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 the Spherical Law of Cosines as just:
- the cosine-formula
- the cosine law
- the cosine rule
- the rule of cosines
- the law of cosines
but all of these are also used to refer to the plane version, so on $\mathsf{Pr} \infty \mathsf{fWiki}$ this is not recommended.
Some sources carefully refer to it as the law of cosines for spherical triangles.
Also see
Historical Note
The Spherical Law of Cosines was first stated by Regiomontanus in his De Triangulis Omnimodus of $1464$.
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: $5.97$
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): cosine rule (law of cosines): 2.
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): cosine rule (law of cosines): 2.
- 2008: Ian Stewart: Taming the Infinite ... (previous) ... (next): Chapter $5$: Eternal Triangles: Early trigonometry
- ... where he misattributes it to Georg Joachim Rhaeticus