From the inner (dot) product, employing the synthetic method, in this online textbook, we develop the elementary geometries (including the trigonometry), in a pseudo-metrizable topology. The hyperbolic and circular trigonometric functions on a complex space are defined in terms of the exponential function. Their inverses are obtained. The addition theorems are presented and inverted. The derivatives, integrals, and infinite expansions are obtained. The Raphson, Newton, and Horner methods of finding the roots of an equation are presented. The binomial theorem and Gamma function are presented, also.
Synthetic Geometry does not employ a coordinate system; it develops the congruence and similarity theorems. Analytic Geometry employs coordinate systems. It is happy to make use of the congruence or similarity theorems; but usually does not prove them. We employ the Synthetic Geometry, exclusively. The same distinction may be made between Analytic Vector Analysis vs. Synthetic Vector Analysis.
Extensive cross-links are provided to related or pre-requisite material. To make the equations clearer, we omit any punctuation at their ends. Alternatively, see the index. Spherical Trigonometry is vital to geodesy, navigation, and celestial mechanics.
We presuppose a familiarity with the concepts of points and vectors. The trigonometric functions are convenient combinations of the exponential function. We already have discussed both the circular and the hyperbolic trigonometric functions.
Tacitly, we assume that the geometry is a plane geometry (curvature equal to zero). Later we will show the generalization to the two remaining elementary geometries, namely, a spherical geometry (strictly positive curvature) and a hyperbolic geometry (strictly negative curvature).
An alternative approach would be to begin with the Spherical Geometry/Trigonometry. Then to obtain the Plane Geometry/Trigonometry as the limit as the radius of curvature approaches zero. That way, we would employ the inner-product only once. But for pedagogical reasons, we defer to the historical sequence.
The cross product (sometimes called the vector product) of two vectors is a pseudo-vector, and as such, is dependent upon the chirallity of the space. The cross product cannot be defined in less than three-dimensional space. In the mid 1920’s, Cartan generalized the cross product to higher than three-dimensions. However, this generalization is too complicated to be practical. The same functionality may be achieved easily by the use of matrices. Thus since the cross product is practical only in three-dimensions, we will not consider the cross product at all.
Observe that the cross product of a pseudo-vector by a vector is, once again, a vector.
However, one must not underestimate the utility of the cross product. Maxwell’s electromagnetic theory, either in its elegant Hamilton’s quaternion or vulgar Gibb’s vector form, expresses the magnetic pseudo-vector as a cross product of the electric and velocity vectors. And closes the relationships by expressing the electric vector as the cross product of the magnetic pseudo-vector and current vector. Each of these is a first-order differential equation.
The Einstein’s relativity theory owes its difficulty to its avoidance of the cross product. By eliminating the magnetic pseudo-vector, we obtain a second-order differential equation.
As the only product of vectors that we consider, we will omit the dot in the dot product. The dot product (sometimes called the scalar or inner product) of two vectors (emanating from a point) is a scalar, equal to the product of the norms of each of these two vectors times the cosine of the included angle. We turn this definition around to define the angle, in the closed interval [0, pi]. Then the norm of a vector (which is written ||vector||) is equal to the square root of its square; that is of the dot product of the vector by itself.
Given three vectors a, b, and c, the equation of a triangle is
a + b + c = 0
Solve for a and square to obtain
a^2 = b^2 + c^2 – 2 (-b) c
Evaluation of this dot product yields the first version of the Law of Cosine
a^2 = b^2 + c^2 – 2 ||b|| ||c|| cos(A) = (||b|| – ||c||)^2 + 2 ||b|| ||c|| versine(A)
where the capital letter indicates the angle opposite the like-named vector. Cyclical permutation yields the remaining two equations. Then, solve for the angle to obtain the second version of the Law of Cosine
cos(A) = ((b^2 + c^2) – a^2) / (2 ||b|| ||c||))
versine(A) = (||a|| – (||b|| – ||c||)) (||a|| + (||b|| – ||c||)) / (2 ||b|| ||c||))
Again, cyclical permutation yields the remaining two equations. Despite their popularity, because of their excessive round-off errors; these equations involving the cosine should not be employed for numerical calculation. The equations involving the obscure function, versine, however, circumvent the round off problem. Indeed, these equations are the motivation for the original definition of the versine function. I do not know any application for the coversine function. Probably, its only reason for being is as the co-function to the versine function.
Definition: A circle is the locus of point at the end of vectors, whose norm is the radius from the point of the center.
Definitions: A circumscribing circle passes through the vertices of the triangle. An inscribed circle passes through the foot of a perpendicular dropped from the center to each of the sides of the triangle.
These definitions generalize to n-dimensional space. Please see the discussion of the hyper-spheres.
Before we proceed, let us consider two preliminary algebraic problems: Finding square-roots and solving polynomial equations. These techniques will be used -- or at least implied -- throughout the following exposition, without any links. Three interesting topics -- otherwise without a link -- are progressions (including interest, annuity, and interpolation/extrapolation), evaluation of a power series, and linear regression.
