Cosine 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:
- $\cos \dfrac A 2 = \sqrt {\dfrac {\sin s \, \map \sin {s - a} } {\sin b \sin c} }$
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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \cos b \cos c + \sin b \sin c \paren {2 \cos^2 \dfrac A 2 - 1}\) | Double Angle Formula for Cosine: Corollary $1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \cos {b + c} + 2 \sin b \sin c \cos^2 \dfrac A 2\) | Cosine of Sum | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \cos a - \map \cos {b + c}\) | \(=\) | \(\ds 2 \sin b \sin c \cos^2 \dfrac A 2\) | rearranging | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2 \sin \dfrac {a + \paren {b + c} } 2 \sin \dfrac {\paren {b + c} - a} 2\) | \(=\) | \(\ds 2 \sin b \sin c \cos^2 \dfrac A 2\) | Cosine minus Cosine | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \sin {\dfrac {a + b + c} 2} \map \sin {\dfrac {a + b + c} 2 - a}\) | \(=\) | \(\ds \sin b \sin c \cos^2 \dfrac A 2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sin s \, \map \sin {s - a}\) | \(=\) | \(\ds \sin b \sin c \cos^2 \dfrac A 2\) | setting $s = \dfrac {a + b + c} 2$ and simplifying |
The result follows.
$\blacksquare$
Also see
- The other Half Angle Formulas for Spherical Triangles:
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: $5.99$
- 1976: W.M. Smart: Textbook on Spherical Astronomy (6th ed.) ... (previous) ... (next): Chapter $\text I$. Spherical Trigonometry: $5$. The cosine-formula.