Analogue Formula for Spherical Law of Cosines/Corollary

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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:

\(\ds \sin A \cos b\) \(=\) \(\ds \cos B \sin C + \sin B \cos C \cos a\)
\(\ds \sin A \cos c\) \(=\) \(\ds \cos C \sin B + \sin C \cos B \cos a\)


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 $\triangle A'B'C'$.


We have:

\(\ds \sin a' \cos B'\) \(=\) \(\ds \cos b' \sin c' - \sin b' \cos c' \cos A'\) Analogue Formula for Spherical Law of Cosines
\(\ds \leadsto \ \ \) \(\ds \map \sin {\pi - A} \, \map \cos {\pi - b}\) \(=\) \(\ds \map \cos {\pi - B} \, \map \sin {\pi - C} - \map \sin {\pi - B} \, \map \cos {\pi - C} \, \map \cos {\pi - a}\) Side of Spherical Triangle is Supplement of Angle of Polar Triangle
\(\ds \leadsto \ \ \) \(\ds \map \sin {\pi - A} \paren {-\cos b}\) \(=\) \(\ds \paren {-\cos B} \, \map \sin {\pi - C} - \map \sin {\pi - B} \, \paren {-\cos C} \, \paren {-\cos a}\) Cosine of Supplementary Angle
\(\ds \leadsto \ \ \) \(\ds \sin A \, \paren {-\cos b}\) \(=\) \(\ds \paren {-\cos B} \sin C - \sin B \, \paren {-\cos C} \, \paren {-\cos a}\) Sine of Supplementary Angle
\(\ds \leadsto \ \ \) \(\ds \sin A \cos b\) \(=\) \(\ds \cos B \sin C + \sin B \cos C \cos a\) simplifying

$\blacksquare$


and:

\(\ds \sin a' \cos C'\) \(=\) \(\ds \cos c' \sin b' - \sin c' \cos b' \cos A'\) Analogue Formula for Spherical Law of Cosines
\(\ds \leadsto \ \ \) \(\ds \map \sin {\pi - A} \, \map \cos {\pi - c}\) \(=\) \(\ds \map \cos {\pi - C} \, \map \sin {\pi - B} - \map \sin {\pi - C} \, \map \cos {\pi - B} \, \map \cos {\pi - a}\) Side of Spherical Triangle is Supplement of Angle of Polar Triangle
\(\ds \leadsto \ \ \) \(\ds \map \sin {\pi - A} \paren {-\cos c}\) \(=\) \(\ds \paren {-\cos C} \, \map \sin {\pi - B} - \map \sin {\pi - C} \, \paren {-\cos B} \, \paren {-\cos a}\) Cosine of Supplementary Angle
\(\ds \leadsto \ \ \) \(\ds \sin A \, \paren {-\cos c}\) \(=\) \(\ds \paren {-\cos C} \sin B - \sin C \, \paren {-\cos B} \, \paren {-\cos a}\) Sine of Supplementary Angle
\(\ds \leadsto \ \ \) \(\ds \sin A \cos c\) \(=\) \(\ds \cos C \sin B + \sin C \cos B \cos a\) simplifying

$\blacksquare$


Sources

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