Spherical Trigonometry Addendum

First, to make the printed version of this document self-contained, we reproduce certain theorems, etc.

Description of a spherical triangle ---
The angle, by convention usually constrained to be in the semi-open interval [0, pi), between the normals to each of two intersecting planes is called the dihedral-angle.
The plane of a side of a triangle is that plane defined by the two adjacent vertices (angles) and the center of the sphere.
It is the dihedral-angle between two adjacent sides of a triangle that constitutes the included angle; i.e., the angle between these adjacent sides.
The angle. at the center of the sphere, subtended by a great-circle arc between two points on a sphere is taken as the measure of that side.  It properly is written as ||a||; but usually as just the a.  Depending upon the convention, at most one side of a triangle may exceed 2 pi.  Though, usually, each of the sides is constrained to be in the semi-open interval [0, pi).
The three points of the intersection of the inward-pointing normal to each of the planes of the sides of a triangle constitute the vertices of the polar-triangle.

Spherical Law of Cosines spherical law of cosines for the cos(a)
    cos(a) = sin(b) sin(c) cos(A) + cos(b) cos(c)
Within the proof of the first version of the Law of Half-tangent spherical law of half-tangent, we have
    (1 - cos(A)) (sin(b) sin(c)) = 2 sin(s - c) sin(s - b) and
    (1 + cos(A)) (sin(b) sin(c)) = 2 sin(s) sin(s - a)
Second #4 formula spherical Mollweide of Mollweide = Karl Brandan Mollweide (3 II 1774 - 10 III 1825)
    sin(c) = tan(b) cot(B)
where the angle A = pi/2 is the right angle
Sums or differences sums or differences
    sin(u) + sin(v) = 2 sin((u + v) / 2) cos((u - v) / 2)
    cos(v) - cos(u) = 2 sin((u + v) / 2) sin((u - v) / 2)
Definition: S = (A + B + C) / 2 and s = (a + b + c) / 2.four definitions.
Definition: The spherical excess, E = (A + B + C) - pi. In the context of the duality, we employ T, rather than E, for the spherical excess.
Hence: E = 2 S - pi.

Second, some preliminaries:
Consider the set of three pairs {(a,A), (b,B), (c,C)}. From Group Theory group theory, we see that the six permutations of these three pairs, taken three at a time, constitute the symmetric group of order six. The subset of even permutations constitutes a subgroup of order three. These permutations are cyclic. The set of the three odd permutations do not constitute a group; because they lack the identity element.  However, a product of any two odd permutations is an even permutation.  For completeness, we mention that the product of any odd by any even permutation is an odd permutation. If these pairs are the elements of a spherical triangle, an odd permutation inverts the sphere, that is, it turns the sphere inside-out.

In the following, we consider the spherical triangle ABC, with the angles (A B C) and sides opposite (a b c). Also RR is the circum-radius and rr is the in-radius.  [Regrettably, modern textbooks have entrenched the usage of R and r as definitions -- closely related to; but distinct from -- these radii.  (Since the status of limitations has expired long ago, it is too late to reclaim the old usage of the single-letters for the radii.)  The relation will be provided, once these radii are derived.]
We postpone the discussion of the constraints of the sizes of the angles and sides to the end of this page; because that discussion refers back to most of the following theorems, etc.  Here, however, we do not impose any such constrains, except as stated explicitly otherwise.

We have alluded to it several times.  Allan Cortzen [private communications via e-mail during April 2009] has suggested that we elevate it to a formal theorem.  Gramercy for your tour de force.  Individual contributions will be acknowledged in the following exposition.

Theorem:   Duality.  Any of the six permutations carries a valid equation or theorem into another valid one.
Proof:  The duality of the polar triangle maps a side into the supplementary angle a --> pi - A and an angle into the supplementary side A --> pi - a.
Proof:  Get yourself a sphere and draw a typical triangle, with the sides extended to become great circles.  Then, observe.  Alternatively, see this discussion of duality.
Corollary:  We have s --> 3 pi/2 - S = pi - T/2 and S --> 3 pi/2 - s as well as (s - a) --> pi/2 - (S - A) and (S - A) --> pi/2 - (s - a).
Proof is obvious from the definitions of s and S four definitions. Since we do not impose any constraints upon the domain of the angles or sides, we have had to introduce +- for each square-root.

Theorem:  Duality of the in-circle vs. circum-circle.  This duality maps an in-radius rr into the complementary circum-radius rr --> pi/2 - RR and the circum-radius RR into the complementary in-radius RR --> pi/2 - rr.
Proof:  We have a triangle ABC and the in-circle with center o and radius rr = oP.  Pray, see the illustration, provided in several formats.  GIF JPEG PNG TIFF  The content is identical, view whichever one displays better on your system.
Now A* is pole for the side a implies pi/2 = PA*. (indeed for any point P on a).  By subtraction oA* = pi/2 - rr.  Likewise are oB* = pi/2 - rr and oC* = pi/2 - rr.
Hence the circle with center o and radius pi/2 - rr passes through A*, B*, C*.  Therefore this circle is the circum-circle for the polar triangle and RR* = pi/2 - rr.
Conversely as the polar triangle to A* B* C* is the triangle ABC we have RR = pi/2 - rr* or rr* = pi/2 - RR.
This proof was provided by Allan Curtzen, in an e-mail on 12-VII-2009.  Gramercy.

Theorem:   Half-tangent Law, second version.  Consider a triangle ABC, with an inscribed circle of radius rr whose center is at P. Then we have
    tan(A/2) = tan(rr) / sin(s - a),
where. as is usual. s = (a + b + c) / 2 and rr is the in-radius.  The first version of this law is here.
Proof:  Let the tangent point of the circle and the side AB be designated Pc, etc.  The angle P-Pc-A is a right-angle.  The triangles P-Pc-A and P-Pb-A are mirror-image congruent by ASS.  Hence ||A-Pc|| = ||A-Pb||.  Call this length alpha, etc.  Hence alpha + beta + gamma = s.  Thus alpha = s - (beta + gamma).  Since a = beta + gamma, we have alpha = s - a, etc.  Consider the triangle A-Pb-P.  From Mollweide #4 second spherical follows that sin(s - a) = tan(rr) cot(A/2).  QED.
This proof has been expanded from that supplied by Allan Cortzen.  Gramercy.

