Spherical Law of Sines

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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

\(\ds \sin b \sin c \cos A\) \(=\) \(\ds \cos a - \cos b \cos c\) Spherical Law of Cosines
\(\ds \leadsto \ \ \) \(\ds \sin^2 b \sin^2 c \cos^2 A\) \(=\) \(\ds \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c\)
\(\ds \leadsto \ \ \) \(\ds \sin^2 b \sin^2 c \paren {1 - \sin^2 A}\) \(=\) \(\ds \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c\) Sum of Squares of Sine and Cosine
\(\ds \leadsto \ \ \) \(\ds \sin^2 b \sin^2 c - \sin^2 b \sin^2 c \sin^2 A\) \(=\) \(\ds \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c\) multiplying out
\(\ds \leadsto \ \ \) \(\ds \paren {1 - \cos^2 b} \paren {1 - \cos^2 c} - \sin^2 b \sin^2 c \sin^2 A\) \(=\) \(\ds \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c\) Sum of Squares of Sine and Cosine
\(\ds \leadsto \ \ \) \(\ds 1 - \cos^2 b - \cos^2 c + \cos^2 b \cos^2 c - \sin^2 b \sin^2 c \sin^2 A\) \(=\) \(\ds \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c\) multiplying out
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \sin^2 b \sin^2 c \sin^2 A\) \(=\) \(\ds 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c\) 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 $(1)$:

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


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 $\cos C$:

\(\ds \sin a \sin c \cos B\) \(=\) \(\ds \cos b - \cos a \cos c\)
\(\ds \sin a \sin b \cos C\) \(=\) \(\ds \cos c - \cos a \cos b\)

we arrive at the same point:

\(\ds X\) \(=\) \(\ds \dfrac {\sin B} {\sin b}\)
\(\ds \) \(=\) \(\ds \dfrac {\sin A} {\sin a}\)

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

Spherical-Cosine-Formula-2.png

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:

\(\text {(1)}: \quad\) \(\ds PQ\) \(=\) \(\ds OP \sin b\)
\(\text {(2)}: \quad\) \(\ds PR\) \(=\) \(\ds OP \sin a\)
\(\text {(3)}: \quad\) \(\ds OQ\) \(=\) \(\ds OP \cos b\)
\(\text {(4)}: \quad\) \(\ds OR\) \(=\) \(\ds OP \cos a\)


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

\(\ds PS\) \(=\) \(\ds PS \sin \angle PRS\)
\(\ds \) \(=\) \(\ds PQ \sin A\)
\(\ds PS\) \(=\) \(\ds PR \sin \angle PRS\)
\(\ds \) \(=\) \(\ds PR \sin B\)
\(\ds \leadsto \ \ \) \(\ds OP \sin b \sin A\) \(=\) \(\ds OP \sin a \sin B\) from $(1)$ and $(2)$
\(\ds \leadsto \ \ \) \(\ds \dfrac {\sin a} {\sin A}\) \(=\) \(\ds \dfrac {\sin b} {\sin B}\)


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

$\blacksquare$


Also known as

Some sources refer to the Spherical Law of Sines as just:

the sine-formula
the sine law
the sine rule
the rule of sines
the law of sines

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 sines for spherical triangles.


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

... where he misattributes it to Georg Joachim Rhaeticus
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