Talk:Spherical trigonometry

Radians or degrees

this artical seems over complicated by use of both degrees and radians. I propose to a change to make it entirely radians.
so for example
"E=A+B+C-π radians" —Preceding unsigned comment added by 138.251.30.10 (talk) 15:52, 3 October 2007 (UTC)[reply]

I'm not sure. This whole article contradicts itself in the way that it is now. In the example you gave, for example, it says the spherical excess formula calculates how much the triangle exceeds 180 degrees. But in the formula, it subtracts pi, not 180 degrees. On the other hand, geometry is often performed using degrees. I would agree to your proposition, but I do still have reservations. I'll begin switching it over now. EDIT: Actually, I've just thought now to check the page for triangles. That whole page is in degrees. Should this page align to that? (Buggy793 (talk) 17:22, 27 June 2011 (UTC))[reply]

Explicit Radius?

It seems necessary to say something about the radius in the sin law? —Preceding unsigned comment added by 71.230.115.205 (talk) 20:07, 7 May 2009 (UTC)[reply]

Valid theorems?

Please try to answer whether these questions are known:

Is SSS a valid congruence theorem for spherical triangles??
Is SAS a valid congruence theorem for spherical triangles??
Is ASA a valid congruence theorem for spherical triangles??
Is AAS a valid congruence theorem for spherical triangles??

66.245.7.28 18:33, 6 Nov 2004 (UTC)

Yes, there is a SSS congruence for spherical triangles such that AAA is also true where Angle A = Angle B = Angle C = 270 Degrees (implying that each angle is 90 Degrees). The distance of each "side" in this "spherical" model would be the Circumference (measurement 360 Degrees) divided by 4. For a sphere, each Side then equals 2*Pi*r/4. For a Unit Sphere, this Equilateral Triangle has it's sides equa to 1.570796327 == 1.57. "I Think" [[Rashaad Jamal Frazier ali as Bump P Johnson | 85019 2011.1212.013101] — Preceding unsigned comment added by Bumppjohnson (talk • contribs) 08:33, 12 December 2011 (UTC)[reply]

These are good questions--do SSS, SAS, ASA, and/or AAS imply congruence of spherical triangles? The answer deserves to be in the article. Does anyone know? Duoduoduo (talk) 17:26, 5 June 2010 (UTC)[reply]

Certain degenerate triangles (those with antipodal vertices, e.g.,) make these congruences not generally valid, even if they are generically valid. ASA is the only one that actually holds: since "congruence" is upto an action of O(3) we can situate one of the vertices at a given angle at the south pole and run the side with length given, up the prime meridian. Knowing both angles to either end of the segment of fixed length ensures that the other 2 sides emanate with a uniquely determined trajectory, and thus will meet each other at a uniquely determined point, so ASA is valid. On the other hand, take side lengths of pi, pi/2, pi/2. One has a continuous family of non-congruent triangles with such measurements.

As an aside, the sum of angles - pi = area formula should probably be included. 128.84.234.59 (talk) 20:11, 8 March 2011 (UTC)[reply]

Napier's circle

I corrected the image for Neper's circle

Eroica 08:30, 1 November 2005 (UTC)[reply]

I corrected the image a bit more. The original figure was mismatched with the description of the equations given in the text. The figure, as it was previously drawn was similar to the one on the referred page: http://www.rwgrayprojects.com/rbfnotes/trig/strig/strig.html. The equations as stated in the wiki article are the co-equations of this page. So either the equations need to be corrected, or the picture needed to have the complementary angles drawn. I chose to fix the picture.

Arif Zaman 16:29, 19 December 2005 (UTC)[reply]

the image labeled "Spherical Triangle" is actually of a "right spherical triangle", as indicated by its German title: rechtkugel dreieck, literally "right-angle three-corner". --Don 23:56, 13 August 2007 (UTC)[reply]

Since the image is a projection into R^2 and none of the angles are labelled with an "L" type symbol indicating an angle of pi/2, i think that the figure can be used as a general spherical triangle on S^2 without much problem... Boud (talk) 16:57, 4 March 2008 (UTC)[reply]

The choice to change the picture (made by Arif Zaman above) was an unfortunate one, because it completely loses the incredibly clever mnemonic developed by Napier for the right spherical triangle formulae; I will attempt to restore it. I also corrected the title above. Ted Sweetser (talk) 06:44, 5 June 2010 (UTC) Okay, I created a new version of the figure in .svg format with the complements reversed, uploaded that to Commons, and changed the link (old and new figures are in the "spherical trigonometry" category). I then edited the text to give the correct corresponding formulae and added a description of Napier's mnemonic. Ted Sweetser (talk) 22:04, 5 June 2010 (UTC)[reply]

