Spherical Law of Sines/Proof 1

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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:

$\dfrac {\sin a} {\sin A} = \dfrac {\sin b} {\sin B} = \dfrac {\sin c} {\sin C}$


Proof

\(\displaystyle \sin b \sin c \cos A\) \(=\) \(\displaystyle \cos a - \cos b \cos c\) Spherical Law of Cosines
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sin^2 b \sin^2 c \cos^2 A\) \(=\) \(\displaystyle \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sin^2 b \sin^2 c \paren {1 - \sin^2 A}\) \(=\) \(\displaystyle \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c\) Sum of Squares of Sine and Cosine
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sin^2 b \sin^2 c - \sin^2 b \sin^2 c \sin^2 A\) \(=\) \(\displaystyle \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c\) multiplying out
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {1 - \cos^2 b} \paren {1 - \cos^2 c} - \sin^2 b \sin^2 c \sin^2 A\) \(=\) \(\displaystyle \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c\) Sum of Squares of Sine and Cosine
\(\displaystyle \leadsto \ \ \) \(\displaystyle 1 - \cos^2 b - \cos^2 c + \cos^2 b \cos^2 c - \sin^2 b \sin^2 c \sin^2 A\) \(=\) \(\displaystyle \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c\) multiplying out
\((1):\quad\) \(\displaystyle \leadsto \ \ \) \(\displaystyle \sin^2 b \sin^2 c \sin^2 A\) \(=\) \(\displaystyle 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c\) rearranging and simplifying


Let $X \in \R_{>0}$ such that:

$X^2 \sin^2 a \sin^2 b \sin^2 c = 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c$


Then from $(1)$:

\(\displaystyle \dfrac {X^2 \sin^2 a \sin^2 b \sin^2 c} {\sin^2 b \sin^2 c \sin^2 A}\) \(=\) \(\displaystyle \dfrac {1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c} {1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle X^2\) \(=\) \(\displaystyle \dfrac {\sin^2 A} {\sin^2 a}\)


In a spherical triangle, all of the sides are less than $\pi$ radians.

The same applies to the angles.

From Shape of Sine Function:

$\sin \theta > 0$ for all $0 < \theta < \pi$

Hence the negative root of $\dfrac {\sin^2 A} {\sin^2 a}$ does not apply, and so:

$X = \dfrac {\sin A} {\sin a}$


Similarly, from applying the Spherical Law of Cosines to $\cos B$ and $\cos C$:

\(\displaystyle \sin a \sin c \cos B\) \(=\) \(\displaystyle \cos b - \cos a \cos c\)
\(\displaystyle \sin a \sin b \cos C\) \(=\) \(\displaystyle \cos c - \cos a \cos b\)

we arrive at the same point:

\(\displaystyle X\) \(=\) \(\displaystyle \dfrac {\sin B} {\sin b}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\sin A} {\sin a}\)

where:

$X^2 \sin^2 a \sin^2 b \sin^2 c = 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c$

as before.

Hence we have:

$\dfrac {\sin a} {\sin A} = \dfrac {\sin b} {\sin B} = \dfrac {\sin c} {\sin C}$

$\blacksquare$


Sources