Analogue Formula for Spherical Law of Cosines/Proof 2

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:

$\ds \sin a \cos B$

$=$

$\ds \cos b \sin c - \sin b \cos c \cos A$

$\ds \sin a \cos C$

$=$

$\ds \cos c \sin b - \sin c \cos b \cos A$

Suppose $c$ is less than $\dfrac \pi 2$.

Let $BA$ be produced to $D$ so that $BD = \dfrac \pi 2$.

Then:

$AD = \dfrac \pi 2 - c$

and:

$\angle CAD = pi - A$

Let $C$ and $D$ be joined by an arc of a great circle, denoted $x$.

From the triangle $\sphericalangle DAC$, using the Spherical Law of Cosines:

$\ds \cos x$

$=$

$\ds \map \cos {\dfrac \pi 2 - c} \cos b + \map \sin {\dfrac \pi 2 - c} \sin b \, \map \cos {\pi - A}$

$\ds $

$=$

$\ds \sin c \cos b - \cos c \sin b \cos A$

From the triangle $\sphericalangle DBC$, using the Spherical Law of Cosines:

$\ds \cos x$

$=$

$\ds \cos \dfrac \pi 2 \cos a + \sin \pi 2 \sin a \cos B$

$\ds $

$=$

$\ds \sin a \cos B$

Hence:

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

The case where $c > \dfrac \pi 2$ is worked similarly, but by making $D$ the point between $A$ and $B$ such that $BD$ is $\dfrac \pi 2$.

$\blacksquare$

1976: W.M. Smart: Textbook on Spherical Astronomy (6th ed.) ... (previous) ... (next): Chapter $\text I$. Spherical Trigonometry: $7$. The analogue formula.