Spherical trigonometry

thumb|The [[octant of a sphere is a spherical triangle with three right angles.]]

Spherical trigonometry is the branch of spherical geometry that deals with the metrical relationships between the sides and angles of spherical triangles, traditionally expressed using trigonometric functions. On the sphere, geodesics are great circles. Spherical trigonometry is of great importance for calculations in astronomy, geodesy, and navigation.

The origins of spherical trigonometry in Greek mathematics and the major developments in Islamic mathematics are discussed fully in History of trigonometry and Mathematics in medieval Islam. The subject came to fruition in Early Modern times with important developments by John Napier, Delambre and others. Since then, significant developments have been the application of vector methods, quaternion methods, and the use of numerical methods.

Preliminaries

right|thumb|200px|Eight spherical triangles defined by the intersection of three great circles.

Spherical polygons

A spherical polygon is a polygon on the surface of the sphere. Its sides are arcs of great circles—the spherical geometry equivalent of line segments in plane geometry.

Such polygons may have any number of sides greater than 1. Two-sided spherical polygons—lunes, also called digons or bi-angles—are bounded by two great-circle arcs: a familiar example is the curved outward-facing surface of a segment of an orange. Three arcs serve to define a spherical triangle, the principal subject of this article. Polygons with higher numbers of sides (4-sided spherical quadrilaterals, 5-sided spherical pentagons, etc.) are defined in similar manner. Analogously to their plane counterparts, spherical polygons with more than 3 sides can always be treated as the composition of spherical triangles.

One spherical polygon with interesting properties is the pentagramma mirificum, a 5-sided spherical star polygon with a right angle at every vertex.

From this point in the article, discussion will be restricted to spherical triangles, referred to simply as triangles.

Notation

thumb|right|200px|The basic triangle on a unit sphere.

Both vertices and angles at the vertices of a triangle are denoted by the same upper case letters , , and .
Side lengths on a unit-radius sphere are denoted by lower-case letters: , , and . The side lengths and lower case angles are equivalent when the latter are measured in radians (see arc length). By convention, the sides of proper spherical triangles are less than radians, and <math display=block>0 < a + b + c < 2\pi

</math>(Todhunter, Art.22,32).

:In particular, the sum of the angles of a spherical triangle is strictly greater than the sum of the angles of a triangle defined on the Euclidean plane, which is always exactly radians.

The sphere's radius is taken as unity. For specific practical problems on a sphere of radius the measured lengths of the sides must be divided by before using the identities given below. Likewise, after a calculation on the unit sphere the sides , , and must be multiplied by .

Polar triangles

right|thumb|200px|The polar triangle

The polar triangle associated with a triangle is defined as follows. Consider the great circle that contains the side . This great circle is defined by the intersection of a diametral plane with the surface. Draw the normal to that plane at the centre: it intersects the surface at two points and the point that is on the same side of the plane as is (conventionally) termed the pole of and it is denoted by . The points and are defined similarly.

The triangle is the polar triangle corresponding to triangle . The angles and sides of the polar triangle are

given by (Todhunter, and spherical astronomy give different proofs and the online resources of MathWorld provide yet more. There are even more exotic derivations, such as that of Banerjee who derives the formulae using the linear algebra of projection matrices and also quotes methods in differential geometry and the group theory of rotations.

The derivation of the cosine rule presented above has the merits of simplicity and directness and the derivation of the sine rule emphasises the fact that no separate proof is required other than the cosine rule. However, the above geometry may be used to give an independent proof of the sine rule. The scalar triple product, evaluates to in the basis shown. Similarly, in a basis oriented with the -axis along , the triple product , evaluates to . Therefore, the invariance of the triple product under cyclic permutations gives which is the first of the sine rules. See curved variations of the law of sines to see details of this derivation.

Differential variations

When any three of the differentials da, db, dc, dA, dB, dC are known, the following equations, which are found by differentiating the cosine rule and using the sine rule, can be used to calculate the other three by elimination:

<math display=block>\begin{align}

da = \cos C \ db + \cos B \ dc + \sin b \ \sin C \ dA, \\

db = \cos A \ dc + \cos C \ da + \sin c \ \sin A \ dB, \\

dc = \cos B \ da + \cos A \ db + \sin a \ \sin B \ dC. \\

\end{align}</math>

Identities

Supplemental cosine rules

Applying the cosine rules to the polar triangle gives (Todhunter, (Art 45) derives the half angle formulas for the angles and sides in terms of the sides and angles, respectively. His book is available as an ebook in the public domain from Project Gutenberg. The first equation may be proved by using the law of cosines for side a in terms of sides b and c and angle A, by using the identity <math>2\sin^2\!\tfrac{A}{2} = 1 - \cos A,</math> and by expressing the product of two sines as half the difference of the cosine of their angle difference angle minus the cosine of their angle sum (See sum-to-product identities). In detail:

