Talk:Spherical law of cosines

Variant for angles?

An anonymous contributor just added, without sources:

The law of cosines could be also used to solve for angles, in this case it states:

$\cos(A)=-\cos(B)\cos(C)+\sin(B)\sin(C)\cos(a)\,$

I haven't seen this before, and since no sources were provided and I don't have time to check it right now, I've removed it from the article. Feel free to provide a reputable source and add it back. Or maybe there is some trivial argument to derive it from the ordinary form of the law of cosines that I'm missing right now?

—Steven G. Johnson 15:48, 3 July 2006 (UTC)

This formula should be restored. Its proof is very simple: one should take a spherical triangle dual to the given one. For it the following equalities hold $A'=\pi -a,\quad B'=\pi -b,\quad C'=\pi -c$ $a'=\pi -A,\quad b'=\pi -B,\quad c'=\pi -C$ Taking $a,b,c,A,B,C$ from here and dropping primes, one obtains the mentioned result. In the Russian literature this is known as the second spherical cosine theorem, in contrast to the first one.Aburov 22:02, 7 March 2007 (UTC)

We still need a reference for this. —Steven G. Johnson (talk) 04:44, 26 August 2008 (UTC)

The formula is entirely legit; it's probably in Bowditch (I'll confirm that) and it's certainly in other collections of spherical-trig formulas. Tim Zukas (talk) 16:49, 24 August 2010 (UTC)

I already added a reference for this formula some time ago. — Steven G. Johnson (talk) 09:25, 25 August 2010 (UTC)

Mnemonic

 "The sea is sissy and crass."
 1)    c  =  c  c   +  c s s
 2)   cos = cos cos + cos sin sin
 3)     a       b     c       A     b     c
 4) cos a = cos b cos c + cos A sin b sin c

Q.E.D.

For angles, just change the sign and cases:

 cos A = cos B cos C - cos a sin B sin C

(Invented by me, Marshall Price of Miami, while sailing to Newport, RI, circa 1983. All rights abandoned.) D021317c 23:42, 23 March 2007 (UTC)

(My username is now "Unfree".) Unfree (talk) 19:16, 31 July 2009 (UTC)

Incidentally, the mnemonic for angles can be thought of as "The sea is sissy, not crass". Unfree (talk) 19:18, 31 July 2009 (UTC)

Unit sphere?

All the "unit sphere" stuff is utterly off-topic. The formula works for _all_ spheres, including the celestial sphere, which theoretically has an infinite radius, or none at all. The same goes for "radians." You can use degrees, grads, radians -- any angular measure your heart desires. The law only involves sines and cosines anyway, and they are dimensionless. It's great for navigation, in which it's the latitude and longitude (angles) that matter, and degrees are always used. If you need to convert to distances, just multiply the degrees by the circumference of the earth and divide by 360 degrees. Occasionally, you might need the law of sines for spherical triangles, but it's so simple it can't be forgotten. It simply says that the ratios of the sides' sines to the angles' sines are all equal. D021317c 00:08, 24 March 2007 (UTC)

Well, by immediate inspection the second law can't work for arbitrary radii. It has

cos(A) = -cos(B)cos(C) + sin(B)sin(C)cos(a)

Imagine inflating the sphere by some amount, but projecting the triangle onto the new sphere. So A, B, and C are exactly as before (since they are the dihedral angles at the origin, and thus don't depend on the radius), but a is defined here as the 'arc-length' which does depend on the radius, which is clearly impossible unless a is zero. So either the article defines 'a' incorrectly, or the law only holds for the unit sphere (incidentally Weisstein's world of math also includes this problem because it explicitly says that a is the arc-length). Idmillington (talk) 18:47, 10 October 2009 (UTC)

The usual assumption when reading the formulas of spherical trigonometry is that the sides of a spherical triangle (a, b and c) are measured not by their length but by the angle they subtend at the center of the sphere. If you take another look at the law of cosines for sides (the subject of this article) you'll notice that it too makes no sense if you consider the lengths of the sides to be actual lengths rather than angles. Tim Zukas (talk) 16:47, 24 August 2010 (UTC)

Broken links

Some of the links on the page referred to in "Romuald Ireneus 'Scibor-Marchocki, Spherical trigonometry, Elementary-Geometry Trigonometry web page (1997)" are broken, but by changing the extensions of the URLs from ".htm" to ".html", most of them work.D021317c 00:31, 24 March 2007 (UTC)

Origins of formula

Is there any one person (or group) that originated the spherical law of cosines? —Preceding unsigned comment added by 146.227.1.12 (talk) 12:45, 22 August 2009 (UTC)

Consistency

The first paragraph defines lowercase letters as sides, uppercase letters as angles. Then we present a formula that takes the sine/cosine of both types of variables, only later adding that for the special case of the unit sphere, the lengths of the sides are numerically identical (in radians) to the central angles subtended by those sides.