There is an error with the statement "the sine of the middle angle is equal to the product of the cosines of the opposite angles." As a counter-example, consider the case of a great triangle, where all corners are 90 degreees (I presume this also counts as a right spherical triangle by Napier's definition). If this is the case, a, b, and c will all be 90 degrees as well, and the statement sin c = cos a cos b evaluates to 1=0 although it is implied by the given drawing of Napier's pentagon. As further proof, if the right spherical triangle in question is not quite a great triangle, a, b, and c may still be very close to 90 degrees, sin c would be close to 1, and cos a cos b would still be very close to 0. I think the statement should probably read "the cosine of the middle angle is equal to the product of the cosines of the opposite angles." This makes more sense in the case of the angles opposite c, however, I cannot verify it. Planet sputter (talk) 20:42, 26 November 2010 (UTC)[reply]

The counter-example you give is in error because you forgot to use the complement of c (indicated by a bar over the c) in the formula. The sine of c_bar is the cosine of c, so it comes out as 0 = 0, and the formulae are correct. Ted Sweetser (talk) 03:02, 14 February 2011 (UTC)[reply]

Minor vandalism

No idea why someone would want to remove large sections of this page as has been done recently. The last edit (which i reverted) was from someone at an IP in Kowloon, P.R.China. Maybe just experimenting to see if wikipedia is available. Or a Chinese censor trying to be disruptive? Who knows, this is just speculation... Let's just revert to the full version. Boud (talk) 16:57, 4 March 2008 (UTC)[reply]

Some Work Needed

This entry needs some work before it is explanatory. It uses a lot of terms without defining them. For example, the formula sin A/sin a etc is stated without explaining what a, A, b, B etc refer to. The diagram is pretty but the letters in it do not correspond to those used in the entry. —Preceding unsigned comment added by Gnomon (talk • contribs) 23:06, 21 November 2008 (UTC)[reply]

Also, the gamma label has been lost from the first illustration. It would also be nice if there were some way to accentuate the triangle in the figure, e.g. by heavier lines defining the triangle of interest, or by color. —Preceding unsigned comment added by 98.222.58.129 (talk) 13:56, 5 February 2010 (UTC)[reply]

Credit for Girard's Theorem

The Hyperbolic Geometry page credits J.H. Lambert with Girard's Theorem; this page Harriot. (At least you agree it wasn't Girard! :-) Perhaps there was a Harriot-->Girard-->Lambert progression? Anyway, I think the two pages should be consistent.Jamesdowallen (talk) 14:24, 20 January 2010 (UTC)[reply]

I don’t know the facts of the case, but Hyperbolic Geometry no longer mentions Girard’s Theorem at all AFAICT, and Thomas Harriot#Legacy agrees with this article—with a citation, yet! —Odysseus1479 (talk) 07:49, 21 September 2012 (UTC)[reply]

Figure

The figure showing a spherical triangle can be improved to display the projection of the circles as ellipses. Bo Jacoby. —Preceding unsigned comment added by 89.239.205.120 (talk) 22:05, 27 January 2011 (UTC)[reply]

Inconsistency in notation

The spherical sine law as given under Spherical trigonometry#Islamic world uses Greek letters for angles and italics for sides, mathematician-style as at Law of sines#Spherical case, but under Spherical trigonometry#Identities they’re respectively large and small letters, as is more usual among navigators and consistent with the others in the section. Moreover, FWIW, the first is ‘upside down’ WRT the other two. Do we need to provide this formula twice in one article, anyway? —Odysseus1479 (talk) 07:40, 21 September 2012 (UTC) Edited 07:56, 21 September 2012 (UTC)[reply]

Major tidy up

I started out with the intention of adding Napier's rules but I have branched out to a general tidy up, re-ordering material where necessary. The section on identities will get some more added. The section on derivations will be condensed a little. Please check for typos. Please let me have any adverse reactions. There is still more to come. Peter Mercator (talk) 22:47, 23 July 2013 (UTC)[reply]

Yes! the page definitely needs an overhaul and has several gaps that need filling. One request: L'Huillier's theorem is the only formula offered to calculate the spherical excess and this is used by some software packages to compute the area of a general spherical polygon. However a triangle is sometimes badly characterized by its sides so the resulting area can be inaccurate. A better alternative for area calculations is one of Napier's analogies applied to the quadrangle bounded by a segment of a great circle, two meridians, and the equator for which (

\phi ,\lambda

are latitude and longitude)

\tan {\frac {E}{2}}={\frac {\sin {\frac {1}{2}}(\phi _{2}+\phi _{1})}{\cos {\frac {1}{2}}(\phi _{2}-\phi _{1})}}\tan {\frac {\lambda _{2}-\lambda _{1}}{2}}.