Square the foregoing equation
(cos(A))^2 = a^2 ((a^4 + b^4 + c^4) - 2 (b^2 c^2 + c^2 a^2 + a^2 b^2)) / (4 a^2 b^2 c^2) + 1
Then, by use of the identity
1 = (sine(A))^2 + (cos(A))^2
we obtain
(sine(A))^2 = a^2 (2 (b^2 c^2 + c^2 a^2 + a^2 b^2) – ( a^4 + b^4 + c^4)) / (4 a^2 b^2 c^2)
And taking square roots, we obtain the Law of Sine
sine(A) / ||a|| = 1 / (2 R)
where R, the radius of the circumscribed circle, is the positive square root of
R^2 = (a^2 b^2 c^2) / (4 (b^2 c^2 + c^2 a^2 + a^2 b^2) – (a^2 + b^2 + c^2)^2)
As it should, R is a symmetric form in the three vectors. That R indeed is the radius of the circumscribed circle will be shown later. By cyclic permutation of the Law of Sine, we obtain the Law of Sines
2 R = ||a|| / sin(A) = ||b|| / sin(B) = ||c|| / sin(C)
In this form, the Law of Sines may be employed to solve a triangle whose three sides (SSS) are known. Alternatively, the Law of Tangents may be employed for the SSS. If either two sides and an angle opposite one of them (SSA) or two angles and a side opposite one of them (AAS) is know, the Law of Sines yields the remaining element of the other pair. In either case, we have obtained two sides and the angles opposite them. The third angle follows from the equation
pi + area / rho^2 = A + B + C
which we will prove later. Now that we have all three angles and at least one side, the remaining sides may be obtained from the Law of Sines.
The only situation, which the Law of Sines does not solve, is that of two sides and the included angle (SAS). Most people succumb to the temptation of the Law of Cosine for this purpose. Instead, one should employ its variant, the Law of Versine.
The final situation of two (AA) or of three angles (AAA) yields only a similar triangle; because we cannot find any of the sides. In a non-flat space (rho not equal to zero), however, we can solve the AAA problem.
There are constraints upon the sides and angles, for instance, the sum of the norms of the two sides with the smaller norms has to exceed the norm of the remaining side. Other constraints are left for the reader to discover. The SSA, with the angle being acute, has two solutions; or none, if the applicable constraint is not satisfied.
Basket winding presents an interesting application of congruent triangles.
Theorem: Pythagorean. The sum of the squares of the sides (called legs) including a right angle is equal to the square of the other side (called hypotenuse), opposite the right angle.
Proof: Set the angle in the Law of Cosine equal to pi / 2. Q.E.D.
Note: This is the trigonometric statement of the Pythagorean theorem; the geometric has “square(s) on”.
Theorem: The sum of the exterior angles of a polygon is an integral (dependent upon the winding) multiple of two-pi minus the quotient of enclosed area divided by the radius of curvature, rho, squared. Each of the formulae is written with the assumption that the winding is non-negative.
Proof follows from the modulus of periodicity of the sine and cosine functions. Q.E.D.
Corollary: The sum of the interior angles of a triangle is pi minus the quotient of its enclosed area divided by the radius of curvature, rho, squared.
Proof: Obviously the winding of a triangle can be only either plus or minus one. Let the angles of a triangle be A, B, and C. Then the exterior angles are pi – A, pi – B, and pi – C. Substitute them into the foregoing theorem (pi – A) + (pi – B) + (pi – C) = 2 pi – area / rho^2. Rearrange to yield A + B + C = pi + area / rho^2. Q.E.D.
This equation may be solved for the radius of curvature, rho, to yield
rho^2 = area / (A + B + C – pi)
Definition: Two triangles are said to be similar iff (= if and only if) each of their respective angles is equal to each other.
Definition: Two triangles are said to be congruent iff (= if and only if) they are similar and each of their respective sides is equal to each other.
Theorem: The similarity and the congruence are equivalence relationships.
The proof is obvious.
Definition of mirror-image. If triangle B is the mirror-image of, but is not congruent to, triangle A and triangle C is the mirror-image of triangle B; then, triangle A is congruent to triangle C
Theorem: Two congruent triangles, located upon concentric spheres of unequal radii, may be moved (translated or rotated) so that they project corresponding parts by means of straight lines from the common center. If the radii are equal (including the special case of plane triangles); the corresponding parts may be superimposed upon each other.
The proof is obvious
Any condition for which a unique solution of a triangle exists is a condition for the congruence of a pair of triangles. We provide an
exhaustive table of these conditions. For example, we have the
Theorem: SSS. Two triangles are congruent if each of their respective sides is equal to each other.
Proof follows from the Law of Cosine. Q.E.D.
Corollary: The hypotenuse of a right triangle is a diagonal of the circumscribing circle.
Proof is obvious from the penultimate corollary. Q.E.D.
Theorem: Two similar triangles have their respective sides proportional to each other and proportional to the respective radius of the circumscribing circles.
Proof follows from the Law of Sines. Q.E.D.
Definition. If triangle ABC is congruent to triangle ACB; the triangle is said to be isosceles.
Axioms: The metric of our space is homogeneous, orthogonal, and isotropic. At first, we tacitly assume that the radius of curvature, rho, is zero; that is, that the space is a plane. Presently, we will show that the only modification required is that the norm of a vector be the radius of curvature times the absolute value of the sine of the angle from the center of curvature.
Definition: A pair of lines is said to be parallel, if there exists a transverse, which is perpendicular to each of the given lines.
Observe that the winding of a quadrilateral is at most one, in magnitude.
Definition: A trapezoid is a quadrilateral, of non-zero winding, with one pair of opposite-sides parallel.
Definition: A parallelogram is a trapezoid with the other pair of opposite-sides parallel, also.
Theorems: Alternate angles of a parallelogram are equal. Opposite sides of a parallelogram are equal, in magnitude. The diagonals of a parallelogram bisect each other.
Proof: Construct a diagonal, to partition the parallelogram into two triangles, which are congruent. Q.E.D.
Definition: A rectangle is a parallelogram, with one right angle.
Corollary: The other three angles also are right angles.
Proof follows from the penultimate theorem.
Definition: A rhombus is a parallelogram, with one pair of adjacent-sides equal, in norm.
