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Spherical Trigonometry
 Introductory Theorems
 Right-Angled Triangles
 Oblique-Angled Triangles
  Napier's Formula
  Gauss's Formula
 Spherical Triangle and Circle
 Spherical Areas
 Area of Spherical Polygon
  Cagnoli's Theorem
  Llhuillier's Theorem
 The Five Regular solids
 Sources and References

Spherical Trigonometry

Introductory Theorems



Planes through the centre of a sphere intersect the surface in great circles; other planes intersect it in small circles. Unless otherwise stated, all arcs are measured on great circles.
The poles of a great circle are the extremities of the diameter perpendicular to its plane.
The sides 𝑎, 𝑏, 𝑐 of a spherical triangle are the arcs of great circles 𝐵𝐶, 𝐶𝐴, 𝐴𝐵 on a sphere of radius unity; and the angles 𝐴, 𝐵, 𝐶 are the angles between the tangents to the sides at the vertices, or the angles between the planes of the great circles. The centre of the sphere will be denoted by 𝑂.
The polar triangle of a spherical triangle 𝐴𝐵𝐶 has for its angular points 𝐴′, 𝐵′, 𝐶′, the poles of the sides 𝐵𝐶, 𝐶𝐴, 𝐴𝐵 of the primitive triangle in the directions of 𝐴, 𝐵, 𝐶 respectively (since each great circle has two poles). The sides of 𝐴′, 𝐵′, 𝐶′ are denoted by 𝑎′, 𝑏′, 𝑐′. 871 image The sides and angles of the polar triangle are respectively the supplements of the angles and sides of the primitive triangle; that is, 𝑎′+𝐴=𝑏′+𝐵=𝑐′+𝐶=𝜋 𝑎+𝐴′=𝑏+𝐵′=𝑐+𝐶′=𝜋 Let 𝐵𝐶 produced cut the sides 𝐴′𝐵′, 𝐶′𝐴′ in 𝐺, 𝐻. 𝐵 is the pole of 𝐴′𝐶′, therefore 𝐵𝐻=𝜋2. Similarly 𝐶𝐺=𝜋2, therefore, by addition, 𝑎+𝐺𝐻=𝜋 and 𝐺𝐻=𝐴′, because 𝐴′ is the pole of 𝐵𝐶.
The polar diagram of a spherical polygon is formed in the same way, and the same relations subsist between the sides and angles of the two figures.
Rule: Hence, any equation between the sides and angles of a sphrical triangle porduces a supplementary equation by changing 𝑎 into 𝜋−𝐴 and 𝐴 into 𝜋−𝑎, ⋯ 872 The centre of the inscribed circle, radius 𝑟, is also the centre of the circumscribed circle, radius 𝑅′, of the polar triangle, and 𝑟+𝑅′=12𝜋.
Proof: In the last figure, let 𝑂 be the centre of the inscribed circle of 𝐴𝐵𝐶; then 𝑂𝐷, the perpendicular on 𝐵𝐶, passes through 𝐴′, the pole of 𝐵𝐶. Also, 𝑂𝐷=𝑟; therefore 𝑂𝐴′=12𝜋−𝑟. similarly 𝑂𝐵′=𝑂𝐶′=12𝜋−𝑟; therefore 𝑂 is the centre of the circumscribed circle of 𝐴′𝐵′𝐶′, and 𝑟+𝑅′=12𝜋. 873 The sine of the arc joining a point on the circumference of a small circle with the pole of a parallel great circle, is equal to the ratio of the circumferences or corresponding arcs of the two circles.
For it is equal to the radius of the small circle divided by the radius of the sphere; that is, by the radius of the great circle. 874 Two sides of a triangle are greater than the third. (By XI. 20) 875 The sides of a triangle are together less than the circumference of a great circle. (By XI. 21) 876 The angles of a triangel are together greater than two right angles.
For 𝜋−𝐴+𝜋−𝐵+𝜋−𝐶 is <2𝜋, by (875) and the polar triangle. 877 If two sides of a triangle are equal, the opposite angles are equall. (By the geometrical proof in (894)) 878 If two angles of a triangle are equal, the opposite sides are equal. (By the polar triangle and (877)) 879 The greater angle of a triangle has the greater side opposite to it.
Proof: If 𝐵 be >𝐴, draw the arc 𝐵𝐷 meeting 𝐴𝐶 in 𝐷, and make ∠𝐴𝐵𝐷=𝐴, therefore 𝐵𝐷=𝐴𝐷; but 𝐵𝐷+𝐷𝐶>𝐵𝐶, therefore 𝐴𝐶>𝐵𝐶. 880 The greater side of a triangle has the greater angle opposite to it. (By the polar triangle and (879))

