# Spherical Law of Sines

## 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:

- $\dfrac {\sin a} {\sin A} = \dfrac {\sin b} {\sin B} = \dfrac {\sin c} {\sin C}$

## Proof 1

\(\displaystyle \sin b \sin c \cos A\) | \(=\) | \(\displaystyle \cos a - \cos b \cos c\) | Spherical Law of Cosines | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \sin^2 b \sin^2 c \cos^2 A\) | \(=\) | \(\displaystyle \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \sin^2 b \sin^2 c \paren {1 - \sin^2 A}\) | \(=\) | \(\displaystyle \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c\) | Sum of Squares of Sine and Cosine | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \sin^2 b \sin^2 c - \sin^2 b \sin^2 c \sin^2 A\) | \(=\) | \(\displaystyle \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c\) | multiplying out | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \paren {1 - \cos^2 b} \paren {1 - \cos^2 c} - \sin^2 b \sin^2 c \sin^2 A\) | \(=\) | \(\displaystyle \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c\) | Sum of Squares of Sine and Cosine | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 1 - \cos^2 b - \cos^2 c + \cos^2 b \cos^2 c - \sin^2 b \sin^2 c \sin^2 A\) | \(=\) | \(\displaystyle \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c\) | multiplying out | |||||||||

\(\text {(1)}: \quad\) | \(\displaystyle \leadsto \ \ \) | \(\displaystyle \sin^2 b \sin^2 c \sin^2 A\) | \(=\) | \(\displaystyle 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c\) | rearranging and simplifying |

Let $X \in \R_{>0}$ such that:

- $X^2 \sin^2 a \sin^2 b \sin^2 c = 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c$

Then from $(1)$:

\(\displaystyle \dfrac {X^2 \sin^2 a \sin^2 b \sin^2 c} {\sin^2 b \sin^2 c \sin^2 A}\) | \(=\) | \(\displaystyle \dfrac {1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c} {1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c}\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle X^2\) | \(=\) | \(\displaystyle \dfrac {\sin^2 A} {\sin^2 a}\) |

In a spherical triangle, all of the sides are less than $\pi$ radians.

The same applies to the angles.

From Shape of Sine Function:

- $\sin \theta > 0$ for all $0 < \theta < \pi$

Hence the negative root of $\dfrac {\sin^2 A} {\sin^2 a}$ does not apply, and so:

- $X = \dfrac {\sin A} {\sin a}$

Similarly, from applying the Spherical Law of Cosines to $\cos B$ and $\cos C$:

\(\displaystyle \sin a \sin c \cos B\) | \(=\) | \(\displaystyle \cos b - \cos a \cos c\) | |||||||||||

\(\displaystyle \sin a \sin b \cos C\) | \(=\) | \(\displaystyle \cos c - \cos a \cos b\) |

we arrive at the same point:

\(\displaystyle X\) | \(=\) | \(\displaystyle \dfrac {\sin B} {\sin b}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {\sin A} {\sin a}\) |

where:

- $X^2 \sin^2 a \sin^2 b \sin^2 c = 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c$

as before.

Hence we have:

- $\dfrac {\sin a} {\sin A} = \dfrac {\sin b} {\sin B} = \dfrac {\sin c} {\sin C}$

$\blacksquare$

## 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:

\(\displaystyle \angle COB\) | \(=\) | \(\displaystyle a\) | |||||||||||

\(\displaystyle \angle COA\) | \(=\) | \(\displaystyle b\) | |||||||||||

\(\displaystyle \angle AOB\) | \(=\) | \(\displaystyle c\) |

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:

\(\text {(1)}: \quad\) | \(\displaystyle PQ\) | \(=\) | \(\displaystyle OP \sin b\) | ||||||||||

\(\text {(2)}: \quad\) | \(\displaystyle PR\) | \(=\) | \(\displaystyle OP \sin a\) | ||||||||||

\(\text {(3)}: \quad\) | \(\displaystyle OQ\) | \(=\) | \(\displaystyle OP \cos b\) | ||||||||||

\(\text {(4)}: \quad\) | \(\displaystyle OR\) | \(=\) | \(\displaystyle OP \cos a\) |

From the right triangles $\triangle PQS$ and $\triangle PRS$, we have:

\(\displaystyle PS\) | \(=\) | \(\displaystyle PS \sin \angle PRS\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle PQ \sin A\) | |||||||||||

\(\displaystyle PS\) | \(=\) | \(\displaystyle PR \sin \angle PRS\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle PR \sin B\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle OP \sin b \sin A\) | \(=\) | \(\displaystyle OP \sin a \sin B\) | from $(1)$ and $(2)$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \dfrac {\sin a} {\sin A}\) | \(=\) | \(\displaystyle \dfrac {\sin b} {\sin B}\) |

The result follows by applying this technique mutatis mutandis to the other angles of $ABC$.

$\blacksquare$

## Also known as

Some sources refer to this as just the **sine-formula**.

## Also see

## Historical Note

The Spherical Law of Sines was first stated by Abu'l-Wafa Al-Buzjani.

It was rediscovered by Regiomontanus, and published in his *De Triangulis Omnimodus* of $1464$.

## Sources

- 1968: Murray R. Spiegel:
*Mathematical Handbook of Formulas and Tables*... (previous) ... (next): $\S 5$: Trigonometric Functions: $5.96$ - 1998: David Nelson:
*The Penguin Dictionary of Mathematics*(2nd ed.) ... (previous) ... (next): Entry:**sine rule (law of sines)**:**2.** - 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next): Entry:**sine rule (law of sines)**:**2.** - 2008: Ian Stewart:
*Taming the Infinite*... (previous) ... (next): Chapter $5$: Eternal Triangles: Early trigonometry

*... where he misattributes it to Georg Joachim Rhaeticus*