Theorem: The diagonals of a rhombus are perpendicular bisectors of each other.
Proof: Construct a diagonal, to partition the rhombus into two triangles, which are congruent. Q.E.D.
Definitions: A square is a rhombus with one right angle. A square is a rectangle with one pair of adjacent sides equal, in norm.
Corollary: These two definitions are equivalent.
Proofs are obvious. Q.E.D.
Definition: A cord is a line segment between a pair of points on a circle. Extend this cord indefinitely in one direction. The resulting ray is a secant of the circle.
Theorem: Consider a pair of intersecting circles and the mutual cord between the two points of intersection. The line segment between the centers of these circles is a perpendicular bisector of the cord. This theorem is the basis of all compass and straight-edge constructions. Furthermore, since the proof depends only upon SAS -- which is true of either the spherical or hyperbolic case -- the construction carries over to the non-plane cases. Of course, the straight-edge has to be replaced by a geodesic -- a great circle in the spherical case.
Proof: Draw the four radii to the ends of the cord. The pair of triangles formed by the segment between the centers and these radii are congruent by SSS. The pair of triangles with an apex at one of the centers and bounded by the cord are congruent by SAS. The conclusion of the theorem follows. Q.E.D.
Corollary: If the circles have the same radius, the two segments of the theorem are mutual perpendicular bisectors.
Proof is obvious. Q.E.D.
Theorem: A pair of parallel lines, which interset a circle, bounds equal arcs.
Proof: Construct a perpendicular bisector of one of the lines. Since the lines are parallel, it (or its extension) is perpendicular to the other line. Draw the four radii to the intersections of the lines with the circle. By the previous theorem, these radii, together with the perpendicular, bound the pairs of congruent triangles. By subtraction, the central angles of the two arcs of the present theorem are equal. Q.E.D.
Theorem: The R of the Law of Sines indeed is the radius of the circumscribing circle.
Proof: Solve the original triangle and call its sides a, b, and c and the angles opposite A, B, and C, respectively. Take a point P and draw a circle of radius R with P as its center. Take the central angels 2 A, 2 B, and 2 C, whose sum, of necessity, is two pi. By the Law of Sine, the lengths of the chords are a, b, and c, respectively. By SSS, this triangle is congruent to the original triangle. Q.E.D.
This proof fails in either the spherical or the hyperbolic case, Why? Because A + B + C = pi only in the plane case.
Corollary: Each central angle of the circumscribing circle is twice the corresponding vertical angle of the triangle.
Proof was demonstrated in the foregoing construction. Q.E.D.
Corollary: The angle between two lines, which intersect in the interior (exterior) of a circle, is equal to one-half of the sum (difference, respectively) of the central angles.
Proofs are obvious from the previous corollary. Hint: Draw a line parallel to one of those which passes through either of the intersections of the other line with the circle. Then employ the preceding theorem. Q.E.D.
From the theory of measure, we know that: (1) The area of a rectangle is the product of the norms of a pair of adjacent sides. This statement is true only in the differential sense, if the curvature of the space is non-zero. (2) The area of a partitioned region is equal to the sum of the areas of its partitions.
Definition: In a triangle, the line from a vertex perpendicular to the side opposite, called the base, is called the corresponding altitude.
Definition: The cardinality of a set is the collection of sets, which may be placed in one-to-one correspondence with the given set. Any one of these sets may be singled out as the representative.
Definition: The geometric mean is defined as the exponential of the arithmetic mean (provided that it exists) of the logarithms of a set of numbers.
Corollary: If the cardinality of this set is two, then the geometric mean is the square root of the product of the two numbers.
Proof: This is an obvious special case. Q.E.D.
Theorem: The altitude to its hypotenuse partitions a right triangle into two similar triangles, similar to the original triangle.
Corollary: In a right triangle, the norm of the altitude to its hypotenuse is the geometric mean of the norms of the segments
of the hypotenuse, into which this altitude has partitioned its base.
Lemma: The area of a parallelogram is the product of the norm of a base and the norm of the corresponding altitude. This statement
is true only in the differential sense, if the curvature of the space is non-zero.
Proof: Partition off a triangular end and place it at the opposite end. Q.E.D.
Theorem: The area of a triangle is equal to one-half of the product of the norm of a base and the norm of the corresponding altitude. This statement is true only in the differential sense, if the curvature of the space is non-zero.
Proof: Consider the triangle as obtained by the partitioning, into a congruent pair of triangles, of a parallelogram by one of its diagonals. Q.E.D.
Lemma: The area of a triangle is one-half of the product of the norms of two adjacent sides multiplied by the sine of the included angle
area = ||b|| ||c|| sin(A) / 2
and its cyclical permutations..This formula is true only in the differential sense, if the curvature of the space is non-zero.
Proof follows from the Law of Sines. Q.E.D.
Corollary: The area of a triangle is one fourth of the product of the norms of its sides divided by the radius of the circumscribed circle:
area = ||a|| ||b|| ||c|| / (4 R)
This formula is true only in the differential sense, if the curvature of the space is non-zero. This formula is, as it should be, a symmetric function of the sides.
Proof follows from the Law of Sine. Q.E.D.
Theorem: The area of a triangle is equal to one-half of the product of the radius of the inscribed circle and the sum of the norms of the sides.
area =r (||a|| + ||b|| + ||c||) / 2
This formula is true only in the differential sense, if the curvature of the space is non-zero.
Proof is obvious from the definition of an inscribed circle. Q.E.D.
Theorem: The radius of the inscribed circle is
r = ||a|| ||b|| ||c|| / (2 R (||a|| + ||b|| + ||c||))
Proof: Equate the preceding two formulae for the area of a triangle. Q.E.D.
