# Analogue Formula for Spherical Law of Cosines/Corollary

Jump to navigation Jump to search

## 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:

 $\ds \sin A \cos b$ $=$ $\ds \cos B \sin C + \sin B \cos C \cos a$ $\ds \sin A \cos c$ $=$ $\ds \cos C \sin B + \sin C \cos B \cos a$

## Proof

Let $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$.

Let the sides $a', b', c'$ of $\triangle A'B'C'$ be opposite $A', B', C'$ respectively.

From Spherical Triangle is Polar Triangle of its Polar Triangle we have that:

not only is $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$
but also $\triangle ABC$ is the polar triangle of $\triangle A'B'C'$.

We have:

 $\ds \sin a' \cos B'$ $=$ $\ds \cos b' \sin c' - \sin b' \cos c' \cos A'$ Analogue Formula for Spherical Law of Cosines $\ds \leadsto \ \$ $\ds \map \sin {\pi - A} \, \map \cos {\pi - b}$ $=$ $\ds \map \cos {\pi - B} \, \map \sin {\pi - C} - \map \sin {\pi - B} \, \map \cos {\pi - C} \, \map \cos {\pi - a}$ Side of Spherical Triangle is Supplement of Angle of Polar Triangle $\ds \leadsto \ \$ $\ds \map \sin {\pi - A} \paren {-\cos b}$ $=$ $\ds \paren {-\cos B} \, \map \sin {\pi - C} - \map \sin {\pi - B} \, \paren {-\cos C} \, \paren {-\cos a}$ Cosine of Supplementary Angle $\ds \leadsto \ \$ $\ds \sin A \, \paren {-\cos b}$ $=$ $\ds \paren {-\cos B} \sin C - \sin B \, \paren {-\cos C} \, \paren {-\cos a}$ Sine of Supplementary Angle $\ds \leadsto \ \$ $\ds \sin A \cos b$ $=$ $\ds \cos B \sin C + \sin B \cos C \cos a$ simplifying

$\blacksquare$

and:

 $\ds \sin a' \cos C'$ $=$ $\ds \cos c' \sin b' - \sin c' \cos b' \cos A'$ Analogue Formula for Spherical Law of Cosines $\ds \leadsto \ \$ $\ds \map \sin {\pi - A} \, \map \cos {\pi - c}$ $=$ $\ds \map \cos {\pi - C} \, \map \sin {\pi - B} - \map \sin {\pi - C} \, \map \cos {\pi - B} \, \map \cos {\pi - a}$ Side of Spherical Triangle is Supplement of Angle of Polar Triangle $\ds \leadsto \ \$ $\ds \map \sin {\pi - A} \paren {-\cos c}$ $=$ $\ds \paren {-\cos C} \, \map \sin {\pi - B} - \map \sin {\pi - C} \, \paren {-\cos B} \, \paren {-\cos a}$ Cosine of Supplementary Angle $\ds \leadsto \ \$ $\ds \sin A \, \paren {-\cos c}$ $=$ $\ds \paren {-\cos C} \sin B - \sin C \, \paren {-\cos B} \, \paren {-\cos a}$ Sine of Supplementary Angle $\ds \leadsto \ \$ $\ds \sin A \cos c$ $=$ $\ds \cos C \sin B + \sin C \cos B \cos a$ simplifying

$\blacksquare$