<math display="block">\begin{align} 2 \, \sin(s-b)\,\sin(s-c)&= \cos((s-b)-(s-c))-\cos((s-b)+(s-c)) \\ &=\cos(c-b)-\cos(a)\\ &=\cos(b)\cos(c)+\sin(b)\sin(c)-(\cos(b)\cos(c)+\sin(b)\sin(c)\cos(A)) \\ &=\sin(b)\sin(c)(1-\cos(A))\\ &=\sin(b)\sin(c)\,\bigg(2\,\sin^2\bigg(\frac{A}{2}\bigg)\bigg)\end{align}</math>

The second formula uses the identity <math>2\cos^2\!\tfrac{A}{2} = 1 + \cos A,</math> the third is a quotient and the remainder follow by applying the results to the polar triangle.

Delambre analogies

The Delambre analogies (also called Gauss analogies) were published independently by Delambre, Gauss, and Mollweide in 1807–1809.

\begin{align}

\frac{\sin{\tfrac{1}{2(A+B)}

{\cos{\tfrac{1}{2C}

=\frac{\cos{\tfrac{1}{2(a-b)}

{\cos{\tfrac{1}{2c}

&\qquad\qquad

\frac{\sin{\tfrac{1}{2(A-B)}

{\cos{\tfrac{1}{2C}

=\frac{\sin{\tfrac{1}{2(a-b)}

{\sin{\tfrac{1}{2c}

\\[2ex]

\frac{\cos{\tfrac{1}{2(A+B)}

{\sin{\tfrac{1}{2C}

=\frac{\cos{\tfrac{1}{2(a+b)}

{\cos{\tfrac{1}{2c}

&\qquad

\frac{\cos{\tfrac{1}{2(A-B)}

{\sin{\tfrac{1}{2C}

=\frac{\sin{\tfrac{1}{2(a+b)}

{\sin{\tfrac{1}{2c}

\end{align}

</math>

Another eight identities follow by cyclic permutation.

Proved by expanding the numerators and using the half angle formulae. (Todhunter,)

Napier's analogies

<math display="block">\begin{align}

\tan\tfrac{1}{2}(A+B) = \frac{\cos\tfrac{1}{2}(a-b)}{\cos\tfrac{1}{2}(a+b)} \cot\tfrac{1}{2}C

&\qquad&

\tan\tfrac{1}{2}(a+b) = \frac{\cos\tfrac{1}{2}(A-B)}{\cos\tfrac{1}{2}(A+B)}\tan\tfrac{1}{2}c

\\[2ex]

\tan\tfrac{1}{2}(A-B) = \frac{\sin\tfrac{1}{2}(a-b)}{\sin\tfrac{1}{2}(a+b)} \cot\tfrac{1}{2}C

&\qquad&

\tan\tfrac{1}{2}(a-b) =\frac{\sin\tfrac{1}{2}(A-B)}{\sin\tfrac{1}{2}(A+B)} \tan\tfrac{1}{2}c

\end{align}</math>

Another eight identities follow by cyclic permutation.

These identities follow by division of the Delambre formulae. (Todhunter, provided an elegant mnemonic aid for the ten independent equations: the mnemonic is called Napier's circle or Napier's pentagon (when the circle in the above figure, right, is replaced by a pentagon).

First, write the six parts of the triangle (three vertex angles, three arc angles for the sides) in the order they occur around any circuit of the triangle: for the triangle shown above left, going clockwise starting with gives . Next replace the parts that are not adjacent to (that is , , and ) by their complements and then delete the angle from the list. The remaining parts can then be drawn as five ordered, equal slices of a pentagram, or circle, as shown in the above figure (right). For any choice of three contiguous parts, one (the middle part) will be adjacent to two parts and opposite the other two parts. The ten Napier's Rules are given by

sine of the middle part = the product of the tangents of the adjacent parts
sine of the middle part = the product of the cosines of the opposite parts

The key for remembering which trigonometric function goes with which part is to look at the first vowel of the kind of part: middle parts take the sine, adjacent parts take the tangent, and opposite parts take the cosine.

For an example, starting with the sector containing we have:

<math display=block>\begin{align}

\sin a &= \tan(\tfrac{\pi}{2} - B)\,\tan b \\[2pt]

&= \cos(\tfrac{\pi}{2} - c)\, \cos(\tfrac{\pi}{2} - A) \\[2pt]

&= \cot B\,\tan b \\[4pt]

&= \sin c\,\sin A.

