Sine 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:

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

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:

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


Then:

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


and:

\(\ds s'\) \(=\) \(\ds \dfrac {\paren {\pi - A} + \paren {\pi - B} + \paren {\pi - C} } 2\) Side of Spherical Triangle is Supplement of Angle of Polar Triangle
\(\ds \) \(=\) \(\ds \dfrac {3 \pi} 2 - \dfrac {A + B + C} 2\) simplifying
\(\ds \) \(=\) \(\ds \dfrac {3 \pi} 2 - S\) where $S = \dfrac {A + B + C} 2$
\(\ds \leadsto \ \ \) \(\ds \sin s'\) \(=\) \(\ds \map \sin {\dfrac {3 \pi} 2 - S}\)
\(\ds \leadsto \ \ \) \(\ds \sin s'\) \(=\) \(\ds -\map \sin {\dfrac \pi 2 - S}\) Sine of Angle plus Straight Angle
\(\ds \) \(=\) \(\ds -\cos S\) Sine of Complement equals Cosine

The result follows.

$\blacksquare$


Also see


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