We have obtained the Euclidean plane geometry, easily and concisely and without drawing any pictures. In this case, trigonometry is worth a thousand pictures.
s = (||a|| + ||b|| + ||c||) / 2
Then we have
These are for a right triangle, with the right angle being A = pi / 2.
||b|| = ||a|| sin(B) and ||c|| = ||a|| sin(C)
B + C = pi / 2
Proof: The first is immediate from the Law of Sines.
(||b|| – ||c||) / (||b|| + ||c||) = tangent((B – C) / 2) / tangent((B + C) / 2)
Proof: Henceforth, for brevity of notation, we omit the magnitude bars. Write the Law of Sines as
a / b = sin(A) / sin(B)
and add or subtract one to each side to obtain, respectively,
(a + b) / b = (sin(A) + sin(B)) / sin(B)
(a - b) / b = (sin(A) - sin(B)) / sin(B)
Then, by the use of the Sums or Differences, the quotient is
(a + b) / (a - b) = 2 sin((A + B) / 2) cos((A - B) / 2) / (2 cos((A + B) / 2) sin((A - B) / 2))
which, by the definition of the tangent, simplifies to the Law of Tangents. QED.
The Law of Half-Tangent is proven in the (more difficult) spherical case
tan(A / 2) = r / (s - ||a||).
The area of a triangle is
area^2 = s (s – ||a||) (s – ||b||) (s – ||c||) = (||a|| ||b|| ||c|| / (4 R))^2
r = area / s
r^2 = (s – ||a||) (s – ||b||) (s – ||c||) / s
R = ||a|| ||b|| ||c||| / (4 area)
R^2 = (||a|| ||b|| ||c||)^2 / (16 s (s – ||a||) (s – ||b||) (s – ||c||))
Proof: Substitute s and multiply-out the first formula for the area. Substitute the circum-radius R. We had proven the formula for the area in terms of the circum-radius R and of the in-radius r.
One of the motivations for this presentation of Trigonometry is to document -- in a readily available form -- the Spherical Trigonometry, before it vanishes from common view.
Herbert Busemann Theorem, which first was promulgated in his Geometry of Geodesics. In any space which supports the concept of a geodesic, if some two distinct geodesics emanating from a given point intersect at another point, which we call the antipodal point; then all the geodesics emanating from that given point also pass through the antipodal point. The distance along each geodesic between these two points is the same. Furthermore, there are only these two points where these geodesics intersect. The proof is obvious. We would like to point out that on a sphere, the geodesics are called great-circles and that a straight-line from the given point to its antipodal point is a diameter of the sphere and hence passes through the center of the sphere.
Since the sides of a triangle are great circles, we express them in terms of the sine of the angle, measured from the center of curvature. Thus, we observe that in the Law of Sines, the norm of a side becomes the sine.
Elementary textbooks confuse the issue by a multitude of drawings. It takes a thousand words to dispel the confusion caused by one picture. Advanced textbooks assume that the reader knows how to derive these formulae. Thus their formal derivation never is shown, except here.
For generality and easier verification of consistency, as we already did for the plane triangle, we begin with the dot product. Consider a sphere, centered on the origin. Let u, v, w be three distinct non-coplanar unit-vectors from the origin. Their extensions will intersect the sphere at three points.
Since, by definition, these vectors have a unit norm, u . u = v . v = w . w = 1. Define the angles a, b, c in the open interval (0, pi) by the implicit equations
Then R = v - v . u u is in the u-v plane. R . u = 0 proves that R is normal to u.
Likewise, S = w - w . u u is in the w-v plane. Again S . u = 0 proves that S is normal to u.
Since R . R = 1 - (v . u)^2 = 1 - (cos(c))^2 = (sin(c))^2, where we have employed the cosine identity, we have || R || = sin(c). Then the unit vectors are
We have their dot product
cos(A) = (cos(a) - cos(c) cos(b)) / (sin(c) sin(b))
where A, B, C are in the open interval (0, pi). Then rearranging, we obtain the Law of Cosines
cos(a) = sin(b) sin(c) cos(A) + cos(b) cos(c)
The Law of Cosines is not suitable for numerical calculations, because of the excessive round off errors. Rearrange once again
sin(b) sin(c) cos(A) = cos(a) - cos(b) cos(c)
Square and replace each square of a cosine, using its identity,
- (sin(b) sin(c) sin(A))^2 = 2 - ((sin(a))^2 + (sin(b))^2 + (sin(c))^2) - 2 cos(a) cos(b) cos(c).
Divide by -(sin(a) sin(b) sin(c))^2 and take its square root, to obtain the Law of Sines
sin(A) / sin(a) = sin(B) / sin(b) = sin(C) / sin(c) =
sqrt((2 cos(a) cos(b) cos(c) + ((sin(a))^2 + (sin(b))^2 + (sin(c))^2)^2 - 2)) / (sin(a) sin(b) sin(c))
whose last expression is a symmetric function in {a, b, c}.
The foregoing Law of Cosines may be written in terms of the haversine as the Law of Haversines
haversine(a) = haversine(b – c) + sin(b) sin(c) haversine(A)
where we have employed the real addition theorem for the cosine. Finally, the haversine has its own law. As it stands, this law may be employed to solve the SAS triangle. Solved for A, this law may be employed to solve the SSS triangle. In each case, the Law of Sines would be employed to obtain the rest of the elements of the triangle.
In passing, we observe that the SAS is the usual navigation problem. Let B be the current position and C be the destination. Then b is the current latitude and c is the latitude of the destination. The angle A is the difference of the longitudes. The angle B is the heading, while C is the heading on the return journey. The distance between the two points is a.
