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

\(\displaystyle \sin b \sin c \cos A\) \(=\) \(\displaystyle \cos a - \cos b \cos c\) 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

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:

\(\displaystyle \angle COB\) \(=\) \(\displaystyle a\)
\(\displaystyle \angle COA\) \(=\) \(\displaystyle b\)
\(\displaystyle \angle AOB\) \(=\) \(\displaystyle c\)


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

... where he misattributes it to Georg Joachim Rhaeticus