Each of the following two remarks is obvious; but, useful to be stated explicitly.

• The point Pc is at a distance alpha from the vertex A and beta from B.  The sum of these two distances is (alpha + beta) = ((s - a) + (s - b)) = c; gratifyingly so.

• The inscribed circle passes through each of the three points {Pa, Pb, Pc}.  These three points may be considered to be the vertices of a triangle, which we henceforth will call the in-triangle.

Corollary:  The circum-circle of the in-triangle is the inscribed circle of the original triangle (ABC).
Proof follows from the fact that any three non-collinear points define a unique circle.

Two more remarks:

• The construction algorithm for a circum-circle suffices for our repertoire.  We do not need the more-difficult algorithm for the construction of the inscribed-circle.

• All of these discussions, theorems, formulae, etc. (beginning at the second version theorem) carry over -- except for the requirement of taking the usual limits -- to the plane triangle.

Lemma:  We have
    cot(B/2) cot(C/2) = +-sin(s) / sin(s - a).
Proof:  Within the proof of the first version of the Law of Half-tangent spherical half-tangent, we have
    (1 - cos(A)) (sin(b) sin(c)) = 2 sin(s - c) sin(s - b) and
    (1 + cos(A)) (sin(b) sin(c)) = 2 sin(s) sin(s - a).
Permute the first of these cyclically once, that is, a --> b and the second twice, that is, a --> c, to yield
    (1 - cos(B)) (sin(c) sin(a)) = 2 sin(s - a) sin(s - c)
    (1 + cos(C)) (sin(a) sin(b)) = 2 sin(s) sin(s - c).
Their quotient (the second by the first) is
    ((1 + cos(C)) sin(b)) / ((1 - cos(B)) sin(c)) = sin(s) / sin(s - a).
Swap the pair (b,B) versus (c,C) (since this is an odd permutation, it inverts the sphere), to obtain
    ((1 + cos(B)) sin(c)) / ((1 - cos(C)) sin(b)) = sin(s) / sin(s - a).
The product of these two equations is
    ((1 + cos(C))./.(1 - cos(C))) ((1 + cos(B)) / (1 - cos(C))) = (sin(s) / sin(s - a)^2.
Employ the half-angle formula for the cotangent half angle, to obtain
    (cot(C/2) cot(B/2))^2 = (sin(s)/sin(s - a)^2.
Its square-root is cot(C/2) cot(B/2) = +- sin(s) / sin(s - a).  QED.
This proof has been expanded from that supplied by Allan Cortzen.  Gramercy.
Theorem:  The in-radius rr is given by
    (tan(rr))^2 = | ((sin(s - a) sin(s - b) sin(s-c)) / sin(s) |.
Proof:  Take sin(s - a) from the Half-tangent Law (the previous theorem) and substitute it into the current lemma, to obtain
    cot(C/2) cot(B/2) = +- sin(s) tan(A/2) / tan(rr).
Solve for tan(rr), to obtain,
    tan(rr) = +- sin(s) / (cot(A/2) cot(B/2) cot(C/2)),
where we have employed the definition of the cotangent circular definitions. Pray, observe that the right-hand side is a symmetric function of the elements of the triangle ABC as it should, since the in-radius (in the left-hand side) is likewise.  Again, from the Half-tangent Law, we have
    tan(rr) = sin(s - a) / cot(A/2),
where we again have employed the definition of the cotangent.  By cyclical permutations,  a --> b and a --> c, respectively, we have
    tan(rr) = sin(s - b) / cot(B/2) tan(rr) = sin(s - c) / cot(C/2).
Multiply these three equations, to obtain
    (tan(rr))^3 = (sin(s - a) sin(s - b) sin(s - c)) / (cot(A/2) cot(B/2) cot(C/2)).
Again, pray, observe that the right-hand side is a symmetric function of the elements of the triangle ABC as it should be, since the in-radius (in the left-hand side) is likewise.  Now divide by the second equation in this proof, to obtain the desired conclusion. QED. We have replaced the +- by the more restrictive absolute-value; because the left side of the equation obviously is positive.
This proof has been expanded from that supplied by Allan Cortzen.  Gramercy.
Corollary:  The circum-radius RR is given by
    (tan(RR))^2 = | - cos(S) / ((cos(S - A) cos(S - B) cos(S - C)) |.
Proof follows from the Duality theorem.   duality. And that of the circles.
Note:  By comparison with the definitions of r and R four definitions four definitions, we now see that r = tan(rr) and R = tan(RR).  This relation is what I had surmised in 1997, when I posted the original version of this Spherical Trigonometry textbook.  However, it is only now (2009) -- twelve years later -- that I have been exonerated by the proofs supplied by Alan Cortzen.  Gramercy.
Special case of equilateral/equiangular triangle:
    (tan(rr))^2 = (sin(a/2))^2 | sin(a/2) / sin(3 a/2) | and
    (tan(RR))^2 = | - cos(3 A/2) / cos(A/2) | / (cos(A/2))^2.

Lemma:  The dual of tan(rr) is given by tan(rr) --> cot(RR).
Proof:  We already have shown that the dual of rr is given by rr --> pi/2 - RR.  Take the tangent of each side, to obtain tan(rr) --> tan(pi/2 - RR).  Then use the properties of the trigonometric functions, to obtain, tan(pi/2 - RR) = cot(RR) = 1 / tan(RR).
Theorem:  The dual of r is given by r --> 1 / R.
Proof follows from the relation between r and rr and that between R and RR.
Corollary:  The dual of the Law of Half-tangent (first version), namely, tan(A / 2) = +- r / sin(s - a) is cot(a / 2) = +- (1 / R) / (- cos(S - A), which may be written as tan(a / 2) = -+ R cos(S - A) .
Proof follows from the duality theorem and the properties of the trigonometric functions.