(Of course, this requires a generalization of the definition of spherical excess to polygons; this is also needed!) cffk (talk) 12:23, 24 July 2013 (UTC)[reply]

Upgrade is proceeding. This article is obviously fairly heavy on mathematics. Only recently have I discovered the great improvement made the Preference->Appearance->mathJax. Is there a nice way of flagging an article which encourages readers to set this preference? Peter Mercator (talk) 21:51, 26 July 2013 (UTC)[reply]

The end of my upgrade is now in sight. The History section had been lifted almost verbatim from History of trigonometry. I shall therefore replace it with a link and brief comments. The section on Conruent triangles didn't really sit well here so I shall moved it to Congruence (geometry). I shall also move the Spherical excess to the end of the article (and include the above expression) although Spherical excess probably deserves a page of its own where its applications could be discussed. Perhaps a stub should be created? Peter Mercator (talk) 11:43, 28 July 2013 (UTC)[reply]

A fairly major limitation to the suggestion to use Mathjax rendering is that it can only be used by people who are logged into Wikipedia. I've no idea how many people this leaves out, possibly it's 90% of readers. In any case, I've started lobbying for a way to expand the use of the Mathjax renderer; see Village pump (technical)#How can an user without a Wikipedia account get the benefits of Mathjax rendering?. Please consider adding to this discussion. cffk (talk) 00:33, 6 August 2013 (UTC)[reply]

Polar Triangle Question

The image of the Polar Triangle shows references to A',B',C',A,B,C. Yet, the text formulas that seem to be associated with it include a,b,c,a',b',c' but these lower-case points are not shown on the image. It may be confusing for the easily confused such as me. Tesseract501 (talk) 06:02, 1 April 2014 (UTC)[reply]

By convention, where vertices are labelled with capital letters, the corresponding small letters refer to the sides opposite, respectively—as shown in the illustration immediately above. Adding all six labels to this drawing could make it look rather cluttered and impede visualization in three dimensions.—Odysseus 1479 06:31, 1 April 2014 (UTC)[reply]

Formatting trig formulas

I've just reverted a bunch of reformats of trig formulas. These made the formulas less easy to understand, e.g., by using built-up fractions instead of a 1/2 multiplier for half-angle formulas and by introducing unnecessary parentheses. I recommend that such blanket changes be discussed on this talk page first. cffk (talk) 12:42, 23 November 2015 (UTC)[reply]

These were no blanket changes but corrections. As stated now (without the brackets), these formulas are formally incorrect. The parentheses are not optional here as there is no universally accepted or understood rule giving implied multiplication a higher priority than explicit multiplication, therefore we should avoid this formally incorrect jargon where it causes difficulties for the readers to correctly interprete the formulas.

I am open in regard to using a 1/2 fore-factor or using fractions instead, it is only that by using proper fractions we can avoid the parentheses.

Regarding readability, is there any particular reason to use the ^(1/2) form instead of the much easier to read square root form (which, again, made parentheses obsolete)?

--Matthiaspaul (talk) 15:27, 23 November 2015 (UTC)[reply]

I disagree with your assertion that the inclusion of parentheses is necessary to make the formulas "formally correct" (see my comment on Meridian arc talk page). (And I don't understand the point you are making about implied and explicit multiplications.) I can go either way on the square root signs; the editor that used the 1/2 power presumably thought that this looked better. I tend to agree (the large square root sign is distracting); in any case, I would normally defer to the preferences of the editor. cffk (talk) 16:25, 23 November 2015 (UTC)[reply]

It’s a bit late for this discussion, but note that notation of the form

{\textstyle \tan {\tfrac {1}{2}}a}

or

{\textstyle \tan {\tfrac {1}{2}}{\bigl (}{\tfrac {1}{2}}\pi -\theta {\bigr )}}

has been widely adopted in spherical trigonometry for centuries, and is found in most of the best and most historically important textbooks. It is not ambiguous, just differs slightly from conventions found in (some) modern sources which don’t typically need nearly so many half-angle trig functions. –jacobolus (t) 01:45, 19 October 2022 (UTC)[reply]

