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:

$\cos a = \cos b \cos c + \sin b \sin c \cos A$


Proof

Spherical-Cosine-Formula.png

Let $A$, $B$ and $C$ be the vertices of a spherical triangle on the surface of a sphere $S$.

By definition of a spherical triangle, $AB$, $BC$ and $AC$ are arcs of great circles on $S$.

By definition of a great circle, the center of each of these great circles is $O$.

Let $AD$ be the tangent to the great circle $AB$.

Let $AE$ be the tangent to the great circle $AC$.

Thus the radius $OA$ of $S$ is perpendicular to $AD$ and $AE$.

By construction, $AD$ lies in the same plane as $AB$.

Thus when $OB$ is produced, it will intersect $AD$ at $D$, say.

Similarly, $OC$ can be produced to intersect $AE$ at $E$, say.

The spherical angle $\sphericalangle BAC$ is defined as the angle between the tangents $AD$ and $AE$.

Thus:

$\sphericalangle BAC = \angle DAE$

or, denoting that spherical angle $\sphericalangle BAC$ as $A$:

$A = \angle DAE$

In the (plane) triangle $OAD$, we have that $\angle OAD$ is a right angle.

We also have that $\angle AOD = \angle AOB$ is equal to $c$, by definition of the length of a side of a spherical triangle.

Thus:

\(\ds AD\) \(=\) \(\ds OA \tan c\)
\(\ds OD\) \(=\) \(\ds OA \sec c\)

and by similar analysis of $\triangle OAE$, we have:

\(\ds AE\) \(=\) \(\ds OA \tan b\)
\(\ds OE\) \(=\) \(\ds OA \sec b\)


From consideration of $\triangle DAE$:

\(\ds DE^2\) \(=\) \(\ds AD^2 + AE^2 - 2 AD \cdot AE \cos \angle DAE\) Law of Cosines
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds OA^2 \paren {\tan^2 c + \tan^2 b - 2 \tan b \tan c \cos A}\)


From consideration of $\triangle DOE$:

\(\ds DE^2\) \(=\) \(\ds OD^2 + OE^2 - 2 OD \cdot OE \cos \angle DOE\) Law of Cosines
\(\text {(2)}: \quad\) \(\ds \) \(=\) \(\ds OA^2 \paren {\sec^2 c + \sec^2 b - 2 \sec b \sec c \cos a}\) as $\angle DOE = \angle BOC$


Thus:

\(\ds \sec^2 c + \sec^2 b - 2 \sec b \sec c \cos a\) \(=\) \(\ds \tan^2 c + \tan^2 b - 2 \tan b \tan c \cos A\) from $(1)$ and $(2)$
\(\ds \leadsto \ \ \) \(\ds \paren {1 + \tan^2 c} + \paren {1 + \tan^2 b} - 2 \sec b \sec c \cos a\) \(=\) \(\ds \tan^2 c + \tan^2 b - 2 \tan b \tan c \cos A\) Difference of Squares of Secant and Tangent
\(\ds \leadsto \ \ \) \(\ds 1 - \sec b \sec c \cos a\) \(=\) \(\ds \tan b \tan c \cos A\) simplifying
\(\ds \leadsto \ \ \) \(\ds \cos b \cos c - \cos a\) \(=\) \(\ds \sin b \sin c \cos A\) multiplying both sides by $\cos b \cos c$

and the result follows.

$\blacksquare$


Sources

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