For the following two lemmata, the theorem, and its corollary, we proscribe against any reflex (that is, greater than pi) angles.  Because otherwise there would be ambiguities as to the signs.

We will employ the strategy of computing the cosine from the relation
    cos(x) = cot(x) sin(x)
Then,
    (tan(x/2))^2 = (1 - cos(x)) / (1 + cos(x)).
Lemma:  We have the cot(T/2) as
    cot(T/2) = cot(A) + cot(b/2) cot(c/2) / sin(A).
Proof: Employ the addition theorem for the sine real addition in the equation of the lemma of the in-radius, to obtain
    cot(B/2) cot(C/2) = sin(s) / (sin(s) cos(a) - cos(s) sin(a)).
Employ the definition of the cotangent circular definitions and clear of fractions, to obtain
    cos(a) - cot(s) sin(a) = tan(B/2) tan(C/2).
Employ the definition of the cotangent on the cot(a), then solve for cot(s), to obtain
    - cot(s) = - cot(a) + tan(B/2) tan(C/2) / sin(a).
Its dual is the equation of the lemma.  QED.
This proof has been expanded from that supplied by Allan Cortzen.  Gramercy.
Lemma:  We have the sin(T/2) as
    sin(T/2) = sin(b/2) sin(c/2) sin(A) / cos(a/2).
Proof: Multiply the second equation within the proof of the lemma for the cot(B/2) cot T/2 by the fourth equation therein and divide by third equation therein, to obtain
    ((1 + cos(A)) (sin(b) sin(c))) ((1 + cos(C)) (sin(a) sin(b))) / ((1 - cos(B)) (sin(c) sin(a))) = (2 sin(s) sin(s - a)) (2 sin(s) sin(s - c)) / (2 sin(s - a) sin(s - c)).
Cancel (sin(a) sin(c)) on the left-hand side and (2 sin(s - a) sin(s - c)) on the right-hand side, to obtain
    (1 + cos(A) (1 + cos(C) (sin(b))^2 / (1 - cos(B) = 2 (sin(s))^2.
Employ the half-angle formula of the cosine upon cos(A/2) and cos(C/2) and that of the sine upon sin(B/2).  Then cancel the factor of 2 from each side of the equation, to obtain
    (cos(A/2) cos(C/2) sin(b) / sin(B/2))^2 = (sin(s))^2.
Take the square-root of each side, to obtain
    cos(A/2) cos(C/2) sin(b) / sin(B/2) = +- sin(s))^2.
Its dual is
    sin(a/2) sin(c/2) sin(B) / cos(b/2) = +- sin(T/2).
Swap the pair (a,A) versus (b,B) (since this is an odd permutation, it inverts the sphere).  Then transpose across the equals-sign, to obtain the equation in the statement of this lemma.  QED.
This proof has been expanded from that supplied by Allan Cortzen.  Gramercy.  Also, a pair of back-and-forth double-angle formula upon the sin(B) has been eliminated.
Theorem:  We have the cos(T/2) as
    cos(T/2) = (1 + cos(a) + cos(b) + cos(c)) / (4 cos(a/2) cos(b/2) cos(c/2)).
Proof:  Using the definition of the cotangent circular definitions, multiply the product of the two lemmata by the cos(a/2), to obtain
    cos(T/2) cos(a/2) = (cot(A) + cot(b/2) cot(c/2) / sin(A)) (sin(b/2) sin(c/2) sin(A)).
Expand the right-hand side, and again use the definition of the cotangent, to obtain
    = cos(A) sin(b/2) sin(c/2) + cos(b/2) cos(c/2).
Solve the spherical Law of Cosines spherical Law of Cosines of the cos(a) for the cos(A), to obtain
    cos(A) = (cos(a) - cos(b) cos(c)) / (sin(b) sin(c)).
Substitute it into the previous equation, to obtain
    = sin(b/2) sin(c/2) ((cos(a) - cos(b) cos(c)) / (sin (b) sin (c))) + cos(b/2) cos(c/2).
Employ the double-angle formula for the sine double angle and place the result over a common denominator, to obtain
    = (cos(a) - cos(b) cos(c) + 4 (cos(b/2))^2 cos(c/2))^2) / (4 cos(b/2) cos(c/2)).
In the numerator, employ the half-angle formula for the cosine half angle, to obtain
    = (cos(a) - cos(b) cos(c) + (1 + cos(b)) (1 + cos(c))) / (4 cos(b/2) cos(c/2)).
Finally, multiply-out the numerator, to obtain
    = (1 + cos(a) + cos(b) + cos(c)) / (4 cos(b/2) cos(c/2)).
Division of both sides of this equation by cos(a/2) yields the formula in the statement of the theorem.  QED.
This proof has been expanded from that supplied by Allan Cortzen.  Gramercy.
Corollary:   l'Huilier's (Swiss mathematician Simon Antoine Jean l'Huilier (24.4.1750-28.3.1840)) We have tan(T/4) as
    (tan(T/4))^2 = tan(s/2) tan((s - a) / 2) tan((s - b) / 2) tan((s - c) / 2).
Proof: Let
    us = sin(s/2) sin((s - a)/2) sin((s - b)/2) sin((s - c)/2)).
Substitute the definition of s four definitions four definitions and write the foregoing as
    = sin(((b + c) + a)/2) sin(((b + c) - a)/2) sin((a - (b - c))/2) sin((a + (b - c))/2).
Employ the sums or differences sums or differences of sines, to obtain
    = (- cos((b + c)/2) + cos(a/2)) (- cos(a/2) + cos((b - c)/2)) / 4.
Multiply it out, to obtain
    = (- (cos(a/2))^2 - cos((b + c)/2) cos((b - c)/2) + cos(a/2) (cos((b + c)/2) + cos((b - c)/2)) / 4.
Likewise, let
    uc = cos(s/2) cos((s - a)/2) cos((s - b)/2) cos((s - c)/2).
Substitute the definition of s four definitions four definitions and write the foregoing as
    = cos(((b + c) + a)/2) cos(((b + c) - a)/2) cos((a - (b - c))/2) cos((a + (b - c))/2).
Employ the sums or differences sums or diferences of cosines, to obtain
    = (cos((b + c)/2) + cos(a/2)) (cos(a/2) + cos((b - c)/2)) / 4.
Multiply it out, to obtain
    = ((cos(a/2))^2 + cos((b + c)/2) cos((b - c)/2) + cos(a/2) (cos((b + c)/2) + cos((b - c)/2)) / 4.
Add and subtract, to obtain
    uc + us = cos(a/2) (cos((b + c)/2) + cos((b - c)/2)) / 2.
    uc - us = ((cos(a/2))^2 + cos((b + c)/2) cos((b - c)/2) / 2.
Employ the addition theorem upon the cosines addition, to obtain
    uc + us = cos(a/2) cos(b/2) cos(c/2).
Employ the double-angle formula double angle and the product formula products, to obtain
    uc - us = (1 + cos(a) + cos(b) + cos(c)) / 4.
From the theorem, it is evident that
    cos(T/2) = (uc - us) / (uc + us).
By the half-angle formula for the tangent half angle, we have
    (tan(T/4))^2 = (1 - cos(T/2)) / (1 + cos(T/2)).
Substitute the penultimate equation into the previous equation, and multiply the numerator and the denominator by (us + uc), to obtain
    = us / uc.
which is the equation in the statement of this corollary.  QED.
This proof has been expanded from that supplied by Allan Cortzen.  Gramercy.
Pray, observe that the right-hand side of each the theorem and its corollary is a symmetric function of the elements of the triangle ABC as it should be, since the spherical excess T (in the left-hand side of each) is likewise.  This same comment also pertains to the following special cases.
Pray, also observe that the proof of this corollary does not employ any spherical trigonometry -- it only uses trigonometric identities and addition theorems.  Hence, the corollary is a logical equivalence, however obscure.

