Cosine of Half Angle for Spherical Triangles

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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:

$\cos \dfrac A 2 = \sqrt {\dfrac {\sin s \, \map \sin {s - a} } {\sin b \sin c} }$

where $s = \dfrac {a + b + c} 2$.


Proof

\(\ds \cos a\) \(=\) \(\ds \cos b \cos c + \sin b \sin c \cos A\) Spherical Law of Cosines
\(\ds \) \(=\) \(\ds \cos b \cos c + \sin b \sin c \paren {2 \cos^2 \dfrac A 2 - 1}\) Double Angle Formula for Cosine: Corollary $1$
\(\ds \) \(=\) \(\ds \map \cos {b + c} + 2 \sin b \sin c \cos^2 \dfrac A 2\) Cosine of Sum
\(\ds \leadsto \ \ \) \(\ds \cos a - \map \cos {b + c}\) \(=\) \(\ds 2 \sin b \sin c \cos^2 \dfrac A 2\) rearranging
\(\ds \leadsto \ \ \) \(\ds 2 \sin \dfrac {a + \paren {b + c} } 2 \sin \dfrac {\paren {b + c} - a} 2\) \(=\) \(\ds 2 \sin b \sin c \cos^2 \dfrac A 2\) Cosine minus Cosine
\(\ds \leadsto \ \ \) \(\ds \map \sin {\dfrac {a + b + c} 2} \map \sin {\dfrac {a + b + c} 2 - a}\) \(=\) \(\ds \sin b \sin c \cos^2 \dfrac A 2\)
\(\ds \leadsto \ \ \) \(\ds \sin s \, \map \sin {s - a}\) \(=\) \(\ds \sin b \sin c \cos^2 \dfrac A 2\) setting $s = \dfrac {a + b + c} 2$ and simplifying

The result follows.

$\blacksquare$


Also see



Sources

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