The solution of the spherical SSA triangle resembles that of the plane triangle; but is more difficult. Use the Law of Sines to obtain the angle opposite the other side, to obtain ASSA. In the plane case, the remaining angle is the supplement of the sum of these two angles. And the Law of Sines yields the remaining element -- the third side.
However, in the spherical case, the Law of Sines has to be solved for the third side. It is a quadratic equation in the square of the sine of that side. Once this side is obtained, the Law of Sines yields the third angle. There is an easier -- but still not easy -- method. Solve the Law of Sines for a in terms of, say, b and B and square the result. Also square the Law of Cosines and add them. Simplify employing the identity. Now, we have a quadratic equation in the cosine of A. It cannot get any easier; because the SSA triangle has two solutions. Of course, the easiest method is to employ one of the Napier's Analogies, then finish it with the Law of Sines. Napier -- the inventor of logarithms -- discovered a rational (in terms of trigonometric functions) solution of the aforementioned quadratic equation.
Because the spherical SSA case is more difficult, we will consider the special case of an isosceles triangle..Let -- by definition -- side c equal side b. Then by the Law of Sines, angle C is equal to angle B. Then the square-root of the Law of Haversines becomes the Isosceles Law of Haversine
sin(a / 2) = sin(b) sin(A / 2) when c = b
Dual. If we consider u, v, w to be the vertices; then a, b, c are the sides and A, B, C are the angles of the spherical triangle. (A side is the central angle of the great-circle arc it subtends. Thus the length of the side a (it is the arc) is rho times a, where rho is the radius of the sphere. The angle opposite A is the dihedral angle between the adjacent planes. This angle is that of the intersection of the planes of these adjacent sides with any plane (for instance, the plane tangent to the sphere, at the vertex) normal to the vector from the center of the sphere to the vertex of the triangle. The angle A also is equal to the angle between the normals to the planes of the two adjacent sides, b and c.) On the other hand, if we consider u, v, w to be the normals to the planes; then a, b, c are the angles and A, B, C are the sides. Because of this duality, we did not commit ourselves to either interpretation at the outset of this derivation. These triangles are called polar; their corresponding elements are supplementary. Thus the trigonometric functions have to be adjusted accordingly. Specifically, the mapping is a --> pi - A and A --> pi - a and cyclical permutations thereof. The resulting equations are said to be the dual of the original.
Observe that in the spherical case -- by the duality -- we actually can solve the AAA triangle, which we could not solve in the plane case.
The limit of rho times the sine of a side, as rho, the radius of curvature, increases without bound, is the side, itself. The limit of the cosine of a side, as rho increases without bound, is one. The limit of any angle is the angle itself. Thus, gratifyingly, the spherical Law of Sines goes over to the plane Law of Sines. The Law of Cosine is the only one which does not have a limit; but it may be obtained from the limit of the Law of Haversines. As a consequence, each of the following equations also provides the corresponding plane equation, provided that the sides are involved homogeneously exclusively as their sines, haversines (each of which counts as two), or tangents. However, except for the Law of Sines itself, where the duality is trivial, we loose the duality in the plane equations.
Since the proof of the spherical Law of Tangents is identical to that of the correspoinding plane Law of Tangents, we state the spherical Law of Tangents without proof.
tan((b - c) / 2) / tan((b + c) / 2) = tan((B - C) / 2) / tan((B + C) / 2)
While everything up to this point has been my own, and the proofs of both the plane and spherical Law of Cosines and Law of Sines strictly my own, the balance of the Spherical Trigonometry largely is based upon Essentials of Trigonometry with Applications by David Raymond Curtiss and Elton James Moulton, D C. Heath and Company, N.Y., 1942. Why such an old book? Spherical Trigonometry has become not fashionable.
s = (a + b + c) / 2
S = (A + B + C) / 2
r^2 = sin(s - a) sin(s - b) sin(s - c) / sin(s)
R^2 = - cos(S) / (cos(S - A) cos(S - B) cos(S - C))
Does anybody know, are r and R related to the radius of the in-circle and the radius of the circum-circle? Perhaps being the tangent of these respective radii? Or how would one find these radii? A vector-analytical derivation of these radii is provided. Pray, see the recently-added in-radius and circum-radius for the formulae of the radii and their derivation employing Spherical Trigonometry.
Evaluate from the Law of Cosines
(1 - cos(A)) (sin(b) sin(c)) = sin(b) sin(c) + cos(b) cos(c) - cos(a)
Using the addition theorem for the cosine
= cos(b - c) - cos(a)
And using the sums or differences
= 2 sin((a + b - c) / 2) sin((a + c - b) / 2) = 2 sin(s - c) sin(s - b)
Again, using theaddition theorem for the cosine
(1 + cos(A)) (sin(b) sin(c)) = sin(b) sin(c) - cos(b) cos(c) + cos(a)
= - (cos(b + c) - cos(a))
And using the sums or differences and that the sine is an odd function
= sin((a + b + c) / 2) sin((b + c - a) / 2) = 2 sin(s) sin(s - a)
Their quotient is
(1 - cos(A)) / (1 + cos(A)) = sin(s - c) sin(s - b) / (sin(s) sin(s - a))
Using the half-angle formula for the tangent, the square root of the foregoing equation is the Law of Half-Tangent
tan(A / 2) = +- r / sin(s - a)
The remaining equations have to be derived from the foregoing laws; but it is a bear to do so.