Special case of equilateral/equiangular triangle:
    cos(T/2) = (1 + 3 cos(a)) / (4 (cos(a/2))^3) and
    (tan(T/4))^2 = (tan(a/4))^2 |tan(a/4) tan(3 a / 4)|.

A summary of duality and the plane limits, in tabular-form.  The limit of any angle is the angle, itself.  The duality is reflexive; hence, we only state it in one direction. Also, we only state the first of the three cyclical permutation.  The proofs have been provided in several places, previously:  duality duality of circles.  In an equation, the limit is valid only if the sides within each term are raised to the same degree; i.e. , the equation is homogeneous in the degree of the sides.  A haversine of a side counts as two.

Projection Construction

In Plane Geometry, we know how to construct a circle through any three given points:   Construct the perpendicular bisectors of the line-segment between each pair of these points.  They will intersect in a common point, which is called the circum-center of the triangle of those points.  Construct the circle, centered at the circum-center and passing through any one of the given points.  It will pass through each of the other two points.

[[We pause for a brief editorial comment from the Web master.  Inconsistencies reign!  In navigation, the latitude is measured positive from the equator towards the north pole.  In mathematics, the latitude just as well may be measured positive from the north pole towards the equator.  Then either of these may be called the latitude and the other the co-latitude.  Furthermore, the roles of phi and theta may be interchanged.  Reader beware.  End of note.]]

From the properties of the stereographic projection, we know that it is a one-to-one projection from the sphere onto the plane tangent at the north pole, which projection has the property that any spherical circle projects onto a circle.  This projection is conformal.  However, regrettably, the center of the circle does not project to the center of the projected circle.  We present an algorithm for constructing the center of three important projected circles, namely:  The great-circle representing the side of a triangle.  The circum-circle of a triangle.  The in-circle of a triangle.  We also provide the construction of the antipodal point to a plane given by two points - usually the side of a triangle.  [[We place the plane of projection as tangent at the north pole.  This is the traditional placement, for instance, in the constructions of the Astrolabe.  Modern usage often places the plane of the projection as that of the equator of the sphere.  It is the same stereographic projection; but, the radius becomes r=tan(phi/2).]]

We define the two-argument inverse tangent function atan(y,x) as atan(y,x)=atan(y/x) if x is in [0,oo) otherwise atan(y,x)=atan(y/x)+pi.  Any integral multiple of 2*pi may be added, as convenient.  Pray observe that we place the y before the x, in deference to the usage in many computer programming languages.  [[In older programming libraries, this function is written as atan2(y,x).  However, the 2 has been dropped in the current libraries.]]

For simplicity, as usual, we take the radius rho of the sphere as one.  The angle phi is the co-latitude -- measured positive from the north pole towards the equator -- and the angle theta is the longitude.  They are the spherical coordinates.  On the plane, r is the radius and theta is the angle of the polar coordinate system.  We use neither the two-dimensional Cartesian coordinate system on the plane nor the three-dimensional Cartesian coordinates.  However, for the record, here are the transformations:  In the plane, we have

• x=r*cos(theta)

• y=r*sin(theta)

It inverts as

• r=sqrt(x^2+y^2)

• theta=atan(y, x).