Alternative derivations

I have added a paragraph on alternative derivations of the fundamental formulae and at the same time I have given a reduced description of the proof added by recent editors. I hope the article will not be expanded to include more of the alternative derivations and thereby lose its concision. I would be grateful if experienced editors can comment. Peter Mercator (talk) 20:23, 10 May 2017 (UTC)[reply]

I like your edit as it provides a sensible approach to the alternative derivation "problem". I say problem because I have seen too many articles where some editors have felt that every alternative should be reported on. This may be alright if the subject of an article is the proof of a result, but outside of that limited scenario I don't see that it has much value. --Bill Cherowitzo (talk) 21:16, 10 May 2017 (UTC)[reply]

I agree—avoid multiple derivations and keep the article concise. cffk (talk) 01:11, 11 May 2017 (UTC)[reply]

Sadly I am again reverting the edits of new wikipedian User:Joao Nemmers. There is absolutely no need to bring the parallelpiped into the discussion. The scalar triple product is simply a descriptive name of a scalar product in which one of the vectors is a vector product of two others: it is not defined as a volume of a parallelpiped although it can certainly be interpreted as such. Similarly the cyclic symmetry follows from algebraic definitions and it does not need a proof by the basis independence of volume. Furthermore I know of no problems using spherical trig in which the volume interpretation is of relevance. Peter Mercator (talk) 18:10, 11 May 2017 (UTC)[reply]

I am not sure if concision is necessarily a virtue. If someone seeks concision (s)he can simply stop reading beyond the statement and the section on Alternate derivations. Students (even in high school) and researchers often seek more than simply the statement of the results. Vector methods are fundamental in modern geometries and, perhaps more importantly, very accessible. Of course purely algebraic methods are fine, but Wikipedia pages on the scalar triple product and on the parallelepiped clearly talk about their application to volumes. Given that the cosine rule has been accorded a formal proof, I see no harm in using the same basis to derive the sine law. Relating to the volume is a simple extension that does not affect the concision in my opinion. This relation to the volume is not unusual and given in http://mathworld.wolfram.com/SphericalTrigonometry.html and the book by Jennings on Modern Geometry. I did not want the page to simply be a collection of identities and not offer accessible insight into some of the mathematics of spherical trigonometry. For example, there is no mention of dual triangles on the sphere that could also provide alternative proofs to the spherical law of sines. Again, the book by Jennings is worth mentioning. Joao_Nemmers 16:08, 12 May 2017.

Limits in sine and cosine rules

I have recently made some changes in this section that I think need some additional explanation. The issue has to do with the comparison of the spherical laws with their planar counterparts. The phrasing "in the limit" that had been used is ambiguous without referring to the changing variable, in this case an increasing radius. An editor objected to my inclusion of this information with the rationale that the page was written from the viewpoint of a fixed radius for the sphere (in fact, unity), so an increasing radius should not be mentioned. I am not sold on this rationale, but fortunately an alternative phrasing was given (that I have now adopted) that made the point moot. Unfortunately, another editor, for the sake of parallel construction, changed this back to a limit statement (without a variable). My recent edit keeps the parallel construction but removes the limit statement. In this setting, one can either talk about limits or about approximations, but if you phrase things in terms of limits you need to specify the variable in an article at this level, otherwise you are falling into the "math jargon" trap. Yes, it is easy for an experienced reader to figure out what "in the limit" is going to mean in this setting, but this is not the intended audience. Also, the terms, "reduces to" and "is a special case" which had been used here are inaccurate and misleading, so had to be removed. There is one more instance of the use of "in the limit" occurring in the last section. I have left this alone for the time being (assuming that only more sophisticated readers would get that far), but the same reasoning could be applied there.--Bill Cherowitzo (talk) 18:10, 23 July 2017 (UTC)[reply]

The image of the polar triangle

The image of the polar triangle (upper right) is wrong. For all corners of the polar triangle to lie on the visible half of the sphere the origin in the image must lie inside ABC. As AB lies SE of the origin, the corner C' lies on the backside. The image with the poles as shown in the figure implies that AB is the other side of the great circle (i.e. ellipse in the figure) that lies to the NW of the origin (also opposing poles should be at the same distance from the rim in the image, which they are not, and the polar axises be perpendicular to the major axises of the ellipses, which looks doubtful). Compare my figure to the lower right (note that the angles between the "equatorial planes" are not the same in both images). Episcophagus (talk) 08:13, 16 July 2019 (UTC)[reply]