I believe that I have seen these formulae thus ascribed; but, please correct me. These are for a right triangle, with the right angle being A = pi / 2. They are the ten combinations of the five remaining parts of the triangle, taken three at a time.
sin(b) = sin(a) sin(B) and sin(c) = sin(a) sin(C)
cos(B) = cos(b) sin(C) and cos(C) = cos(c) sin(B)
cos(a) = cos(b) cos(c)
sin(b) = tan(c) ctan(C) and sin(c) = tan(b) ctan(B)
cos(B) = tan(c) ctan(a) and cos(C) = tan(b) ctan(a)
cos(a) = ctan(B) ctan(C)
The third is the Law of Cosines. The first are the Laws of Sines. The sixth is the dual of the Law of Cosines. Hint: take the side a = pi / 2, then find the dual -- which will have the angle A = pi / 2. The second is a cyclical permutation of the dual of the Law of Cosines. Hint: cyclically permute the dual, then set A = pi / 2. The sixth divided by the third yields the product of the two equations designated four. The quotient of the right by the left of the fifth multiplied by the quotient of the right by the left of the first yields the quotient of the left by the right of the two equations designated four. Thus establishing both of the fourth. The left of the second divided by the product of the right of the first by the third yields the left of the fifth. The right of the second divided by the product of the left of the first by the third yields the right of the fifth. Thus establishing both of the fifth. Furthermore, each of a pair may be obtained from the other by a renaming. There proofs are my own.
tan((a – b) / 2) / tan(c / 2) = sin((A – B) / 2) / sin((A + B) / 2)
tan((a + b) / 2) / tan(c / 2) = cos((A – B) / 2) / cos((A + B) / 2)
tan((A – B) / 2) / ctan(C / 2) = sin((a – b) / 2) / sin((a + b) / 2)
tan((A + B) / 2) / ctan(C / 2) = cos((a – b) / 2) / cos((a + b) / 2)
The Scottish mathematician John Napier (1550 - 1617) probably invented these formulae; but, a reference says, "it seems likely that the English mathematician Henry Briggs [(1561 - 1630)] had a share in these.". They also are known as Delambre's Analogies, who seems to have popularized them. Here is a reference to Jean Baptiste Joseph Delambre (1749 - 1822). It is not clear who actually originated them.
Observe that these equations each contain five -- one more than the minimum requirement of four -- elements of the triangle. Hence, they are suited particularly for checking the solution of a triangle. However, the solution of either the SSA or AAS triangle by the Law of Sines yields an ASSA triangle, which is easiest solvable by the Napier's Analogies.
The second analogy may be obtined from the first by the division by the Law of Tangents. The fourth from the third by the multiplication by the Law of Tangents. The third is the dual of the first. Also, the fourth is the dual of the second. Hence, the only puzzle is the derivation of the first
Proof of the first Napier's Analogy: Take the expression
(tan(A / 2) - tan(B / 2) / (tan(A / 2) + tan(B / 2))
Use the definition of the tangent and clear the secondary denominators. Then the addition theorem for the sines yields the right-hand side of the first Napier's Analogy. Evaluate the foregoing expression employing the Law of Half tangent and clear the secondary denominators. The sine is an odd function, as stated in its definition. Employ the addition theorem for the sine. Then the definition of the tangent yields the left-hand side of the first Napier's Analogy. Observe that the tangents from the Law of Half Tangent yields the sines, while the sines from that Law yield the tangents. Tricky?
Congruence of spherical triangles is the same as that of plane triangles, which please see.
In general, area is a measure, one of whose axioms is that the measure of the union (of necessity measurable) of disjoint measurable sets is the sum of their measures. We compute the cumulative exterior angle as the Lebesque-Stieltjes integral of the exterior angle while traversing around a given region, with the region on our left-side and remaining on the outside of the surface if the radius of curvature is positive (spherical) or the inside if negative (hyperbolic). "Left" is defined as the non-dominant side of the majority of the Homo sapiens sapiens, absent the possibility of a Mathematical definition.
Consider a capital 'T'. Obviously the exterior angle is the same (actually zero) as one either traverses from the upper-right to the middle, down to the bottom, turns right to return to the top and goes to the upper-left; or straight from the upper-right to the upper-left. Then, for a contiguous simply-connected region (fissures and bridges being permitted), the
spherical excess = E = 2 pi - cumulative exterior angle
adds in the same manner as the area. Hence they are proportional. Since the spherical excess of a hemi-sphere is 2 pi and its area is 2 pi rho^2 [sorry, this is not a textbook of the Calculus -- you will have to find this area yourself], the proportion is
area = E rho^2
In particular, for a spherical triangle
E = A + B + C – pi = area / rho^2
Without proof (any suggestions? Pray, see the recently-added proof -- first, second--of both versions of this theorem.), we provide l'Huilier's (Swiss mathematician Simon Antoine Jean l'Huilier (24.4.1750-28.3.1840)) Theorem:
tan(E / 4) = sqrt((tan(s / 2)) (tan((s - a) / 2)) (tan((s - b) / 2)) (tan((s - c) / 2)))
As a check, the limit, as the curvature rho approaches zero, of any spherical equation is a valid plane-equation. For instance, the foregoing equation becomes Heron’s formula for the area of a plane triangle.
area = sqrt(s (s - a) (s - b) (s - c)) = a b c / ( 4 R)
Since the area obviously is equal to r s, r the radius of the inscribed circle is
r = a b c / (4 R s)
On the other hand, a strictly negative curvature yields a hyperbolic space. As it should, each of the spherical excess, l'Huilier's theorem, Heron's formula, and the radii of the circumscribed and inscribed circles is a symmetric function of the sides (or of the angles) of the triangle. Furthermore, for certain specific triangles, we can verify l'Huilier's theorem. For an isosceles triangle, with two right-angles, it is easy to do so. Could an ingenuous proof be built around such an approach?
