Analogue Formula for Spherical Law of Cosines/Proof 1

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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

\(\ds \sin c \sin a \cos B\) \(=\) \(\ds \cos b - \cos c \cos a\) Spherical Law of Cosines
\(\ds \) \(=\) \(\ds \cos b - \cos c \paren {\cos b \cos c + \sin b \sin c \cos A}\) Spherical Law of Cosines
\(\ds \) \(=\) \(\ds \cos b \paren {1 - \cos^2 c} - \sin b \sin c \cos c \cos A\) rearranging
\(\ds \) \(=\) \(\ds \sin^2 c \cos b - \sin b \sin c \cos c \cos A\) Sum of Squares of Sine and Cosine
\(\ds \leadsto \ \ \) \(\ds \sin a \cos B\) \(=\) \(\ds \sin c \cos b - \sin b \cos c \cos A\) simplifying


\(\ds \sin a \sin b \cos C\) \(=\) \(\ds \cos c - \cos a \cos b\) Spherical Law of Cosines
\(\ds \) \(=\) \(\ds \cos c - \cos b \paren {\cos b \cos c + \sin b \sin c \cos A}\) Spherical Law of Cosines
\(\ds \) \(=\) \(\ds \cos c \paren {1 - \cos^2 b} - \sin b \sin c \cos b \cos A\) rearranging
\(\ds \) \(=\) \(\ds \sin^2 b \cos c - \sin b \sin c \cos b \cos A\) Sum of Squares of Sine and Cosine
\(\ds \leadsto \ \ \) \(\ds \sin a \cos C\) \(=\) \(\ds \cos b \sin c - \sin c \cos b \cos A\) simplifying

$\blacksquare$


Sources