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)

Answers to all these questions can be found in the wiki article on Quaternion in general and more specifically in the section on Scalar and vector parts. Yes, "vector" here means 3D vector. Manoguru (talk) 07:34, 18 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)

Your counter "proof" is wrong in this case. In this example, since each quaternion is defined by two vectors 90 degrees apart, the action of each quaternion on a vector is to rotate it by 180 degree, not 90 degrees as you seem to have implied. For example, the first quaternion $q$ defined by the north pole and a point on equator will rotate a vector by 180 degree. So the north pole will end up at the south pole, and the equator point will end up at the opposite end of the equator. If you follow through with other two 180 rotation operations, all the points will end up where they started. Also, the proof that $srq=1$ is already given as one of the steps, so I don't know what you are really trying to "disprove". Manoguru (talk) 07:22, 18 December 2024 (UTC)

"Let us represent the great circle arcs $\mathrm {arc} \;UV,\mathrm {arc} \;VW,\mathrm {arc} \;WU$ by quaternions $q,r,s$ as:" is followed by formulas for the quaternions q, r, and s. The angles listed for the quaternions q, r, and s are, a, b, and c where "$a,b,c$ are the respective arc angles". Or is each of a, b, and c instead equal to 90° or 180°? —Quantling (talk | contribs) 13:31, 18 December 2024 (UTC)

Perhaps by q you mean the rotation by 180° around the axis parallel to u + v. That would take u to v (and vice-versa) and be 180°. Is that what is going on? —Quantling (talk | contribs) 16:22, 18 December 2024 (UTC)

As a courtesy to other editors, please refrain from deleting stuff without further discussion in the talk page, especially something that has been well cited using reputable sources, and does not deviate from the sources. Just because you failed to understand the proof does not mean nobody has understood it. Manoguru (talk) 07:01, 18 December 2024 (UTC)

I am excited for your contribution and am hopeful that it can be made to work. However, sorry, you have that courtesy backwards. Material that is brand new to the article does not get the courtesy of staying in the article without some form of consensus. If it were easy enough to fix this, I would have done so. Especially because this is not an article on quaternions you need to explain how we add the result of a dot product to the result of a cross product. How is it that vu⁻¹uw⁻¹ has the middle two factors canceling to produce vw⁻¹ needs to be explained. Etc. You can do it in text, with footnotes, or wikilinks, but at the moment, especially readers who are not mathematically astute will not be able to understand this text. It needs a lot of work. I think the compromise of working on the text on the talk page is a reasonable way forward and I ask you to join me in that endeavor. —Quantling (talk | contribs) 13:40, 18 December 2024 (UTC)

The quaternion article discusses inverses and conjugation of quaternions but not of 3d-vectors. If in this article we mean inverse of some quaternion(s), let's explicitly indicate which quaternion(s). —Quantling (talk | contribs) 13:48, 18 December 2024 (UTC)

"The ${\mathbf {v}}{\mathbf {u}}^{-1}$ is the Hamilton product of ${\mathbf {v}}$ and ${\mathbf {u}}^{-1}$. " Again if there is some implied quaternion, what is it? The text currently indicates instead that v and u⁻¹ are vectors. —Quantling (talk | contribs) 13:51, 18 December 2024 (UTC)

If you have two "vectors" (pure imaginary quaternions) ⁠$v$⁠ and ⁠$u$⁠, then ⁠$vu^{-1}$⁠ is a quaternion which scales and rotates ⁠$u$⁠ into ⁠$v$⁠ when multiplied on on the left ⁠$(vu^{-1})u=v(u^{-1}u)=v$⁠. But this can't be used directly as a representative of the 3-dimensional rotation, because multiplying ⁠$vu^{-1}$⁠ by vectors which are not in the subspace spanned by ⁠$v$⁠ and ⁠$u$⁠ will not give a correct result. Instead, the 3d rotation is represented by the quaternion ⁠$\textstyle {\sqrt {vu^{-1}}}$⁠, and is performed on an arbitrary other vector ⁠$w$⁠ by the "sandwich product" ⁠$\textstyle {\bigl (}vu^{-1}{\bigr )}{\vphantom {)}}^{1/2}w{\bigl (}vu^{-1}{\bigr )}{\vphantom {)}}^{-1/2}$⁠. –jacobolus (t) 15:21, 18 December 2024 (UTC)

I don't understand:

If you have two "vectors" (pure imaginary quaternions) ⁠$v$⁠ and ⁠$u$⁠, then ⁠$vu^{-1}$⁠ is a quaternion which scales and rotates ⁠$u$⁠ into ⁠$v$⁠ when multiplied on on the left ⁠$(vu^{-1})u=v(u^{-1}u)=v$⁠.

