Sine 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:
- $\sin \dfrac A 2 = \sqrt {\dfrac {\map \sin {s - b} \, \map \sin {s - c} } {\sin b \sin c} }$
where $s = \dfrac {a + b + c} 2$.
Proof
| \(\ds \cos a\) | \(=\) | \(\ds \cos b \cos c + \sin b \sin c \cos A\) | Spherical Law of Cosines | |||||||||||
| \(\ds \) | \(=\) | \(\ds \cos b \cos c + \sin b \sin c \paren {1 - 2 \sin^2 \dfrac A 2}\) | Double Angle Formula for Cosine: Corollary $2$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \cos {b - c} - 2 \sin b \sin c \sin^2 \dfrac A 2\) | Cosine of Difference | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map \cos {b - c} - \cos a\) | \(=\) | \(\ds 2 \sin b \sin c \sin^2 \dfrac A 2\) | rearranging | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 2 \sin \dfrac {a + \paren {b - c} } 2 \sin \dfrac {a - \paren {b - c} } 2\) | \(=\) | \(\ds 2 \sin b \sin c \sin^2 \dfrac A 2\) | Cosine minus Cosine | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 2 \, \map \sin {\dfrac {a + b + c} 2 - c} \, \map \sin {\dfrac {a + b + c} 2 - b}\) | \(=\) | \(\ds 2 \sin b \sin c \sin^2 \dfrac A 2\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map \sin {s - c} \, \map \sin {s - b}\) | \(=\) | \(\ds \sin b \sin c \sin^2 \dfrac A 2\) | setting $s = \dfrac {a + b + c} 2$ and simplifying |
The result follows.
$\blacksquare$
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.