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
| \(\ds \sin b \sin c \cos A\) | \(=\) | \(\ds \cos a - \cos b \cos c\) | Spherical Law of Cosines | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sin^2 b \sin^2 c \cos^2 A\) | \(=\) | \(\ds \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sin^2 b \sin^2 c \paren {1 - \sin^2 A}\) | \(=\) | \(\ds \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c\) | Sum of Squares of Sine and Cosine | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sin^2 b \sin^2 c - \sin^2 b \sin^2 c \sin^2 A\) | \(=\) | \(\ds \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c\) | multiplying out | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {1 - \cos^2 b} \paren {1 - \cos^2 c} - \sin^2 b \sin^2 c \sin^2 A\) | \(=\) | \(\ds \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c\) | Sum of Squares of Sine and Cosine | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 1 - \cos^2 b - \cos^2 c + \cos^2 b \cos^2 c - \sin^2 b \sin^2 c \sin^2 A\) | \(=\) | \(\ds \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c\) | multiplying out | ||||||||||
| \(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \sin^2 b \sin^2 c \sin^2 A\) | \(=\) | \(\ds 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c\) | 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 $(1)$:
| \(\ds \dfrac {X^2 \sin^2 a \sin^2 b \sin^2 c} {\sin^2 b \sin^2 c \sin^2 A}\) | \(=\) | \(\ds \dfrac {1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c} {1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds X^2\) | \(=\) | \(\ds \dfrac {\sin^2 A} {\sin^2 a}\) |
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 $\cos C$:
| \(\ds \sin a \sin c \cos B\) | \(=\) | \(\ds \cos b - \cos a \cos c\) | ||||||||||||
| \(\ds \sin a \sin b \cos C\) | \(=\) | \(\ds \cos c - \cos a \cos b\) |
we arrive at the same point:
| \(\ds X\) | \(=\) | \(\ds \dfrac {\sin B} {\sin b}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {\sin A} {\sin a}\) |
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
- 1976: W.M. Smart: Textbook on Spherical Astronomy (6th ed.) ... (previous) ... (next): Chapter $\text I$. Spherical Trigonometry: $6$. The sine-formula.