Also: the smaller triangle you start with, the greater polar triangle you will get (when the sides are all 90° the triangles are identical, and when one of the triangles is a dot, the other is a great circle). In the current image both the original and the polar triangles are smaller respectively than in my image, which is a contradiction. The sides of the polar triangle does not look like correct ellipses, which should be symmetrical around the origin (look at B'C'in the current image - in no way can that be part of an ellipse centered around the origin with the major axis equal to the diameter of the sphere, and neither can A'B' nor A'C'). Episcophagus (talk) 09:43, 16 July 2019 (UTC)[reply]

suggestion

The text contains

The angles of proper spherical triangles are (by convention) less than π so that π < A + B + C < 3π. (Todhunter,[1] Art.22,32).
The sides of proper spherical triangles are (by convention) less than π so that 0 < a + b + c < 2π. (Todhunter,[1] Art.22,32).

A naive reading suggests that there are SIX independent constraints on the triangle, moreover one may wonder why the sides' sum is 2*pi rather than 3*pi. What about replacing the second sentence by

The sides of proper spherical triangles are less than π and satisfy 0 < a + b + c < 2π; it may be shown that a spherical triangle is proper if and only if all the sides are less than pi . (Todhunter,[1] Art.22,32).

(perhaps another article of Todhunter is needed; I am not absolutely sure about the if-and-only-if statement, if I am wrong please tell me this)Suppongoche (talk) 13:42, 28 December 2020 (UTC)[reply]

Your suggestion makes it confusing for me. What's wrong with the current text?

That the sum of the angles is <3*pi and that the sum of the sides is <2*pi, are both easily explained by the limit case of 3 points on a great circle?

Gollem (talk) 14:29, 28 December 2020 (UTC)[reply]

There is a double implication between the two statements and I feel that is important to make this clear from the very start, in other words to say that are equivalent. The presence of two "by convention" hides this - no problem if one knows this, but may be a problem if he does not. Our disagreement on this evidently reflects a disagreement on the probable nature of the readers.Suppongoche (talk) 17:04, 28 December 2020 (UTC)[reply]

I did not make any assumptions on the nature of other readers. I only mention that your suggestion confuses me. Moreover, it did not make me understand what you explained in your last comment.

I think that the solution is to introduce the definition of a proper triangle elsewhere (not in the notation section) and list the properties of such triangle there.

Gollem (talk) 21:39, 28 December 2020 (UTC)[reply]

Thanks and happy new yearSuppongoche (talk) 07:23, 30 December 2020 (UTC)[reply]

please check

should

(3π for proper angles).

be replaced by

(3π for proper triangles).

? thanks 151.29.137.229 (talk) 21:18, 22 August 2023 (UTC)[reply]

No, but that whole section is confusingly worded and technically sloppy.

By "proper angle" it means angles less than a straight angle (i.e.

0<\theta <\pi

radians), so that a triangle could have almost

3\pi

radians as a sum of such internal angles (spherical excess of almost

2\pi

, i.e. area of nearly a whole hemisphere). If you allow "improper" angles (e.g. take any small spherical triangle and swap your idea of the interior and exterior) then the triangle can have area of nearly the entire surface of the sphere (spherical exces of almost

4\pi

or sum of "internal" angles of almost

5\pi

). –jacobolus (t) 22:17, 22 August 2023 (UTC)[reply]

Add an area formula

On the Lexell's theorem page, appears the formula $\sin {\tfrac {1}{2}}\varepsilon =\tan {\tfrac {1}{2}}c\,\tan {\tfrac {1}{2}}h_{c}$ where $ε$ is the area of the spherical triangle, $c$ is a side length, and $h c$ is the angular difference between two parallel small circles -- one is through the points $A$ , $B$ and the antipodal point $C *$ and the other is through $A *$ , $B *$ and $C$ . I'd like to see this area formula described in this Spherical trigonometry article too (or at the very least a summary and then a link to Lexell's theorem.). I could give it a try but ... anyone know enough about this formula to do it properly? Thank you — $Q$ uantling (talk | contribs) 18:59, 3 June 2024 (UTC)[reply]