# Spherical Law of Cosines

## 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

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

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.

## 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