Spherical Law of Haversines

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Theorem

Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.

Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.


Then:

$\hav a = \map \hav {b - c} + \sin b \sin c \hav A$

where $\hav$ denotes haversine.


Proof

\(\ds \cos a\) \(=\) \(\ds \cos b \cos c + \sin b \sin c \cos A\) Spherical Law of Cosines
\(\ds \leadsto \ \ \) \(\ds 1 - 2 \hav a\) \(=\) \(\ds \cos b \cos c + \sin b \sin c \paren {1 - 2 \hav A}\) Cosine in Terms of Haversine
\(\ds \) \(=\) \(\ds \map \cos {b - c} - 2 \sin b \sin c \hav A\) Cosine of Difference
\(\ds \) \(=\) \(\ds 1 - 2 \map \hav {b - c} - 2 \sin b \sin c \hav A\) Cosine in Terms of Haversine
\(\ds \leadsto \ \ \) \(\ds \hav a\) \(=\) \(\ds \map \hav {b - c} + \sin b \sin c \hav A\) simplifying

$\blacksquare$


Sources