Napier's Cosine Rule for Right Spherical Triangles

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Theorem

Let $\triangle ABC$ be a right 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.

Let the angle $\sphericalangle C$ be a right angle.

Let the remaining parts of $\triangle ABC$ be arranged according to the interior of this circle, where the symbol $\Box$ denotes a right angle.

NapiersRules.png

Let one of the parts of this circle be called a middle part.

Let the two parts which do not neighbor the middle part be called opposite parts.


Then the sine of the middle part equals the product of the cosine of the opposite parts.


Proof

Let $\triangle ABC$ be a right spherical triangle such that the angle $\sphericalangle C$ is a right angle.

Right-spherical-triangle.png

Let the remaining parts of $\triangle ABC$ be arranged according to the interior of the circle above, where the symbol $\Box$ denotes a right angle.


$\sin a$

\(\ds \dfrac {\sin a} {\sin A}\) \(=\) \(\ds \dfrac {\sin c} {\sin C}\) Spherical Law of Sines for side $a$
\(\ds \leadsto \ \ \) \(\ds \dfrac {\sin a} {\sin A}\) \(=\) \(\ds \dfrac {\sin c} 1\) Sine of Right Angle as $C = \Box$
\(\ds \leadsto \ \ \) \(\ds \sin a\) \(=\) \(\ds \sin A \sin c\)
\(\ds \leadsto \ \ \) \(\ds \sin a\) \(=\) \(\ds \map \cos {\Box - A} \map \cos {\Box - c}\) Cosine of Complement equals Sine

$\Box$


$\sin b$

\(\ds \dfrac {\sin b} {\sin B}\) \(=\) \(\ds \dfrac {\sin c} {\sin C}\) Spherical Law of Sines for side $b$
\(\ds \leadsto \ \ \) \(\ds \dfrac {\sin b} {\sin B}\) \(=\) \(\ds \dfrac {\sin c} 1\) Sine of Right Angle as $C = \Box$
\(\ds \leadsto \ \ \) \(\ds \sin b\) \(=\) \(\ds \sin B \sin c\)
\(\ds \leadsto \ \ \) \(\ds \sin b\) \(=\) \(\ds \map \cos {\Box - B} \map \cos {\Box - c}\) Cosine of Complement equals Sine

$\Box$


$\map \sin {\Box - A}$

\(\ds \cos A\) \(=\) \(\ds -\cos B \cos C + \sin B \sin C \cos a\) Spherical Law of Cosines for angle $A$
\(\ds \) \(=\) \(\ds -\cos B \times 0 + \sin B \times 1 \times \cos a\) Cosine of Right Angle and Sine of Right Angle as $C = \Box$
\(\ds \) \(=\) \(\ds \sin B \cos a\)
\(\ds \leadsto \ \ \) \(\ds \map \sin {\Box - A}\) \(=\) \(\ds \map \cos {\Box - B} \cos a\) Sine of Complement equals Cosine, Cosine of Complement equals Sine

$\Box$


$\map \sin {\Box - c}$

\(\ds \cos c\) \(=\) \(\ds \cos a \cos b + \sin a \sin b \cos C\) Spherical Law of Cosines for side $c$
\(\ds \) \(=\) \(\ds \cos a \cos b\) Cosine of Right Angle as $C = \Box$
\(\ds \leadsto \ \ \) \(\ds \map \sin {\Box - c}\) \(=\) \(\ds \cos a \cos b\) Sine of Complement equals Cosine

$\Box$


$\map \sin {\Box - B}$

\(\ds \cos B\) \(=\) \(\ds -\cos A \cos C + \sin A \sin C \cos b\) Spherical Law of Cosines for angle $B$
\(\ds \) \(=\) \(\ds -\cos A \times 0 + \sin A \times 1 \cos b\) Cosine of Right Angle as $C = \Box$
\(\ds \) \(=\) \(\ds \sin A \cos b\)
\(\ds \leadsto \ \ \) \(\ds \map \sin {\Box - B}\) \(=\) \(\ds \map \cos {\Box - A} \cos c\) Sine of Complement equals Cosine, Cosine of Complement equals Sine

$\blacksquare$


Also see


Source of Name

This entry was named for John Napier.


Sources

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