Cosine of Half Side 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:

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

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:

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


Then:

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

and similarly:

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

The result follows.

$\blacksquare$


Also see



Sources

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