An interesting problem of spherical trigonometry is that of finding the area of a spherical cap of either a cone or pyramid, with its apex at the center of the sphere.
A pair of intersecting curves has one point in common. A pair of tangent curves has two coincident points in common. A pair of osculating curves has three coincident points in common.
At a given point of a given curve, construct the osculating circle. The center of this circle is the center of curvature. The radius of this circle is the radius of curvature, whose reciprocal is the curvature. The radius of curvature often it taken as a vector, emanating at the center of curvature, towards the given point on the curve. Then, the loci of these centers of curvature constitute the involute, with the original curve being called the evolute. A variant on this concept employs a circle of fixed radius, for instance, in the study of the path of a reamer or of a cam.
The cross product of two vectors is defined as a pseudo-vector, with the following properties: (1) Its magnitude is the product of their norms multiplied by the sine of the included angle. (2) Its direction is normal to both vectors. (3) Its chirallity is the same as that of the space.
The cross product is skew-symmetric
b x a = – (a x b)
Its norm (magnitude) is the area of the parallelogram. The box product a . (b x c) = - (a x b) . c is the volume (or its negative) of the parallelepipedon. Thus, the box product is zero iff (= if and only if) the three vectors are coplanar. The cross product of a vector and a pseudo-vector is, back again, a vector
a x (b x c) = b (a c) – c (a b)
A curve has the local coordinates: (1) The unit tangent; i.e., a unit vector in the direction of the velocity. (2) The normal, which is a unit vector towards the center of curvature. (3) The binormal, which is a unit vector in the direction of the cross product of the preceding two vectors.
Torsion is defined as the rate of change of the projection on the normal-binormal plane of the direction of the normal, with respect to arc length.
Let us consider a regular-polygon of n sides. By symmetry, its in- and circum- centers coincide. Let a be the circum-radius and b the in-radius. Construct the two radii to a pair of adjacent vertices. This is an isosceles triangle, with the vertex at the center of the polygon. The vertical angle is (2 pi / n). Then, its altitude is b
b = a cos(pi / n).
Its base is
base = 2 a sin(pi / n).
Its area is half the product
b * base / 2 = (a cos(pi / n)) * (2 a sin(pi / n)) / 2 = = a^2 sin(pi / n) cos(pi / n).
The area of the polygon is n times this, namely
area = n a^2 sin(pi / n) cos(pi / n) = n a^2 sin(2 pi / n) / 2,
where we have employed the double-angle formula for the sine. It terms of b, the area is
area = n b^2 sin(pi / n) / cos(pi / n) = n b^2 tan(pi / n),
where we have employed the definition of the tangent. We can turn the problem around. Consider a fixed circle of radius r. Its area is
area = pi r^2.
Construct a regular polygon on the outside. Its area is
area = n r^2 tan(pi / n).
Also, construct a regular polygon on the inside. its area is
area = n r^2 sin(2 pi / n) / 2.
Then, we have the inequality
n r^2 tan(pi / n) > pi r^2 > n r^2 sin(2 pi / n) / 2.
Divide through by r^2, to obtain
n tan(pi / n) > pi > n sin(2 pi / n) / 2.
Let x be the reciprocal of n. Then we have
tan(pi x) / x > pi > sin(2 pi x) / (2 x).
By l'Hospital's rule, the limit, as x approaches zero from the right, is
pi (sec(pi x))^2 > pi > 2 pi cos(2 pi x) / 2 = pi cos(2 pi x),
where we have employed the derivatives of the tangent and sine functions. Its limit is
pi >= pi >= pi.
Hence, we have shown that indeed these formulae may be employed to compute pi. For n=10000000, we have
| n | out = n tan(pi / n) | in = n sin(2 pi / n) / 2 | out - pi | pi - in |
| 10000000 | 3.141592653589900 | 3.141592653589590 | 1.03473E-13 | 2.06501E-13 |
Thus, we have obtained pi as
pi = 3.1415926535898
where for the last place, I cheated. We only know that it is between 6 and 9. Historically, this was the first method for computing pi. There are much better methods known. At present, we know pi to about 2 * 10^11 decimal places.
Embed a surface of one or more dimensions in a flat space of minimal dimensions. For example, consider an ellipsoid, embedded in a flat space of one higher dimension.
A geodesic locally is the shortest path.
Construct a family of geodesic curves drawn through a given point on a surface. Along the tangent of each curve, construct a vector of length equal to the curvature of that curve. The result is an ellipsoid of curvature.
Solve the eigenvalue-eigenvector problem. The eigenvectors -- which may be shown to be mutually orthogonal -- are the principal directions. By convention, the eigenvectors are taken to have a unit norm. The eigenvalues are the principal curvatures. Their respective products are the principal axes of the ellipsoid of curvature. The geometric mean of these eigenvalues is the curvature of the surface, rho. This is the formal definition of the curvature, which we have been employing throughout the previous discussion of Elementary Geometry.
Alternatively, along the tangent of each curve, construct a vector of length equal to the norm of the radius of curvature. The eigenvectors will have the same directions. The eigenvalues are the principal radii of curvature. Their respective products are the principal axes of the ellipsoid of radii of curvature. This ellipsoid osculates the given surface and coincides with the given surface if it had been an ellipsoid.
A set of equal eigenvalues -- and that of their corresponding eigenvectors -- is said to be degenerate. If the original surface had been a sphere, all of the eigenvalues would have been the same and equal to the curvature or radius of curvature, respectively, of the surface. Thus a sphere is fully degenerate. The radius of curvature of a sphere is equal to its radius. The most common example is a two-dimensional sphere embedded in three-dimensional flat space.