There are many rotations that take a unit vector u to a unit vector v. For example take any one you are thinking of and then subsequently rotate around v. So, the notation vu⁻¹ seemingly isn't defining a unique rotation. If we eliminate that ambiguity by taking q as the rotation from u to v that has an axis of rotation of u × v and likewise for r and s — which seems to be what the proof is telling us to do — then we've shown that qrs is not the identity transformation. What am I missing? —Quantling (talk | contribs) 15:40, 18 December 2024 (UTC)

You might want to start by looking at quaternions and spatial rotation (or perhaps some more clearly written source containing the same material). The quaternion ⁠$vu^{-1}$⁠ is the quaternion product, which is clearly uniquely defined. If you multiply this quaternion on the left by any pure imaginary quaternion ⁠$w$⁠ in the 2-dimensional subspace spanned by ⁠$v$⁠ and ⁠$u$⁠ it has the effect of rotating and scaling it by the same amount you would rotate and scale ⁠$u$⁠ to get ⁠$v$⁠. But this doesn't work out correctly for pure imaginary quaternions not in that subspace. To eliminate the extraneous parts of the product you need to break the rotation into two pieces, ⁠$\textstyle {\sqrt {vu^{-1}}}$⁠ and the inverse ⁠$\textstyle {\sqrt {uv^{-1}}}$⁠, and then "sandwich multiply" your target quaternion by one of them on the right and the other on the left. This leaves the desired in-plane rotation while the out-of-plane parts of that product cancel each-other. Because of this the quaternion ⁠$\textstyle {\sqrt {vu^{-1}}}$⁠, normalized to unit magnitude, is commonly used as a representation of the rotation from the direction of ⁠$u$⁠ to the direction of ⁠$v$⁠. (Here we mean the shortest such rotation, along a great-circle arc, with no extra spinning at the end.) –jacobolus (t) 17:50, 18 December 2024 (UTC)

I am starting to understand. Please see the next section where I have made edits. What do you think? —Quantling (talk | contribs) 17:31, 18 December 2024 (UTC)

Thank you all for your patience. I have modified the text to distinguish vectors from vector parts of quaternions, use bold u for a 3-dimensional vector and italics u = (0, u) for the corresponding quaternion, to highlight how quaternions are related to rotations, and similar. And I have re-inserted the text into the article since I was the largest block to that action. I hope that you find my changes to be constructive and I ask that you continue to make updates to it, large or small, to make it the best it can be. Thank you —Quantling (talk | contribs) 18:57, 18 December 2024 (UTC)

As an aside: this proof makes good use of the fact that a 180° rotation about an axis u, followed by a 180° rotation about an axis v produces a rotation that is two applications of the rotation that takes u to v along a great circle. This is cool. In particular, any non-degenerate polygon on the surface of a sphere with edges that are great circles implies a sequence of rotations as the polygon's vertices are traversed in one direction around the polygon. By traversing every edge of the polygon to return to the vertex with which we started, we are sure that this sequence of rotations takes that starting vertex back to itself. However, the total transformation need not be the identity. (In fact, I think it a rotation along the axis parallel to the starting vertex with angle equal to ±area of the polygon. Or something similar to that.) But instead of using each rotation once, R_n...R₂R₁ as applied to a column vector on the right, if we use each rotation twice, R_n²...R₂²R₁² the result is the identity transformation. That's cool. And in the limit as the polygon sides get shorter and shorter (and the number of sides gets more and more numerous) this might all apply to closed curves on a sphere. That's cool too. —Quantling (talk | contribs) 17:58, 19 December 2024 (UTC)

Via Einstein notation

The "second proof" section gives a proof via the Binet–Cauchy identity. That's useful and intuitive for people who know that identity. However, I think that people who don't know that identity but do have a familiarity with Einstein notation would benefit from seeing the proof using the latter. I'd like to see us append to that section's text with text along the lines of:

Restated with Einstein notation, the result follows by contracting the identity for 3-component vectors, $$\epsilon ^{\rho \alpha \beta }\epsilon _{\rho \mu \nu }=\delta _{~\mu }^{\alpha }\delta _{~\nu }^{\beta }-\delta _{~\nu }^{\alpha }\delta _{~\mu }^{\beta }\,,$$ with $u_{\alpha }v_{\beta }u^{\mu }w^{\nu }\,.$

Thoughts? —Quantling (talk | contribs) 14:31, 18 December 2024 (UTC)

Should I boldly reinsert this into the article to spur more comments here? —Quantling (talk | contribs) 15:54, 24 December 2024 (UTC)