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.
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