Tangent of Half Angle for Spherical Triangles
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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:
- $\tan \dfrac A 2 = \sqrt {\dfrac {\map \sin {s - b} \, \map \sin {s - c} } {\sin s \, \map \sin {s - a} } }$
where $s = \dfrac {a + b + c} 2$.
Proof
| \(\ds \tan \dfrac A 2\) | \(=\) | \(\ds \dfrac {\sqrt {\dfrac {\sin \paren {s - b} \sin \paren {s - c} } {\sin b \sin c} } } {\sqrt {\dfrac {\sin s \, \map \sin {s - a} } {\sin b \sin c} } }\) | Sine of Half Angle for Spherical Triangles, Cosine of Half Angle for Spherical Triangles | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {\sqrt {\sin \paren {s - b} \sin \paren {s - c} } } {\sqrt {\sin s \, \map \sin {s - a} } }\) | simplification |
The result follows.
$\blacksquare$
Also presented as
The Tangent of Half Angle for Spherical Triangles formula is also seen presented in the following form:
| \(\ds \tan \dfrac A 2\) | \(=\) | \(\ds \dfrac r {\map \sin {s - a} }\) | ||||||||||||
| \(\ds \tan \dfrac B 2\) | \(=\) | \(\ds \dfrac r {\map \sin {s - b} }\) | ||||||||||||
| \(\ds \tan \dfrac C 2\) | \(=\) | \(\ds \dfrac r {\map \sin {s - c} }\) |
where:
| \(\ds s\) | \(=\) | \(\ds \dfrac {a + b + c} 2\) | ||||||||||||
| \(\ds r\) | \(=\) | \(\ds \sqrt {\dfrac {\map \sin {s - a} \map \sin {s - b} \map \sin {s - c} } {\sin s} }\) |
Also see
- The other Half Angle Formulas for Spherical Triangles:
Sources
- 1976: W.M. Smart: Textbook on Spherical Astronomy (6th ed.) ... (previous) ... (next): Chapter $\text I$. Spherical Trigonometry: $5$. The cosine-formula.