# Cosine of Half Side 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:

$\cos \dfrac a 2 = \sqrt {\dfrac {\map \cos {S - B} \, \map \cos {S - C} } {\sin B \sin C} }$

where $S = \dfrac {A + B + C} 2$.

## Proof

Let $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$.

Let the sides $a', b', c'$ of $\triangle A'B'C'$ be opposite $A', B', C'$ respectively.

From Spherical Triangle is Polar Triangle of its Polar Triangle we have that:

not only is $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$
but also $\triangle ABC$ is the polar triangle of $\triangle A'B'C'$.

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

We have:

 $\displaystyle \sin \dfrac {A'} 2$ $=$ $\displaystyle \sqrt {\dfrac {\sin \paren {s' - b'} \sin \paren {s' - c'} } {\sin b' \sin c'} }$ Sine of Half Angle for Spherical Triangles $\displaystyle \leadsto \ \$ $\displaystyle \sin \dfrac {\pi - a} 2$ $=$ $\displaystyle \sqrt {\dfrac {\map \sin {s' - b'} \, \map \sin {s' - c'} } {\map \sin {\pi - B} \, \map \sin {\pi - C} } }$ Side of Spherical Triangle is Supplement of Angle of Polar Triangle $\displaystyle \leadsto \ \$ $\displaystyle \map \sin {\dfrac \pi 2 - \dfrac a 2}$ $=$ $\displaystyle \sqrt {\dfrac {\map \sin {s' - b'} \, \map \sin {s' - c'} } {\sin B \sin C} }$ Sine of Supplementary Angle $\displaystyle \leadsto \ \$ $\displaystyle \cos \dfrac a 2$ $=$ $\displaystyle \sqrt {\dfrac {\map \sin {s' - b'} \, \map \sin {s' - c'} } {\sin B \sin C} }$ Sine of Complement equals Cosine

Then:

 $\displaystyle s' - b'$ $=$ $\displaystyle \dfrac {\paren {\pi - A} + \paren {\pi - B} + \paren {\pi - C} } 2 - \paren {\pi - B}$ Side of Spherical Triangle is Supplement of Angle of Polar Triangle $\displaystyle$ $=$ $\displaystyle \dfrac {\pi - \paren {A + B + C} } 2 + B$ simplifying $\displaystyle$ $=$ $\displaystyle \dfrac \pi 2 - \paren {S - B}$ where $S = \dfrac {A + B + C} 2$ $\displaystyle \leadsto \ \$ $\displaystyle \map \sin {s' - b'}$ $=$ $\displaystyle \map \sin {\dfrac \pi 2 - \paren {S - B} }$ $\displaystyle$ $=$ $\displaystyle \map \cos {S - B}$ Sine of Complement equals Cosine

and similarly:

$\map \sin {s' - c'} = \map \cos {S - C}$

The result follows.

$\blacksquare$