\end{align}</math>

The full set of rules for the right spherical triangle is (Todhunter,

Solution of triangles

Oblique triangles

The solution of triangles is the principal purpose of spherical trigonometry: given three, four or five elements of the triangle, determine the others. The case of five given elements is trivial, requiring only a single application of the sine rule. For four given elements there is one non-trivial case, which is discussed below. For three given elements there are six cases: three sides, two sides and an included angle, two sides and an opposite angle, two angles and an included side, two angles and an opposite side, or three angles. (The last case has no analogue in planar trigonometry.) The figure below shows the seven non-trivial cases: in each case the given sides are marked with a cross-bar and the given angles with an arc. (The given elements are also listed below the triangle). In the summary notation here such as ASA, A refers to a given angle and S refers to a given side, and the sequence of A's and S's in the notation refers to the corresponding sequence in the triangle.

thumb|center|500px

Case 1: three sides given (SSS). The cosine rule may be used to give the angles , , and but, to avoid ambiguities, the half angle formulae are preferred.
Case 2: two sides and an included angle given (SAS). The cosine rule gives and then we are back to Case 1.
Case 3: two sides and an opposite angle given (SSA). The sine rule gives and then we have Case 7. There are either one or two solutions.
Case 4: two angles and an included side given (ASA). The four-part cotangent formulae for sets () and () give and , then follows from the sine rule.
Case 5: two angles and an opposite side given (AAS). The sine rule gives and then we have Case 7 (rotated). There are either one or two solutions.
Case 6: three angles given (AAA). The supplemental cosine rule may be used to give the sides , , and but, to avoid ambiguities, the half-side formulae are preferred.
Case 7: two angles and two opposite sides given (SSAA). Use Napier's analogies for and ; or, use Case 3 (SSA) or case 5 (AAS).

The solution methods listed here are not the only possible choices: many others are possible. In general it is better to choose methods that avoid taking an inverse sine because of the possible ambiguity between an angle and its supplement. The use of half-angle formulae is often advisable because half-angles will be less than /2 and therefore free from ambiguity. There is a full discussion in Todhunter. The article Solution of triangles#Solving spherical triangles presents variants on these methods with a slightly different notation.

There is a full discussion of the solution of oblique triangles in Todhunter. Nasir al-Din al-Tusi was the first to list the six distinct cases (2–7 in the diagram) of a right triangle in spherical trigonometry.

thumb|100px

Solution by right-angled triangles

Another approach is to split the triangle into two right-angled triangles. For example, take the Case 3 example where , , and are given. Construct the great circle from that is normal to the side at the point . Use Napier's rules to solve the triangle : use and to find the sides and and the angle . Then use Napier's rules to solve the triangle : that is use and to find the side and the angles and . The angle and side follow by addition.

Numerical considerations

Not all of the rules obtained are numerically robust in extreme examples, for example when an angle approaches zero or . Problems and solutions may have to be examined carefully, particularly when writing code to solve an arbitrary triangle.

Area and spherical excess

thumb|[[Lexell's theorem: the triangles of constant area on a fixed base have their free vertex along a small circle through the points antipodal to and .]]

Consider an -sided spherical polygon and let denote the -th interior angle. The area of such a polygon is given by (Todhunter, An earlier proof was derived, but not published, by the English mathematician Thomas Harriot in 1603. On a sphere of radius both of the above area expressions are multiplied by . The definition of the excess is independent of the radius of the sphere.

The converse result may be written as

<math display=block> A+B+C = \pi + \frac{4\pi \times \text{Area of triangle{\text{Area of the sphere.</math>

Since the area of a triangle cannot be negative the spherical excess is always positive. It is not necessarily small, because the sum of the angles may attain 5 (3 for proper angles). For example,

an octant of a sphere is a spherical triangle with three right angles, so that the excess is /2. In practical applications it is often small: for example the triangles of geodetic survey typically have a spherical excess much less than 1' of arc. On the Earth the excess of an equilateral triangle with sides 21.3 km (and area 393 km<sup>2</sup>) is approximately 1 arc second.

There are many formulae for the excess. For example, Todhunter, by a line integral with Green's theorem, or via an equal-area projection as commonly done in GIS. The other algorithms can still be used with the side lengths calculated using a great-circle distance formula.

References

External links

a more thorough list of identities, with some derivation
a more thorough list of identities, with some derivation
TriSph A free software to solve the spherical triangles, configurable to different practical applications and configured for gnomonic
"Revisiting Spherical Trigonometry with Orthogonal Projectors" by Sudipto Banerjee. The paper derives the spherical law of cosines and law of sines using elementary linear algebra and projection matrices.
by Okay Arik
"The Book of Instruction on Deviant Planes and Simple Planes", a manuscript in Arabic that dates back to 1740 and talks about spherical trigonometry, with diagrams
Some Algorithms for Polygons on a Sphere Robert G. Chamberlain, William H. Duquette, Jet Propulsion Laboratory. The paper develops and explains many useful formulae, perhaps with a focus on navigation and cartography.
Online computation of spherical triangles

Preliminaries

Spherical polygons

Notation

Polar triangles

Differential variations

Identities

Supplemental cosine rules

Delambre analogies

Napier's analogies

Solution of triangles

Oblique triangles

Solution by right-angled triangles

Numerical considerations

Area and spherical excess

See also

References

External links