Right-Angled Triangles

881 image Napier's Rules: In the triangle 𝐴𝐵𝐶 let 𝐶 be a right angle, then 𝑎, (12𝜋−𝐵), (12𝜋−𝑐), (12𝜋−𝐴), and 𝑏, are called the five circular parts. Taking any part for middle part, Napier's rules are: I. sine of middle part = product of tangents of adjacent parts. II. sine of middle part = product of cosines of opposite parts. In applying the rules we can take 𝐴, 𝐵, 𝐶 instead of their complements, and change sine into cos or vice versa, for those parts at once. Thus taking 𝑏 for the middle part, sin𝑏=tan𝑎cot𝐴=sin𝐵sin𝑐 Ten equations in all are given by the rules.
Proof: From any point 𝑃 in 𝑂𝐴, draw 𝑃𝑅 perpendicular to 𝑂𝐶, and 𝑅𝑄 to 𝑂𝐵; therefore 𝑃𝑅𝑄 is a right angle; therefore 𝑂𝐵 is perpendicular to 𝑃𝑅 and 𝑄𝑅, and therefore to 𝑃𝑄. Then prove any formula by proportion from the triangles of the tetrahedron 𝑂𝑃𝑄𝑅, which are all right-angled. Otherwise, prove by the formula for oblique-angled triangles.

