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

$\ds \cos a$

$=$

$\ds \cos b \cos c + \sin b \sin c \cos A$

$\ds $

$=$

$\ds \cos b \cos c + \sin b \sin c \paren {2 \cos^2 \dfrac A 2 - 1}$

$\ds $

$=$

$\ds \map \cos {b + c} + 2 \sin b \sin c \cos^2 \dfrac A 2$

$\ds \leadsto \ \ $

$\ds \cos a - \map \cos {b + c}$

$=$

$\ds 2 \sin b \sin c \cos^2 \dfrac A 2$

rearranging

$\ds \leadsto \ \ $

$\ds 2 \sin \dfrac {a + \paren {b + c} } 2 \sin \dfrac {\paren {b + c} - a} 2$

$=$

$\ds 2 \sin b \sin c \cos^2 \dfrac A 2$

$\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.

$\blacksquare$

1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: $5.99$
1976: W.M. Smart: Textbook on Spherical Astronomy (6th ed.) ... (previous) ... (next): Chapter $\text I$. Spherical Trigonometry: $5$. The cosine-formula.
2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 12$: Trigonometric Functions: $12.99.$