Four-Parts Formula/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:

$\cos A \cos c = \sin A \cot B - \sin c \cot b$


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:

\(\ds \cos a' \cos C'\) \(=\) \(\ds \sin a' \cot b' - \sin C' \cot B'\) Four-Parts Formula
\(\ds \leadsto \ \ \) \(\ds \map \cos {\pi - A} \, \map \cos {\pi - c}\) \(=\) \(\ds \map \sin {\pi - A} \, \map \cot {\pi - B} - \map \sin {\pi - c} \, \map \cot {\pi - b}\) Side of Spherical Triangle is Supplement of Angle of Polar Triangle
\(\ds \leadsto \ \ \) \(\ds \paren {-\cos A} \, \paren {-\cos c}\) \(=\) \(\ds \map \sin {\pi - A} \, \map \cot {\pi - B} - \map \sin {\pi - c} \, \map \cot {\pi - b}\) Cosine of Supplementary Angle
\(\ds \leadsto \ \ \) \(\ds \paren {-\cos A} \, \paren {-\cos c}\) \(=\) \(\ds \sin A \, \map \cot {\pi - B} - \sin c \, \map \cot {\pi - b}\) Sine of Supplementary Angle
\(\ds \leadsto \ \ \) \(\ds \paren {-\cos A} \, \paren {-\cos c}\) \(=\) \(\ds \sin A \, \paren {-\cot B} - \sin c \, \paren {-\cot b}\) Cotangent of Supplementary Angle
\(\ds \leadsto \ \ \) \(\ds \cos A \cos c\) \(=\) \(\ds \sin c \cot b - \sin A \cot B\) simplifying and rearranging

$\blacksquare$


Sources

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