Oblique-Angled Triangles

882 image cos𝑎=cos𝑏cos𝑐+sin𝑏sin𝑐cos𝐴 Proof: Draw tangents at 𝐴 to the sides 𝑐, 𝑏 to meet 𝑂𝐵, 𝑂𝐶 in 𝐷 and 𝐸. Express 𝐷𝐸 by (702) applied to each of the triangles 𝐷𝐴𝐸 and 𝐷𝑂𝐸, and subtract.
If 𝐴𝐵 and 𝐴𝐶 are both >𝜋2, produce them to meet in 𝐴′, the pole of 𝐴, and employ the triangle 𝐴′𝐵𝐶.
If 𝐴𝐵 alone be >𝜋2, produce 𝐵𝐴 to meet 𝐵𝐶.
883 The supplementary formula, by (871), is cos𝐴=−cos𝐵cos𝐶+sin𝐵sin𝐶cos𝑎 884 sin𝐴2=sin(𝑠−𝑏)sin(𝑠−𝑐)sin𝑏sin𝑐 885 cos𝐴2=sin𝑠sin(𝑠−𝑎)sin𝑏sin𝑐 886 tan𝐴2=sin(𝑠−𝑏)sin(𝑠−𝑐)sin𝑠sin(𝑠−𝑎), where 𝑠=12(𝑎+𝑏+𝑐) Proof: sin2𝐴2=12(1−cos𝐴). Substitute for cos𝐴 from (872), and throw the numerator of the whole expression into factors by (673). Similarly for cos𝐴2. 887 The supplementary formula are obtained in a similar way, or by the rule in (871). They are cos𝑎2=cos(𝑆−𝐵)cos(𝑆−𝐶)sin𝐵sin𝐶 888 sin𝑎2=cos𝑆cos(𝑆−𝐴)sin𝐵sin𝐶 889 tan𝑎2=cos𝑆cos(𝑆−𝐴)cos(𝑆−𝐵)cos(𝑆−𝐶), where 𝑆=12(𝐴+𝐵+𝐶) 890 Let 𝜎=sin𝑠sin(𝑠−𝑎)sin(𝑠−𝑏)sin(𝑠−𝑐) =121+2cos𝑎cos𝑏cos𝑐−cos2𝑎−cos2𝑏−cos2𝑐 891 Then the supplementary form, by (871), is ∑=cos𝑆cos(𝑆−𝐴)cos(𝑆−𝐵)cos(𝑆−𝐶) =121−2cos𝐴cos𝐵cos𝐶−cos2𝐴−cos2𝐵−cos2𝐶 892 sin𝐴=2𝜎sin𝑏sin𝑐, sin𝑎=2∑sin𝐵sin𝐶 By sin𝐴=2sin𝐴2cos𝐴2 and (884, 885),⋯ 893 The following rules will produce the ten formula (884 to 892):
I. Write sin before each factor in the 𝑠 values of sin𝐴2, cos𝐴2, tan𝐴2, sin𝐴, and △, in Plane Trigonometry (704-707), to obtain the corresponding formula in Spherical Trigonometry.
II. To obtain the supplementary forms of the five results, transpose large and small letters everywhere, and transpose sin and cos everywhere but in the denominators, and write minus before cos𝑆. 894 sin𝐴sin𝑎=sin𝐵sin𝑏=sin𝐶sin𝑐 Proof: By (882). Otherwise, in the figure of 882, draw 𝑃𝑁 perpendicular to 𝐵𝑂𝐶, and 𝑁𝑅, 𝑁𝑆 to 𝑂𝐵, 𝑂𝐶. Prove 𝑃𝑅𝑂 and 𝑃𝑆𝑂 right angles by I. 47 and therefore 𝑃𝑁=𝑂𝑃sin𝑐sin𝐵=𝑂𝑃sin𝑏sin𝐶. 895 cos𝑏cos𝐶=cot𝑎sin𝑏−cot𝐴sin𝐶 To remember this formula, take any four consecutive angles and sides (as 𝑎. 𝐶, 𝑏, 𝐴), and, calling the first and fourth the extremes, and the second and third the middle parts, employ the following rule:
Rule: Product of cosines of middle parts = cot extreme side × sin middle side − cot extreme angle × sin middle angle.
Proof: In the formula for cos𝑎 (882) substitute a similar value for cos𝑐, and for sin𝑐 put sin𝐶sin𝑎sin𝐴.

Napier's Formula

896 tan12(𝐴−𝐵)=sin12(𝑎−𝑏)sin12(𝑎+𝑏)cot𝐶21 tan12(𝐴+𝐵)=cos12(𝑎−𝑏)cos12(𝑎+𝑏)cot𝐶22 tan12(𝑎−𝑏)=sin12(𝐴−𝐵)sin12(𝐴+𝐵)cot𝐶23 tan12(𝑎+𝑏)=cos12(𝐴−𝐵)cos12(𝐴+𝐵)cot𝐶24 Rule: In the value of tan12(𝐴−𝐵) change sin to cos to obtain tan12(𝐴+𝐵). To obtain [3] and [4] from [1] and [2], transpose sides and angles, and change cot to tan.
Proof: In the values of cos𝐴 and cos𝐵 by (883), put 𝑚sin𝑎 and 𝑚sin𝑏 for sin𝐴 and sin𝐵, and add the two equatios. Then put 𝑚=sin𝐴sin𝐵sin𝑎sin𝑏, and transform by (670-672)

Gauss's Formula

897 sin12(𝐴+𝐵)cos12𝐶=cos12(𝑎−𝑏)cos12𝑐1 sin12(𝐴−𝐵)cos12𝐶=sin12(𝑎−𝑏)sin12𝑐2 cos12(𝐴+𝐵)sin12𝐶=cos12(𝑎+𝑏)cos12𝑐3 cos12(𝐴−𝐵)sin12𝐶=sin12(𝑎+𝑏)sin12𝑐4 From any of these formula the others may be obtained by the following rule:
Rule: Change the sign of the letter 𝐵 (large or small) on one side of the equation, and write sin for cos and cos for sin on the other side.
Proof: Take sin12(𝐴+𝐵)=sin12𝐴cos12𝐵+cos12𝐴sin12𝐵 substitute the 𝑠 values by (884, 885), and reduce.