On the sphere, we have

• x=cos(phi)*cos(theta)

• y=cos(phi)*sin(theta)

• z=sin(phi)

 It inverts as

• phi=atan(z/sqrt(x^2+y^2))

• theta=atan(y, x)

The point at (phi, theta) on the sphere projects to the point (r, theta) on the plane.  The radius r is given by r=2*tan(phi/2).  This function inverts to phi=2*atan(r/2).  The angle theta projects as positive, from the vertical-up, clockwise.  This is the complement of the usual two-dimensional polar theta, which is from the horizontal-right, counterclockwise.  [[One more inconsistency!]]  Thus, we can accept the given vectors in either the spherical coordinates on the sphere or the polar coordinates on the plane.

Since we do not know the location of the center of the projected circle, we use plan B:  Project any three non-coincident points on the circle.  Then employ the usual plane geometry method to draw the circum-scribed circle of this triangle.

Two points with known coordinates

Theorem:  A pair of points with known coordinates (on either the sphere or its stereographic projection) determines a unique great-circle arc -- with a unique center and radius -- passing through them.
Proof:  The proof is by means of the following construction.

For any given two points A and B, always place the third vertex C of the triangle at the north pole.  The angle C then becomes C=theta-b - theta-a, provided that C does not lie on the great circle through the points A and B.

We encounter three situations, each is SAS.  We employ the same law of haversines to solve each triangle.

1. Sides b and a and angle C known; find side c and angle A.  We do not need angle B.  Aside from the intrinsic benefit of having solved this triangle, we need the value of the angle A for each of the next two calculations.  Once we have the three sides of this triangle, we may compute its area by means of the l'Huilier's formula.  Alternatively, compute the angle B, as well.  Then the area is given by the spherical excess.

   • c = ahav(hav(phi-b - phi-a) + sin(phi-b) * sin(phi-a) * hav(C)).

   • A = ahav((hav(phi-b) - hav(phi-a - c)) / (sin(phi-a) * sin(c))) * sign(sin(C)).

   • B = ahav((hav(phi-a) - hav(phi-b - c)) / (sin(phi-b) * sin(c))) * sign(sin(C)).

2. We proceed as follows to find any convenient third point on the great circle arc from A to B.  Sides b and c'=alpha and angle A known; find side a and angle C'.  The vertex B' becomes the requisite third point on the great-circle arc which is the side c.  We do not need angle B.  Project the three points A, B, and B'.  Construct the circum-circle through the vertices of this triangle.  The resulting arc from the projection of the point A to that of the point B is the projection of the spherical side c.  For the purpose of drawing this arc the value assigned to alpha is arbitrary, the choice alpha = s - a places the point B' at the point of tangency for the inscribed circle.

   • a = ahav(hav(phi-a - c') + sin(phi-a) * sin(c') * hav(A)).

   • C' = ahav((hav(phi-b) - hav(phi-a - c')) / (sin(phi-a) * sin(c'))) * sign(sin(C)).

   • B = ahav((hav(phi-a) - hav(phi-b - c')) / (sin(phi-b) * sin(c'))) * sign(sin(C)).

3. Sides b and c*=-pi/2 and angle A”=A-pi/2 known; find side a* and angle C*.  The vertex C* becomes the antipodal point to the side c.  For each point C* at (phi*, theta*), there is another one, call it C**, diametrically opposite at (pi-phi*, theta*+pi).  We do not need angle B”.

   • a* = ahav(hav(phi-a - c*) + sin(phi-a) * sin(c*) * hav(A”)) = ahav(hav(phi-a + pi/2) - sin(phi-a) * hav(A - pi/2)).

   • C* = ahav((hav(phi-b) - hav(phi-a - c*)) / (sin(phi-a) * sin(c*))) * sign(sin(C)) = ahav((hav(phi-b) - hav(phi-a + pi/2)) / (- sin(phi-a))) * sign(sin(C)).

   • B = ahav((hav(phi-a) - hav(phi-b - c*)) / (sin(phi-b) * sin(c*))) * sign(sin(C)) = ahav((hav(phi-a) - hav(phi-b + pi/2)) / (- sin(phi-b))) * sign(sin(C)).

Project the points, A, B, and B', using beta = s - b.  Construct the circle through them.  This circle is the projection of the great-circle through the points A and B.

Triangle ABC

In the same manner, using gamma = s - c, construct the point C'.  Project the points B, C, and C'.  Construct the circle through them.  This circle is the projection of the great-circle through the points B and C.  Likewise for C, A, and A'.  Huzzah!  You have drawn the projection of the triangle ABC.  Optionally, you have obtained the antipodal triangle A*B*C*.  You may draw its projection in an analogous manner.

Given the spherical triangle ABC, represented by the three unit-vectors A, B, and C.  Project each of these points.  Construct the circum-circle.  It is the projection of the spherical circum-circle.

From the proof of the Half-tangent law, second version, we know that the distance from the vertex A to the point Pc (on the side c) is alpha=s-a, where, as usual s=(a+b+c)/2.  From that proof, it follows that the circum-circle to the triangle Pa-Pb-Pc is the in-circle to the triangle ABC.  Since we already have learned how to construct a circum-circle, we now can construct an in-circle, as well.

The next step is to draw the circum-circle in the projection plane.  This is just ordinary plane geometry.  We have several obvious choices as to how to do so.

• Straight-edge and compass construction.

• Analytic geometry construction.

• Eyeball it.

• Employ a two-dimensional CAD (=Computer Aided Design) program, a two-dimensional geometry program, or some drawing program.

Now, we describe the first two in detail ---

In Plane Geometry, we know how to construct a circle through any three given points:   Construct the perpendicular bisectors of the line-segment between each pair of these points.  They will intersect in a common point, which is called the circum-center of the triangle of those points.  Construct the circle, centered at the circum-center and passing through any one of the given points.  It will pass through each of the other two points.

In Plane Analytic Geometry, we proceed as follows:

If the vertices of the triangle are given in polar coordinates, compute the Cartesian coordinates:

• For the vertex A:

  • xa = ra * cos(theta-a)

  • ya = ra * sin(theta-a).

