Dual form of the spherical cosine law

We have discussed the spherical cosine law before, namely, given 2 sides {a}, {b} and its included angle {C} of a spherical triangle {\triangle ABC}, we can find the length of the opposite side {c} by

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

Now I would like to ask the dual problem. That is,
Question: on a spherical triangle {\triangle ABC}, given the two angles {A, B} and the length {c} of the side between them, what is the angle {C} which is opposite to the side {c}?

Remark 1 Note that unlike the Euclidean case, two angles of a spherical triangle does not determine the third angles. An example is that {A, B} lie on the equator and {C} is the north pole, clearly we can always move {A, B} along the equator and such that the two angles {A, B} are always {90^\circ}, but the triangle formed can be different.

It turns out that we can use the original cosine law to get the “dual” version of cosine law, or the cosine law for angles (whereas (1) can be called the cosine law for sides ). The formula is given by (note that capital letters are angles and small letters are sides)

\displaystyle  \boxed{ \cos C=-\cos A \cos B+\sin A\sin B \cos c} \ \ \ \ \ (2)

You won’t miss the observation that this is very similar to (1), but is a bit different. (Actually I have worked out this by some lengthy calculation in Euclidean space, and realize this similarity afterwards. ) Let us set up some notations first. We assume that all sides have a given (fixed) orientation. For example {AB} and {BA} are considered to be different when considered as a side (instead of just its length). All great circles are considered to have an orientation too. We define

Definition 1 For each {p\in \mathbb{S}^2}, its polar line to be the (oriented) great circle {C} such that when we rotate {C} to lie on the {x-y} plane with positive orientation (i.e. anticlockwise when viewed from top), then (the rotated) {p} is the north pole. We also say {p} is the polar point of such a great circle {C}. We can also define (in the obvious way) the polar point of a segment {AB} (in that orientation) of a great circle.

Definition 2 For a spherical triangle {\triangle ABC}, a (spherical) triangle {\triangle a'b'c'} is called the dual triangle of {\triangle ABC} if {c'} is the polar point of {AB}, {b'} is the polar point of {CA}, {a'} is the polar point of {BC}, {C} is the polar point of {a'b'}, {B} is the polar point of {c'a'} and {A} is the polar point of {b'c'}.

I will drop the adjective “spherical” from now on. Roughly speaking, the “angle” of the dual triangle corresponds to the “side” of the original triangle (well… not exactly, see Proposition 4) and the side of the dual triangle corresponds to the angle of the original triangle:

Proposition 3 The dual triangle of every triangle exists (and clearly is unique, as the vertices are three fixed points).

Proof: The obvious way to construct the dual triangle of {\triangle ABC} is to take the triangle {\triangle a'b'c'} whose side is the polar lines of the three vertices (with the obvious orientation). The last three conditions are automatically satisfied. For the first three conditions, since {c'} lies in the intersection of {\vec A^\perp} (polar line of {A}!) and {\vec B^\perp}, with our choice of the orientation, it is clear that {c'} is the polar point of {AB}.\Box

Proposition 4 With the notations as above, for {\triangle ABC}, the length {b'c'} of the dual triangle, {\triangle a'b'c'}, is exactly equal to the angle {\pi-A}.

Proof: If {A} is the north pole, {b'c'} is exactly the segment of the equator on the xy plane, such that the endpoint {c'} is perpendicular to the plane {OAB}, and the endpoint {b'} is perpendicular to the plane {OAC}. The angle between these two planes is {A}, so the angle between the two normals is {\pi-A}. \Box

Proposition 5 With the notations as above, for {\triangle ABC}, if the length of {AB} is {c}, then the angle {c'} of the dual triangle, {\triangle a'b'c'}, is {\pi-c}.

Proof: Let me omit the proof first. This is actually the dual statement of the previous proposition. You can either prove it directly, or use the fact that the dual of the dual of a triangle is itself!! So if the angle {\angle a'c'b;} is {c'}, then the length {AB=c=\pi-c'} by Proposition 4, thus we have {c'=\pi-c}. \Box

Remark 2 The dual triangle can also be regarded as a triangle sitting in the dual sphere {S}, which we defines to be the set of oriented great circles on {\mathbb{S}^2}, or, in more fancy language, this is the Grassmannian of oriented two plane in {\mathbb{R}^3}. Anyway, it can be easily seen that {S} is bijective to {\mathbb{S}^2} by the “polar” map which maps a point to its polar line. It is an interesting exercise to visualize how a triangle in {\mathbb{S}^2} corresponds to a “dual triangle” on {S}. (Hint: the “curve” in {S} is a family of continuously moving great circles. )

Proof of (2): Apply (1) to the dual triangle {\triangle a'b'c'} and apply the above two propositions, we have

\displaystyle  \cos (\pi-C)=\cos(\pi- A)\cos(\pi- B)+\sin(\pi- A)\sin(\pi- B) \cos(\pi-c)

This gives (2). \Box

How it will become the Euclidean sine law in the “limit” sense? [Next time]

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