Spherical Triangle and Circle

image 898 Let 𝑟 be the radius of the inscribed circle of 𝐴𝐵𝐶; 𝑟𝑎 the radius of the escribed circle touching the side 𝑎, and 𝑅, 𝑅𝑎 the radii of the circumscribed circles; then tan𝑟=tan12𝐴sin(𝑠−𝑎)=𝜎sin𝑠1  =2sin𝑎sin𝐴sin12𝐴sin12𝐵sin12𝐶3  =2cos12𝐴cos12𝐵cos12𝐶4  =2∑cos𝑆+cos(𝑆−𝐴)+cos(𝑆−𝐵)+⋯ Proof: The first value is found from the right-angled triangle 𝑂𝐴𝐹, in which 𝐴𝐹=𝑠−𝑎. The other values by (884-892). 899 tan𝑟𝑎=tan12𝐴sin𝑠=𝜎sin(𝑠−𝑎)1  =2sin𝑎sin𝐴sin12𝐴cos12𝐵cos12𝐶3  =2cos12𝐴sin12𝐵sin12𝐶4  =2∑cos𝑆−cos(𝑆−𝐴)+cos(𝑆−𝐵)+cos(𝑆−𝐶) Proof: From the right-angled triangle 𝑂′𝐴𝐹, in which 𝐴𝐹′=𝑠. Note: The first two values of tan𝑟𝑎 may be obtained from those of tan𝑟 by interchanging 𝑠 and 𝑠−𝑎. 900 image tan𝑅=tan12𝑎cos(𝑆−𝐴)=cos𝑆1  =sin12𝑎sin𝐴cos12𝑏cos12𝑐3  =2sin12𝑎sin12𝑏sin12𝑐𝜎4  =sin𝑠−sin(𝑠−𝑎)+sin(𝑠−𝑏)+⋯2𝜎  Proof: The first value from the right-angled triangle 𝑂𝐵𝐷, in which ∠𝑂𝐵𝐷=𝑆−𝐴. The other values by the formula (887-802) 901 tan𝑅𝑎=tan12𝑎cos𝑆=cos(𝑆−𝐴)1  =sin12𝑎sin𝐴sin12𝑏sin12𝑐3  =2sin12𝑎cos12𝑏cos12𝑐𝜎4  =sin𝑠−sin(𝑠−𝑎)+sin(𝑠−𝑏)+sin(𝑠−𝑐)2𝜎  Proof: From the right-angled triangle 𝑂′𝐵𝐷, in which ∠𝑂𝐵𝐷=𝜋−𝑆.

Spherical Areas

902 area of 𝐴𝐵𝐶=(𝐴+𝐵+𝐶−𝜋)𝑟2=𝐸𝑟2, where 𝐸=𝐴+𝐵+𝐶−𝜋, the spherical excess Proof: By adding the three lunes 𝐴𝐵𝐷𝐶, 𝐵𝐶𝐸𝐴, 𝐶𝐴𝐹𝐵, and observing that 𝐴𝐵𝐹=𝐶𝐷𝐸, we get 𝐴𝜋+𝐵𝜋+𝐶𝜋2𝜋𝑟2=2𝜋𝑟2+2𝐴𝐵𝐶

Area of Spherical Polygon

903 𝑛 being the number of sides, Area={Interior Angles−(𝑛−2)𝜋}𝑟2  ={2𝜋−Exterior Angles}𝑟2  ={2𝜋−sides of Polar Diagram}𝑟2 The last value holds for curvilinear area in the limit.
Proof: By joining the vertices with an interior point, and adding the areas of the spherical triangles so formed.

Cagnoli's Theorem

904 sin12𝐸={sin𝑠sin(𝑠−𝑎)sin(𝑠−𝑏)sin(𝑠−𝑐)}2cos12𝑎cos12𝑏cos12𝑐 Proof: Expand sin[12(𝐴+𝐵)−12(𝜋−𝐶)] by (628), and transform by Gauss's equations (897i, iii) and (669, 890).