• For the vertex B:

  • xb = rb * cos(theta-b)

  • yb = rb * sin(theta-b).

• For the vertex C:

  • xc = rc * cos(theta-c)

  • yc = rc * sin(theta-c).

Given the three vertices of the triangle ABC in Cartesian coordinates, the circum-center at (xo, yo), and the circum-radius as R.  Then there are three simultaneous equations, in three unknowns, of the form

• (xa - xo)^2 + (ya - yo)^2 = R^2.

Each one expands to

• (xa^2 + ya^2) - 2 * (xa * xo + ya *yo) + (xo^2 + yo^2) = R^2.

We observe that the equation is linear in xo, yo, and R^2.  Thus the system should be easy to solve.  Multiply the first be one and the third by minus one and add to obtain

• 2 * ((xa - xc) * xo + (ya - yc) * yo)) = (xa^2 + ya^2) - (xc^2 + yc^2).

These two equations have the solution:

• det

  • = 2 * ((xa - xc) * (yb - ya) - (ya - yc) * (xb - xa))

  • = 2 * (xa * (yb - yc) + xb * (yc - ya) + xc * (ya - yb))

• xo

  • = ((yb - yc) * ((xa^2 + ya^2) - (xc^2 + yc^2)) - (ya - yc) * ((xb^2 + yb^2) - (xc^2 + yc^2))) / det

  • = ((yb - yc) * (xa^2 + ya^2) + (yc - ya) * (xb^2 + yb^2) + (ya - yb) * (xc^2 + yc^2)) / det

  • = ((yb - yc) * xa^2 + (yc - ya) * xb^2 + (ya - yb) * xc^2+ya^2 * (yb - yc) + yb^2 * (yc - ya) + yc^2 * (ya - yb)) / det

• yo = ((xb - xc) * ya^2 + (xc - xa) * yb^2 + (xa - xb) * yc^2 + xa^2 * (xb - xc) + xb^2 * (xc - xa) + xc^2 * (xa - xb)) / det.

We observe that each of the formulae for xo, yo, and R is symmetric in the three vertices of the triangle, as they should be.

In principle, these values for xo and yo could be substituted back into any one of the original set of three equations and solved for R.  However, it is much easier to go look at the answer that we already had.
From plane trigonometry, we have the circum-radius as

• R = a * b * c / (4 * sqrt(s * (s - a) * (s - b) * (s - c))),

where, as always

• s = (a + b + c) / 2.

The length of each side is

• a = sqrt((xb - xc)^2 + (yb – yc)^2)

• b = sqrt((xc - xa)^2 + (yc – ya)^2)

• c = sqrt((xa - xb)^2 + (ya – yb)^2)

We leave it up to the reader to learn how to use any appropriate computer program or the art of manual construction.

There is one more method for finding the circum-center (xo,yo).  Employ Vector Analysis to perform the geometrical construction.  Then, we obtain

• (xo,yo)=(B+A)/2+-(B-A)xk*sqrt(R^2-((B-A)/2)^2)/sqrt((B-A)^2).

Only four of the sixteen distinct solutions of the spherical analogue make sense in the plane case; because, the interior/exterior and the antipodal transformations are not applicable.

Where we indicate a cross-product by the letter x, k is the unit-vector normal to the xy-plane, and R is the scalar value of the circum-radius.

Lemma:  The sum of the tangents is given by the formula

• tan(a+b)+tan(a-b) =

  • =(tan(a)+tan(b))/(1-tan(a)tan(b))+(tan(a)-tan(b))/(1+tan(a)tan(b))

  • =2tan(a)(1+(tan(b))^2)/(1-(tan(a)tan(b))^2).

Proof:  Hint, employ the addition theorem for the tangent function.
Theorem:  The center of the projected circle is at (xo,yo)=2*tan(phi/2)(1+(tan(RR/2))^2)/(1-(tan(phi/2)tan(RR/2))^2), where RR is the radius of the spherical circle and (phi,theta) is its center.
Proof:  By symmetry, the theta projects to the same value.  The point farest away from the north pole has its phi equal to phi+RR.  The point nearest to the north pole has it phi equal to phi-RR.  Substitute into the lemma as a=(phi+RR)/2 and b=(phi-RR)/2, to obtain the stated result.  QED.

The epicenter of a spherical circle is that on the sphere.  The hypocenter is that on the plane of the circle.  [[The words "epicenter" and "hypocenter" have been usurped by Geologists for use in conjunction with earthquakes.]]

Given a spherical circle with epicenter at (phi,theta) and spherical radius RR.  Its hypocenter is at the radial height of cos(RR).  It projects to r=2*cos(RR)*sin(phi)/(1+cos(RR*cos(phi)).

For reference, each drawing should include the point at the north pole and the mapping of the equator, namely a circle with a radius equal to (2 rho), centered at the north pole, where rho is the radius of the sphere.

Find Spherical Circum-center

 There are various ways to find the spherical circum-center of a triangle.  The choice depends upon how the triangle is given and upon the personal preference or whim.  We describe three of the methods.

1. Given the three unit-vectors - call them {A, B, C} - from the center of the sphere toward the respective vertices.  Employ vector analysis.  Let U = +- (B-A)x(C-A), with the positive sign if either the three vertices are counter-clockwise, as viewed from on top of the circum-center, and the coordinate system is right-handed or clockwise and left-handed; otherwise, with the negative sign.  For an expanded discussion, pray see links to the several GST, at the bottom of this page.  The vector U simplifies to U = +- (AxB + BxC + CxA).  Position this pseudo-vector from the center of the sphere.  Then let u = U / sqrt(U^2).  It obviously is a unit-vector. The epicenter is at the product (u rho), where rho is the radius of the sphere.  The hypo-center at the product (u rho u.A); because the dot-products u.A = u.B = u.C = [A B C] . sqrt(U^2) = cos(RR), where RR is the circum-radius.  Remember that each of the vectors u, A, B, and C is a unit-vector.  Since each dot-product has the same value, their equality confirms that, for sooth, we have located the circum-center.  The notation [A B C] is called the box product of the indicated vectors.  There is no analogue to this construction in the plane.