Given that this is a narrowly focused article written specifically to explain a single formula, this seems sloppy and unnecessarily confusing. I would encourage someone who knows the subject matter to create a better graphic showing and labeling the central angles separately from the sides of the spherical triangle, and listing formulas that use only angles as arguments to trig functions.

Ma-Ma-Max Headroom (talk) 03:06, 11 March 2011 (UTC)

I couldn't agree with you more, the notation in this article was driving me nuts with it using lengths as angles in the arguments to functions which require angles. The article is simply wrong in this regard and requires some a/R, b/R, c/R substitutions for a, b, and c, when used in the trig function arguments. MrInteractive (talk) 00:14, 16 August 2022 (UTC)

Eliminating some proofs

We have four proofs in the article. I propose that we remove the first two. I don't find them to be particularly elegant or intuitive. What do you think? 64.132.59.226 (talk) 15:21, 13 February 2018 (UTC)

I am going ahead and making this edit. If you disagree, please consider this an instance of the BOLD, revert, discuss cycle; please undo the edit to the extent necessary and comment here as to what you would like to see and why. 64.132.59.226 (talk) 15:11, 20 February 2018 (UTC)

Eliminating quaternion proof?

@Manoguru et al.: I do not find the proof with quaternions to be simple. I don't see that it is elegant. I don't see that it gives any insight that the other proofs do not give. I'm not so sure it passes the notability test. The proof that the three transformations combined are the identity transformation is not apparent, so it is even possible that the proof (as currently stated) is incorrect. The proof might be useful in an articled about "see what you can do with quaternions", but this article's topic is not that. In short, I see many reasons to delete the quaternion proof. What do you say? —Quantling (talk | contribs) 21:07, 4 December 2024 (UTC)

At the least, I'd like to see better sourcing and pictures for such proofs. –jacobolus (t) 22:42, 4 December 2024 (UTC)

On the contrary, I think the proof by quaternions is the most elegant one. Quaternions are to spherical trigonometry what complex numbers are to plain trigonometry. Euler's formula for complex numbers and quaternions allows us to reduce the proofs for trigonometric identities into simple algebra, without recourse to clever diagrams or arguments. That is its power. The proof for the spherical law of cosines involves just one step of quaternion multiplication. I have re-written the description concerning the composition of transformations, and I admit that it was open to misunderstanding. Hopefully, it is much clearer and obvious why it should be an identity. @Jacobolus Two sources are given. I'm not sure what more you need. Manoguru (talk) 23:28, 4 December 2024 (UTC)

The composition of the three rotations is not the identity transformation. For example consider v at the north pole and u some point on the equator. Let w = v × u. The rotations that move v to u, then u to w, then w to v do take v back to itself. However, u ends up at w and w ends up at −u. This argues that the third proof should be removed because it is actually false! —Quantling (talk | contribs) 02:52, 5 December 2024 (UTC)

Kuipers does a bit clearer job explaining. The idea here is to rotate along one side of the triangle, then rotate around the vertex to align with the next side, then rotate along the next side, etc., so it's 6 rotations altogether. –jacobolus (t) 07:49, 5 December 2024 (UTC)

So it should be ⁠$q_{1}q_{2}q_{3}q_{4}q_{5}q_{6}=1$⁠ in the article? The current wording is, eh, not at all indicative of that. —Quantling (talk | contribs) 13:29, 5 December 2024 (UTC)

Yeah, I think the version in the article is wrong. What Kuipers does has 6 quaternions which he then rewrites as matrices and multiplies together. Which is frankly not much for a "quaternion proof" if you ask me. –jacobolus (t) 14:45, 5 December 2024 (UTC)

Ah, I see your point now. Thank you for the clarification. Yes, you are right in pointing out that the proof as it stands is not correct. However the spirit of it is not wrong. Check the version of the proof given by Brand. I guess it was wrong for me to mix the Brand's version of proof with the approach given by Kuipers. As both of you have pointed out, Kuipers' approach is needless tedious. I will update the article as per the Brand's version, whenever time permits me to do so. Manoguru (talk) 07:12, 11 December 2024 (UTC)

@Manoguru: Thank you for the new proof. For vector u, what does u⁻¹ mean? What does it mean to multiply vectors v and u⁻¹? Why should I believe that srq = 1? I think that this version of the proof is not yet ready for inclusion in the article. —Quantling (talk | contribs) 01:41, 12 December 2024 (UTC)

The ${\mathbf {u}}^{-1}$ means quaternion inverse of pure vector. The two vectors ${\mathbf {v}}$ and ${\mathbf {u}}^{-1}$ are to be multiplied via Hamilton product. In case of unit vectors, ${\mathbf {u}}^{-1}={\mathbf {u}}^{*}=-{\mathbf {u}}$. Thus, we have ${\mathbf {v}}{\mathbf {u}}^{-1}=-{\mathbf {v}}\cdot {\mathbf {u}}^{-1}+{\mathbf {v}}\times {\mathbf {u}}^{-1}={\mathbf {u}}\cdot {\mathbf {v}}+{\mathbf {u}}\times {\mathbf {v}}=\cos a+{\mathbf {w}}'\sin a$. Lastly, since $s={\mathbf {u}}{\mathbf {w}}^{-1}$, we have $s^{-1}=({\mathbf {u}}{\mathbf {w}}^{-1})^{-1}={\mathbf {w}}{\mathbf {u}}^{-1}$. Since we had shown that $rq=s^{-1}$, pre-multiplying both sides by $s$ will give $srq=ss^{-1}=1.$ I hope this answers your questions. Manoguru (talk) 07:41, 12 December 2024 (UTC)