Llhuillier's Theorem

905 tan14𝐸={tan12𝑠tan12(𝑠−𝑎)tan12(𝑠−𝑏)tan12(𝑠−𝑐)} Proof: Multiply numerator and denominator of the left side by 2cos14(𝐴+𝐵−𝐶+𝜋) and reduce by (667, 668), then eliminate 12(𝐴+𝐵) by Gauss's formula (897 i, iii). Transform by (672, 673), and substitute from (886).


Let the number of faces, solid angles, and edges, of any polyhedron be 𝐹, 𝑆, 𝐸; then 906 𝐹+𝑆=𝐸+2 Proof: Project the polyhedron upon an internal sphere. Let 𝑚=number of sides, and 𝑠=sum of angles of one of the spherical polygons so formed. Then its area={𝑠−(𝑚−2)𝜋}𝑟2, by (903). Sum this for all the polygons, and equate to 4𝜋𝑟2.

The Five Regular solids

Let 𝑚 be the number of sides in each face, 𝑛 the number of plane angles in each solid angle; therefore 907 𝑚𝐹=𝑛𝑆=2𝐸 From these equations and (906), find 𝐹, 𝑆, and 𝐸 in terms of 𝑚 and 𝑛, thus, 1𝐹=𝑚21𝑚+1𝑛12, 1𝑆=𝑛21𝑚+1𝑛12, 1𝐸=1𝑚+1𝑛12 In order that 𝐹, 𝑆, and 𝐸 may be positive, we must 1𝑚+1𝑛>12, a relation which admits of five solutions in whole numbers, corresponding to the five regular solids. The values of 𝑚, 𝑛, 𝐹, 𝑆, and 𝐸 for the five regular solids are exhibited in the following table: Regular Solids𝑚𝑛𝐹𝑆𝐸 Tetrahedron33446 Hexahedron436812 Octahedron348612 Dodecahedron35201230 Icosahedron35201230 908 The sum of all the plane angles of any polyhedron =2𝜋(𝑆−2) Or, Four right angles for every vertex less eight right angles. 909 image If 𝐼 be the angle between two adjacent faces of a regular polyhedron. sin12𝐼=cos𝜋𝑛÷sin𝜋𝑚 Proof: Let 𝑃𝑄=𝑎 be the edge, and 𝑆 the centre of a face, 𝑇 the middle point of 𝑃𝑄, 𝑂 the centre of the inscribed and circumscribed spheres, 𝐴𝐵𝐶 the projection of 𝑃𝑆𝑇 upon a concentric sphere. In this spherical triangle. 𝐶=𝜋2, 𝐴=𝜋𝑛, 𝐴=𝜋𝑚=𝑃𝑆𝑇, Also 𝑆𝑇𝑂=12𝐼 Now, by (881, ii), cos𝐴=sin𝐵cos𝐵𝐶 that is, cos𝜋𝑛=sin𝜋𝑚sin12𝐼Q.e.d. 910 If 𝑟, 𝑅 be the radii of the inscribed and circumscribed spheres of a regular polyhedron, 𝑟=𝑎2tan12𝐼cot𝜋𝑚, 𝑅=𝑎2tan12𝐼tan𝜋𝑛 Proof: In the above figure, 𝑂𝑆=𝑟, 𝑂𝑃=𝑅, 𝑃𝑇=𝑎2; and 𝑂𝑆=𝑃𝑇cot𝜋𝑚tan12𝐼. Also 𝑂𝑃=𝑃𝑇cosec𝐴𝐶, and by (881, i), sin𝐴𝐶=tan𝐵𝐶cot𝐴=cot12𝐼cot𝜋𝑛; therefore, ⋯

Sources and References


ID: 210900014 Last Updated: 9/14/2021 Revision: 0 Ref:



  1. B. Joseph, 1978, University Mathematics: A Textbook for Students of Science &amp; Engineering
  2. Ayres, F. JR, Moyer, R.E., 1999, Schaum's Outlines: Trigonometry
  3. Hopkings, W., 1833, Elements of Trigonometry

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