# 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

 $\ds \cos a$ $=$ $\ds \cos b \cos c + \sin b \sin c \cos A$ 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.

$\blacksquare$