# 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): Entry:**cosine rule (law of cosines)**:**2.** - 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next): Entry:**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*