Assume we are working on a $2$-sphere of radius $1$. Suppose we have a triangle with vertices $A, B, C$ and sides $a, b, c$ opposite to the respective angles.
My starting point is the spherical law of cosines.
I have been following this wiki link with the specific subsection linked.
However, I'm finding that the proof is severely incomplete and lacking, and I've been having trouble filling in a few details.
Starting with the law of cosines and doing some algebra, we find
$$ \Big( \frac{\sin A}{\sin a}\Big)^{2} = \frac{1 – \cos^{2}a – \cos^{2}b – \cos^{2}c + 2\cos a \cos b \cos c}{\sin^{2}\!a \, \sin^{2}\!b \, \sin^{2}\!c}. $$
To take the square root, we need to show that the numerator is nonnegative. As we can see, the left-hand side here is a square and the denominator is a square. Therefore they are nonnegative, and this implies that the numerator is indeed nonnegative.
When we take the square root, we must write
$$ \Big| \frac{\sin A}{\sin a}\Big| = \frac{[1 – \cos^{2}a – \cos^{2}b – \cos^{2}c + 2\cos a \cos b \cos c]^{1/2}}{|\sin a \sin b \sin c|}. $$
This is where I am thinking that the wikipedia page is getting things wrong, because if you allow for triangles with angles $\ge \pi$, then you may run into sign discrepancies unless you include the absolute value signs. I am unable to tell one way or the other at this moment.
I suspect the absolute value signs are necessary actually. However, we can proceed as follows.
The right-hand side of the above formula is invariant under cyclic permutations of the vertices (and the respective sides), so it follows that
$$ \left| \frac{\sin A}{\sin a}\right| = \left| \frac{\sin B}{\sin b}\right| = \left| \frac{\sin C}{\sin c}\right|. $$
This is the spherical law of sines, but it contains the absolute value signs. I conjecture that we can drop the absolute value signs (unlike in the other formula above; this is very tricky), but I am having difficulty proving this step.
I am wondering if anyone can give a quick reason why the signs of $\frac{\sin A}{\sin a}$, $\frac{\sin B}{\sin b}$, $\frac{\sin C}{\sin c}$ all match. This would complete the approach I'm trying to take.
Questions
To summarize, I have the following questions. Thoughts to any one of these would be very helpful.
- Is there a proof of why the signs of $\frac{\sin A}{\sin a}$, $\frac{\sin B}{\sin b}$, $\frac{\sin C}{\sin c}$ all match?
- Am I correct in asserting that the wiki link is committing a fallacy here and the the formula it wrote requires absolute value signs as I wrote it?
- Is my approach unsalvagable? Do I need to restart and an outright different approach?
Best Answer
Okay, I think I have an answer to my own questions, so I wanted to report the findings.
For reference, here is the pic of the wiki section that I referenced (in case it gets edited).
As one of the commenters pointed out, it is conventionally assumed that in all spherical triangles each side and angle is $< \pi$. This means I can drop the absolute value signs by assumption, and everything holds exactly as expected.
However, I think you can still prove the law of sines in the general case. The reader should be careful and check my work for errors here. The answers to my question are as follows.
In the general case, we can prove the law of sines in a different way than the way I presented in my original post. Given our triangle (vertices $A, B, C$ and sides $a, b, c$), let $\vec{u}, \vec{v}, \vec{w}$ be the unit vectors from the origin and pointing to vertices $A, B, C$, respectively. Using rotations and reflections, reorient the triangle so that $$ \vec{u} = (0, 0, 1), \quad \vec{v} = (\sin c, 0, \cos c), \quad \vec{w} = (\sin b\cos A, \sin b\sin A, \cos b). $$ Remember that because we are working on the unit sphere, the angle made between two vectors is the same as the arc length on the sphere from one vector head to the other (this applies even when we make the angle $>\pi$). Then a certain triple product gives us $$ \vec{u}\cdot(\vec{v}\times\vec{w}) = \begin{vmatrix} 0 & 0 & 1 \\ \sin c & 0 & \cos c \\ \sin b\cos A & \sin b\sin A & \cos b \end{vmatrix} = \sin c\sin b\sin A. $$ The triple product is invariant under cyclic permutations of the vectors. By reorienting the coordinate system (keeping the orientation the same) and applying the same identity, we obtain $$ \sin c\sin b\sin A = \sin a\sin c\sin B = \sin a\sin b\sin C. $$ By dividing by $\sin a\sin b\sin c$ (no side length can be equal to $\pi$ or else we wouldn't have a triangle so this is a nonzero quantity), we obtain the law of sines. I think this reasoning works even when some of the angles and sides are $> \pi$.
Concerning the formula in the wiki page: If we're assuming all angles and sides are $< \pi$, then we can ignore the absolute value signs. Otherwise you need them (I may make a counterexample if someone requests it). In any case, I still take issue with the fact that they didn't justify why the numerator is nonnegative before square rooting it, even though showing it is pretty trivial in my original post.
The original approach works only if you assume all angles and sides are $<\pi$. Otherwise, you do need to do a completely different approach taken in the answer to #1. It's just easier that way.