, \cos a), C=(0, 0, 1)

Now the distance {AB}, which is c by definition, is the same as the angle between {OA} and {OB} by the previous remark. So by applying dot product to the vectors {OA} and {OB}, we arrived at

\displaystyle \cos c = \vec{OA}\cdot \vec{OB}=\cos a \cos b + \cos C \sin a \sin b \ \ \ \ \ (3)

This finishes the proof of (1).

We now show that (2) can be viewed as the limiting case of (1). Intuitively, the spherical triangle becomes flatter and flatter when the sides a, b tend to zero, and thus more and more like a Euclidean triangle on a plane. To get (2), note that the Taylor series of {\sin} and {\cos} are given by

\displaystyle \cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots

\displaystyle \sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots

Applying these to (1) except the term {\cos C}, and dropping all the term of order higher than 2, we have

\displaystyle \begin{array}{rcl} 1-\frac{c^2}{2}+\cdots&=&(1-\frac{a^2}{2}+\cdots)(1-\frac{b^2}{2}+\cdots)+(a-\cdots)(b-\cdots)\cos C\\ \Rightarrow 1-\frac{c^2}{2}&=&1-\frac{a^2}{2}-\frac{b^2}{2}+ab\cos C+\text{ third order terms}\\ \Rightarrow c^2&=&a^2+b^2-2ab\cos C+\text{ third order terms}\\ \end{array}

Thus the Euclidean cosine law (2) is the limiting version of the spherical cosine law as the sides tend to zero by dropping the third or higher order terms (which corresponds to the “curvedness” of the sphere).

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4 Responses to Spherical cosine law

  1. mahalakshmi says:

    Really it helps the learners a lot

  2. wonderful put up, very informative. I ponder why
    the opposite experts of this sector don’t notice this.
    You should proceed your writing. I’m sure, you have a great readers’ base already!

  3. It’s very trouble-free to find out any topic on net as compared to books, as I found this piece
    of writing at this website.

  4. I have just developed a 3-D GeoGebra applet (with PDF explanation) on this.
    Here is the link: https://www.geogebra.org/m/rBEEMJG3

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