Napier's Analogies/Tangent of Half Difference of Sides

Napier's Analogies

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:

$\tan \dfrac {a - b} 2 = \dfrac {\sin \frac {A - B} 2} {\sin \frac {A + B} 2} \tan \dfrac c 2$

$\text {(1)}: \quad$

$\ds \cos \dfrac C 2 \sin \dfrac {a - b} 2$

$=$

$\ds \sin \dfrac c 2 \sin \dfrac {A - B} 2$

$\text {(2)}: \quad$

$\ds \cos \dfrac C 2 \cos \dfrac {a - b} 2$

$=$

$\ds \cos \dfrac c 2 \sin \dfrac {A + B} 2$

$\ds \leadsto \ \ $

$\ds \tan \dfrac {a - b} 2$

$=$

$\ds \frac {\sin \frac {A - B} 2} {\sin \frac {A + B} 2} \tan \dfrac c 2$

dividing $(1)$ by $(2)$

$\blacksquare$

This entry was named for John Napier.