2. Given (at least) three elements of the triangle.  Solve the triangle by the methods of Spherical Trigonometry.  Let P be the epicenter.  Consider the triangle ABP.  It is an isosceles-triangle, with each side known:  The side AB is equal to c of the original triangle.  Each of the other two sides is equal to RR, the circum-radius, which may be computed by means of its formula.  Employ the spherical law of haversines to find the angle at P.  Then the short form of the spherical law of sines gives each of the other two angles - they are equal; because the triangle is isosceles, as has been observed already.

3. Given the spherical coordinates of each vertex of the triangle.  Solve the triangle by the method described previously.  Again, let P be the epicenter.  Proceed as in the previous item.  Subtract the angle BAP from the angle BAN, where N is the north-pole.  Solve the triangle NAP for the spherical-coordinates of P, by the method described previously.

Since the stereographic projection requires a specific placement of the spherical triangle upon the sphere, your desire ultimately to project may influence your choice of the method for finding the spherical circum-center.  If you are not given any specific location or orientation for the triangle; you will have to place is somewhere, in an arbitrary position and orientation.  One interesting placement is with the spherical circum-center at the north-pole.  Then, the three vertices of the triangle will lay upon a circle, of radius RR (the circum-radius), centered at the north-pole.  Unless this radius is equal to pi/2, the sides of the triangle will not coincide with the circum-circle, however.

There remains the unanswered question: For a triangle, How do we choose between the diametrically opposed antipodal points in a consistent manner?  This is an especially difficult question to answer in the case where any of the elements of the original triangle ABC exceeds pi.

In closing, I have to extend yet another gramercy to Allan Cortzen for sending me the following e-mail on 05th May 2009:
Hello Romuald
    I just found an article referring to l'Huilier on your page and thought it may interest you how this overlooked theorem can be used in an extremum problem.
    Thank you for having it on your pages!
    The paper has this address: http://www.rose-hulman.edu/mathjournal/archives/2007/vol8-n2/paper11/v8n2-11pd.pdf
Allan  

On Tuesday 09-th June 2009, Allan Cortzen suggested the addition of the following lemma ---

It seems that the basic formulas involving only {a, b, c, A, B, C, s, S, rr, RR, T} are invariant under any permutation of the pairs {(a, A), (b, B), (c, C)} as an orientation of the sphere is not used in the derivation of the formulae - except for the area, where the result anyway is invariant.
Lemma. In a triangle we have following estimations and their analogous
    a, s, s - a, 2 RR in (0, pi)
    a < b + c
    A, T /2, A - T / 2, 2 rr in (0, pi)
    B + C < pi + A
    S in (pi / 2, 3 pi / 2)
    S - A in (- pi / 2, pi / 2)
Proof: For a triangle by definition a, b, c, A, B, C, 2 RR in (0, pi).
From the sign of the factors in formula-type
    (1 + cos(A)) sin(b) sin(c) = 2 sin(s) sin(s - a)
we conclude, that
    sin s , sin(s - a) , sin(s - b) , sin(s - c) (*)
all have the same sign.
A rough estimation using s = (a + b + c) / 2 gives
    0 < s < 3 p / 2, -p ; 2 < s - a < p, etc.
Now if a < b and a < c; then s - a = (-a + b + c) / 2 > 0.
If b or c is the smallest side, we will likewise still have a plus sign in (*), which then is the common sign. This and the rough estimations implies s, s - a in (0, pi).
Finally s - a > 0 iff [= if and only if] a < b + c and the rest comes from duality and definitions.
If r and R is defined positive by
    r = sqrt(sin(s-a) sin(s-b) sin(s-c))
    R = sqrt(- cos(S) / (cos(S - A) cos(S - B) cos(S - C))
then the negative sign in all Law of Half-Tangent formulas can be deleted.
All the double signs can now to be eliminated. E.g.:
    cot(B / 2) cot(C /  2) = sin(s) / sin(s - a) ,
    tan rr = sqrt(sin(s - a) sin(s - b) sin(s - c) / sin(s))
    tan(RR) = sqrt(- cos(S) / (cos(S - A) cos(S- B) cos(S - C)) .

and L'Huliers formula can be stated as
    tan(T / 4) = sqrt(tan(s / 2) tan((s - a) / 2) tan((s - b) / 2) tan((s - c) / 2))

If adequate we can write for the equilateral triangle
    tan(rr) = sqrt((sin(a / 2)^3 / sin(3 a / 2))
and for the equiangular triangle
    tan(RR) = sqrt(- cos(3 A / 2) / (cos(A / 2)^3)

Rejoinder:
In the context of the Spherical Trigonometry, I concur with the first sentence, above.
However, pray consider:
1) vis-a-vis Moibus strip or Klein bottle.
2) volume enclosed by a sphere.
3) possibility of a cross-product (which yields a pseudo-vector) -- thus, is dependent upon the orientation.

All of the ambiguous negative signs disappear if we arbitrarily proscribe against reflex (greater than pi) angles, as proven in this lemma. However, a computer program based upon the formulae in these theorems, etc. would have to contend with the possibility of the user entering an angle or side that violates these constraints.
1) It just could hang or error out. Hardly a desirable behavior.
2) It could catch the violation and provide an error message. Still not very useful.
3) It should re-map the values into those that it can understand and yield the correct answer, together with an explanation of what had happened.
#3 Obviously is the most desirable for the user. However, it would require considerable effort to anticipate the various combinations of possible exceptions in the input.