What does "quaternion inverse of pure vector" mean? Is u a vector or a quaternion? If the latter, is it the rotation with axis along the vector u, but by what angle? Or is it some other quaternion that has to do with the vector u? —Quantling (talk | contribs) 13:39, 12 December 2024 (UTC)

Quaternions are somewhat confusingly used to represent different kinds of objects, in this case vectors (directed magnitudes in 3-space) and rotors (operators which when sandwich-multiplied around a vector will rotate it to a new orientation). It's possible to take the reciprocal of any of these, as a quaternion, by flipping the sign of the "imaginary" portion and taking the reciprocal of the magnitude.

In my opinion the result is always clearer if instead expressed using geometric algebra (Clifford algebra), where a rotor is the sum of a real scalar part and a bivector part. Then there's less confusion between a bivector vs. a vector. –jacobolus (t) 17:51, 12 December 2024 (UTC)

What does −v · u⁻¹ + v × u⁻¹ mean? The first addend −v · u⁻¹ seemingly is a scalar but the second addend v × u⁻¹seemingly is a vector; what does it mean to add a scalar to a vector? Also, by "vector" do you mean 3-dimensional in R³ or something 4-dimensional along the lines of the coefficients of a quaternion? —Quantling (talk | contribs) 17:47, 12 December 2024 (UTC)

As with the originally deleted proof, with the current proof's

q = cos a + w′ sin a

r = cos b + u′ sin b

s = cos c + v′ sin c

I have the easy "north pole, equator, third-point" proof (above) to show that srq is not the identity transformation. Am I missing something?

—Quantling (talk | contribs) 20:16, 12 December 2024 (UTC)

Text of "third proof". Converse above, edit below, until we smooth this out better

The following proof relies on the concept of quaternions and is based on a proof given in Brand: Let u, v, and w denote the unit vectors from the center of the unit sphere to those corners of the triangle. The great circle arc between any two points on the unit sphere can be represented by a unit quaternion. Let us represent the great circle arcs $\mathrm {arc} \;UV,\mathrm {arc} \;VW,\mathrm {arc} \;WU$ by quaternions $q,r,s$ as:

${\begin{aligned}\mathrm {arc} \;UV&\sim q={\mathbf {v}}{\mathbf {u}}^{-1}=\cos a+{\mathbf {w}}'\sin a\\\mathrm {arc} \;VW&\sim r={\mathbf {w}}{\mathbf {v}}^{-1}=\cos b+{\mathbf {u}}'\sin b\\\mathrm {arc} \;WU&\sim s={\mathbf {u}}{\mathbf {w}}^{-1}=\cos c+{\mathbf {v}}'\sin c\end{aligned}}$

where ${\mathbf {w}}',{\mathbf {u}}'$, and ${\mathbf {v}}'$ are the axes of the respective great circle arcs following the right-hand rule and $a,b,c$ are the respective arc angles. The "vector" composition of arcs such that $\mathrm {arc} \;UV+\mathrm {arc} \;VW=\mathrm {arc} \;UW$ corresponds to the quaternion equation

$rq=({\mathbf {w}}{\mathbf {v}}^{-1})({\mathbf {v}}{\mathbf {u}}^{-1})={\mathbf {w}}{\mathbf {u}}^{-1}=s^{-1}.$

Thus, we have $srq=1$. Using the Euler form of the quaternions in the previous identity, we have

$(\cos b+{\mathbf {u}}'\sin b)(\cos a+{\mathbf {w}}'\sin a)=\cos c-{\mathbf {v}}'\sin c.$

Expanding the left-hand side, we obtain

$(\cos a\cos b-{\mathbf {u}}'\cdot {\mathbf {w}}'\sin a\sin b)+({\mathbf {u}}'\cos a\sin b+{\mathbf {w}}'\sin a\cos b+{\mathbf {u}}'\times {\mathbf {w}}'\sin a\sin b).$

Equating the scalar parts on both sides of the identity, we obtain

$\cos a\cos b-{\mathbf {u}}'\cdot {\mathbf {w}}'\sin a\sin b=\cos c.$

The cosine law is obtained by recognizing that ${\mathbf {u}}'\cdot {\mathbf {w}}'=\cos(\pi -C)=-\cos C$. Thus,

$\cos a\cos b+\cos C\sin a\sin b=\cos c.$

—Quantling (talk | contribs) 20:29, 12 December 2024 (UTC)