Each principal direction may be either spherical (positive curvature), flat (zero curvature), or hyperbolic (negative curvature). These individual curvatures need not have the same sign. However, I do not know what implications a mixed-sign curvature has upon the space curvature rho, the spherical excess E, and the area enclosed by a closed curve (a triangle, for instance). If anybody knows and explains to me (or can provide me with a reference to an in-print English-language textbook, which answers this question), I will be grateful. Algebraic topology -- employing denumerable groups -- deals with the structure of spaces which are not simply connected.
In the early 1920's, the problem was posed of finding the necessary and sufficient conditions upon a topology for it to be metrizable. In the early 1950's, the problem was solved that the topology should be measurable. But only the existence theorem was obtained. For a detailed statement of this theorem, see Kelley (sp?) General Topology.
In my dissertation research, a Treatise on the Theory of Information, commencing in 1960, and to be published here: If an additional probability measure is imposed upon the topology
Lebesque-Stieltjes integral of p(x) dx = 1
and if the Information Theoretic integral exists and is bounded
LS integral of p(x) ln(p(x)) dx,
then it is shown that the metrization is unique and that the intrinsic geometry is elementary; If the topology is compact, just locally compact, or not even locally compact; the geometry is spherical, plane, or hyperbolic, respectively. A Riemmanian geometry, if one is constructed, would be an extrinsic geometry.
The inner (that is dot) product is defined in terms of the Information-Theoretic integrals and it is asserted that from this dot product, the elementary geometry may be developed. Well, this page shows how to do so.
For each law, we give the spherical (a, b, c are the sides; A, B, C are the angles), its dual for the polar triangle (A. B. C are the angles; a, b, c are the sides), and the plane (a, b, c are the norms of the sides; A, B, C are the angles) version. Where notable, the plane version of the dual is appended after an italicized "and". For the hyperbolic, replace the circular trigonometric functions of the sides (a, b, c) by the corresponding hyperbolic trigonometric functions, in the first two items in each of the following laws In each case, if the expression is not symmetric in the sides or angles; then, each cyclical permutation also is implied. A computer-friendly version of this table is provided in the computer code page.
For a right triangle, employ the Mollweide’s Formulae: spherical or plane.
For brevity, for the spherical formulae (for hyperbolic, the second formula would employ the hyperbolic sine, instead of the circular sine), let
and, for the plane formulae, r is the radius of the inscribed circle; R of the circumscribed. Except as this limit, I do not know of any geometrical significance of the spherical r or R. If someone knows, please inform me.
In the following laws, the individual elements transform as follows (element, dual, limit):
Law of Cosines (observe that the plane version is not the limit of the spherical version). Furthermore, because of their excessive round-off errors; these are not suitable for numerical calculation.
Law of Haversines Since the haversine function has fallen into obscurity, we incorporate its definition (in terms of the sine function), as an alternative.
Law of Haversines for isosceles triangle. That is, c=b and C=B.
Law of Sines Observe that the short-form of the law of sines is self-dual.
Law of Tangents Observe that the law of tangents is self-dual.
Law of Half-Tangent
Napier's analogies
Area
Algorithm for the solution of a triangle, stated in order of decreasing ease of application. By "solution of a triangle" we mean the process of ascertaining the values of any of the unknown six elements of a triangle. Any situation with leads to a unique solution also is a condition for the congruence of a pair of triangles. We specify the elements of the triangle, in order either counter-clockwise xor clockwise -- each triangle (of the pair) in the same direction. The symbols are: S = side, A = angle; the lower-case indicates that we only need to know its size compared to a multiple of pi/2. (Where we specify "sine", we mean the sort form, unless otherwise indicated.)
The dual is available only in the spherical-trigonometry case -- it is not available in the plane-trigonometry case. Hence, because it is the duals that are required to solve the AAA case; the AAA case cannot be solved in plane-trigonometry case.
The Law of Sines, when employed to find the value of an element of a triangle, has an inherent ambiguity -- it cannot distinguish between an acute and an obtuse angle (or, in the spherical case, side, as well). Some independent method has to be employed to resolve this ambiguity, for instance, the spherical excess. In the plane-trigonometry case, the spherical excess has to be exactly zero. Hence, it is apparent that the angle opposite the longest side is the only one which might need to be assigned as obtuse. In the spherical-trigonometry case, the spherical excess has to be strictly positive; but, there is no a priori method of knowing how large --up to 4 pi -- it might be. Since there might be more than one obtuse -- or even reflex -- angle, the Law of Sines is ill-suited for use in a computer-program to solve for an unknown element.
However, the Law of Haversines, when employed to find an additional angle (or the dual Law to find an additional side) has an ambiguity of its own. You have to find a square-root, namely = 2 * Asin(sqrt(...)). Which sign should you use? The short-form of the Law of Sines comes to the rescue! Already, there will have been one pair of opposite elements known. Thus, one of the ratios will have a known sign. Hence, take such a sign for the square-root as to make each of the new ratios become with the same sign. Please see Where are We for a worked-out example of spherical trigonometry.
Copyright © 1997,8,9, 2000,1,2,3,4,9 R. I. 'Scibor-Marchocki last modified Sunday 07-th June 2009. Three broken links repaired on Tuesday 30-th August 2005. Repairing the syntax of each of the internal hyperlinks on Thursday 01-st September 2005. Another broken link deleted on Monday 05-th September 2005. Three broken links removed Sunday 15-th October 2006, Introduced hyperlinks to the new file entitled Spherical Trigonometry Addendum on Sunday 07-th June 2009. Last modified on Friday 19-VI (June)-2009.