It is obvious that the general -- that is, without the constraints as to the size of the sides or angles -- formula for T is
    (tan(T/4))^2 = |tan(s/2) tan((s - a) / 2) tan((s - b) / 2) tan((s - c) / 2)|.
Its square-root is
    tan(T/4) = +- sqrt(|tan(s/2) tan((s - a) / 2) tan((s - b) / 2) tan((s - c) / 2)|).
Let
    To = 4 * Atan(sqrt(|tan(s/2) tan((s - a) / 2) tan((s - b) / 2) tan((s - c) / 2)|)).
Then, the solution set is
    {T-inside = To, T-outside = 4 * pi - To}.
Consider the triangle ABC.  By definition, its spherical excess is given by T = E = (A + B + C) - pi.  The angles all the way around the outside of each vertex are given by (2 pi - A), etc.  Then, the outside spherical excess becomes T-outside = ((2 pi - A) + (2 pi - B) + (2 pi - C)) - pi = 4 pi - ((A + B + C) - pi) = 4 pi - T-inside.  This result is the same as we just obtained as the T-outside solution of the previous equation.

I will leave it up to Allan to fill-in the details; if he so chooses.

On Friday the twentieth of April, Allan Cortzen sent me this e-mail:
For the time being something to reflect over in an unconstrained theory:

1. It may lead to contradictions.
Assume a triangle with given a, b, c in (0, pi) and a + b + c < 2 pi.
The law of the half tangent, tan(A / 2) = +- r / sin(s - a), gives in the general case for each angle 2 solutions, therefore 8 solutions for the set A, B, C.
An unconstrained theory let us to accept e. g. A in (pi, 2 pi) and B, C in (0, pi).
By the [short-form of the] Law of Sines, sin(A) / sin(a) = sin(B) / sin(b), we get sin(A) > 0 in contradiction to A in (pi, 2 pi).

2. The usual theorems about congruencies does not hold.
It would be nice to have this theorem of congruence:
If the sides in two triangles are pair wise equal the same are true for the angles.
This is true for the constrained theory, because there are at most one triangle with given sides a, b, c in (0, pi) where the angles A, B, C in (0, pi).
But it is not true in an unconstrained theory by the example in the end of the addendum, where an "outside triangle" is considered.

3. Duality have to be reformulated and give rise to peculiar triangles.
Duality modulus 2 pi applied to an "outside" triangle with a, b, c in (0, pi) and A, B, C in (pi, 2 pi) forces us accept a dual triangle with sides in (pi, 2 pi), which always are mutually crossing.

My response:

Forsooth, when one allows angles or sides to exceed pi, extraneous solutions become possible.
One has to exercise caution.  However, even in Plane Trigonometry, the congruence of the SSA (= side side angle) is subject to having an extraneous solution.   But, how interesting that we may have a triangle with intersecting sides?!  Such a triangle would require that two of its sides exceed pi -- in contradiction to my stated constraint for at most one side to exceed pi.  Proceed beyond pi at your own risk.

Definition:  A side or an angle whose magnitude is in the open interval (pi, 2 pi) is said to be a reflex side or reflex angle, respectively.
Theorem:  Allowed domains for the sides and angles.  Each side or angle is constrained to be in the open interval (0, pi), except where explicitly indicated otherwise.  There are five cases, consisting of two or three sub cases each, possible.
Case 0.  Trivial.
Sub case 0a.  The three vertices coincide in a single point.  The sides are a = b = c = 0.  The S = pi / 2.  Arbitrarily, we take A = B = C = pi / 3.  Hence, E = 0; i.e., a null-triangle.
Sub case 0b.  A great-circle. The angles are A = B = C = pi. The s = pi.  Arbitrarily, we take a = b = c = 2 pi / 3.  Hence, E = 2 pi; i.e., a hemi-sphere.
Sub case 0c.  The three vertices coincide in a single point.  The sides are a = b = c = 0.  The S = 5 pi / 2.  Arbitrarily, we take A = B = C = 5 pi / 3.  Hence, E = 4 pi; i.e., the whole sphere.
Case 1.
Sub case 1a.  This is the usual situation.  Hence, E is in the open interval (0, 2 pi).
Sub case 1b.  Each of the angles is reflex.  Hence, E is in the open interval (2 pi, 4 pi).
Case 2.  A single side is reflex.
Sub case 2a.  The angle opposite is reflex.  Hence, E is in the open interval (pi, 3 pi).
Sub case 2b.  Each adjacent angle is reflex.  Hence, E is in the open interval (2 pi, 4 pi).
Case 3.  Two sides are reflex.  There is one crossover.
Sub case 3a.  The included angle is reflex.  Hence, E is in the open interval (pi, 3 pi).
Sub case 3b.  Each opposite angle is reflex.  Hence, E is in the open interval (2 pi, 4 pi).
Case 4.  All three sides are reflex.  There are three crossover's.
Sub case 4a.  Hence, E is in the open interval (0, 2 pi).
Sub case 4b.  All three angles are reflex.  Hence, E is in the open interval (2 pi, 4 pi).
Proof: In the short-form of the Law of Sines, each quotient must be of the same sign.
Pray, see the item #1 for a detailed typical proof.

Pray, see the following GST. in which the foregoing classification is expanded.  A distinction is drawn between E and T, which simplistically was ignored in the preceding classification.
Allan Cortzen has provided the "Generalized Spherical Triangles" in a PDF file-format, which I transcribed to an HTML file format on Tuesday 07-VII-2009.  Gramercy, Allan.  On his own Web site, he has posted an expanded version.  Regrettably, as of now (Thursday 23-VII-2009), it is available only in the proprietary *.PDF format.  This expanded GST includes the denumerable group that governs the mapping from the small domain of the spherical triangle to the large domain.  Intriguing; but, certainly not unexpected.
 

copyright (c) 2009 by R.I. 'Scibor-Marchocki   first posted Sunday 07-VI-2009   revised Wednesday 29-VII-2009.  Minor corrections on Friday 07-VIII-2009.  Last revised on Sunday 09-VIII-2009.