Cosine of Half Angle for Spherical Triangles

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

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

The result follows.

$\blacksquare$