Spherical Law of Cosines

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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:

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


Corollary

$\cos A = -\cos B \cos C + \sin B \sin C \cos a$


Proof 1

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.

$\blacksquare$


Proof 2

Spherical-Cosine-Formula-2.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 $O$ be joined to each of $A$, $B$ and $C$.

Let $P$ be an arbitrary point on $OC$.

Construct $PQ$ perpendicular to $OA$ meeting $OA$ at $Q$.

Construct $PR$ perpendicular to $OB$ meeting $OB$ at $R$.

In the plane $OAB$:

construct $QS$ perpendicular to $OA$
construct $RS$ perpendicular to $OB$

where $S$ is the point where $QS$ and $RS$ intersect.

Let $OS$ and $PS$ be joined.

Let tangents be constructed at $A$ to the arcs of the great circles $AC$ and $AB$.

These tangents contain the spherical angle $A$.

But by construction, $QS$ and $QP$ are parallel to these tangents.

Hence $\angle PQS = \sphericalangle A$.

Similarly, $\angle PRS = \sphericalangle B$.

Also we have:

\(\ds \angle COB\) \(=\) \(\ds a\)
\(\ds \angle COA\) \(=\) \(\ds b\)
\(\ds \angle AOB\) \(=\) \(\ds c\)


It is to be proved that $PS$ is perpendicular to the plane $AOB$.

By construction, $OQ$ is perpendicular to both $PQ$ and $QS$.

Thus $OQ$ is perpendicular to the plane $PQS$.

Similarly, $OR$ is perpendicular to the plane $PRS$.

Thus $PS$ is perpendicular to both $OQ$ and $OR$.

Thus $PS$ is perpendicular to every line in the plane of $OQ$ and $OR$.

That is, $PS$ is perpendicular to the plane $OAB$.

In particular, $PS$ is perpendicular to $OS$, $SQ$ and $SR$

It follows that $\triangle PQS$ and $\triangle PRS$ are right triangles.


From the right triangles $\triangle OQP$ and $\triangle ORP$, we have:

\(\text {(1)}: \quad\) \(\ds PQ\) \(=\) \(\ds OP \sin b\)
\(\text {(2)}: \quad\) \(\ds PR\) \(=\) \(\ds OP \sin a\)
\(\text {(3)}: \quad\) \(\ds OQ\) \(=\) \(\ds OP \cos b\)
\(\text {(4)}: \quad\) \(\ds OR\) \(=\) \(\ds OP \cos a\)


Let us denote the angle $\angle SOQ$ by $x$.

Then:

$\angle ROS = c - x$

We have that:

\(\ds OS\) \(=\) \(\ds OQ \sec x\)
\(\ds OS\) \(=\) \(\ds OR \, \map \sec {c - x}\)
\(\ds \leadsto \ \ \) \(\ds OR \cos x\) \(=\) \(\ds OQ \, \map \cos {c - x}\)
\(\ds \leadsto \ \ \) \(\ds OP \cos a \cos x\) \(=\) \(\ds OP \cos b \, \map \cos {c - x}\) from $(3)$ and $(4)$
\(\ds \leadsto \ \ \) \(\ds \cos a \cos x\) \(=\) \(\ds \cos b \paren {\cos c \cos x - \sin c \sin x}\) Cosine of Difference
\(\text {(5)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \cos a\) \(=\) \(\ds \cos b \cos c + \cos b \sin c \tan x\) dividing both sides by $\cos x$ and multiplying out

But we also have:

\(\ds \tan x\) \(=\) \(\ds \dfrac {QS} {OQ}\)
\(\ds \) \(=\) \(\ds \dfrac {PQ \cos A} {OQ}\)
\(\ds \) \(=\) \(\ds \tan b \cos A\)
\(\ds \leadsto \ \ \) \(\ds \cos a\) \(=\) \(\ds \cos b \cos c + \cos b \sin c \tan b \cos A\) substituting for $\tan x$ from $(5)$
\(\ds \) \(=\) \(\ds \cos b \cos c + \sin b \sin c \cos A\)

Hence the result.

$\blacksquare$


Also known as

Some sources refer to this result as just the cosine-formula.


Also see


Historical Note

The Spherical Law of Cosines was first stated by Regiomontanus in his De Triangulis Omnimodus of $1464$.


Sources

... where he misattributes it to Georg Joachim Rhaeticus