User:Jacobolus/HalfTan


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In mathematics, the half-tangent is a parameter used for representing rotations, angles, or points on a circle. (It is also called various other names; see .) The half-tangent $h$ is the tangent of half the angle measure $\theta$  and it is the stereographic projection of a unit-magnitude complex number $z = \cos\theta + i\sin\theta$  onto the imaginary $$i$$-axis,



h = \tan\tfrac12\theta = -i \frac{z - 1}{z + 1} = \frac{\sin\theta}{1 + \cos\theta}. $$

In the inverse direction, angle measures, unit complex numbers, and unit vectors can be written in terms of the half-tangent,


 * $$\begin{align}

\theta &= 2\arctan h = -i\log z, \\[5mu] z &= \frac{1 + hi}{1 - hi} = \exp \theta i, \\[8mu] \left( \cos\theta,\, \sin\theta \right) &= \left( \frac{1-h^2}{1 + h^2},\, \frac{2h}{1 + h^2} \right). \end{align} $$

The half-tangent $$h$$ is a convenient representation for explicit computation because the transcendental circular functions (sine, cosine, &c.) of the angle measure $$\theta$$ become rational functions of $$h,$$ requiring only elementary arithmetic to compute. The half-tangent $$h$$ is a single real number, unlike the unit complex number $$z$$ which is comprised of two real coordinates.

The hyperbolic half-tangent is the analogous parameter used for representing hyperbolic angles, rotations (Lorentz boosts), points on a hyperbola, or multiplicative scalars. The hyperbolic half-tangent $h$ is the hyperbolic tangent of half the hyperbolic angle measure $\psi$  and it is the stereographic projection of a unit-magnitude hyperbolic number $\zeta = \cosh\psi + j\sinh\psi$  (where $j^2 = 1$) onto the imaginary $$j$$-axis, and also a Cayley transform of a real scalar $$w.$$



h = \tanh\tfrac12\psi = j \frac{\zeta - 1}{\zeta + 1} = \frac{\sinh\psi}{1 + \cosh\psi} = \frac{w - 1}{w + 1}. $$

In the inverse direction, hyperbolic angle measures, unit hyperbolic numbers, and hyperbolic unit vectors can be written in terms of the half-tangent,


 * $$\begin{align}

\psi &= 2\operatorname{artanh} h = \log w = j\log \zeta, \\[5mu] \zeta &= \frac{1 + hj}{1 - hj} = \exp \psi j, \qquad w = \frac{1 + h}{1 - h} = \exp \psi, \\[8mu] \left( \cosh\psi,\, \sinh\psi \right) &= \left( \frac{1+h^2}{1-h^2},\, \frac{2h}{1 - h^2} \right). \end{align} $$

Terminology
The half-tangent is used widely in mathematics, science, and engineering, but does not have a universally established name. In the 16th–19th century in Neo-Latin it typically appeared in descriptive phrases such as tangens dimidiae ("tangent of half") or tangens semi-summae ("tangent of the semi-sum"). This was often translated into similar phrases in other languages but also sometimes given the dedicated name halbe tangente in German; demi-tangente in French; and especially half-tangent or semi-tangent in English, widely adopted in mathematical instruments used for astronomy and navigation starting in the mid 17th century, an era in Britain of high demand for trained navigators. By the late 19th century the name semi-tangent even appeared in general-purpose dictionaries, though it has since fallen out of currency.

Many historical and most modern sources refer to the half-tangent purely descriptively, e.g. as the tangent of half the angle, Examples:              or use it in mathematical expressions such as $$\tan\tfrac12 \theta$$ or $$\tan(\theta/2)$$ without an explicit name. In the context of integral calculus, substituting $$t = \tan \tfrac12x$$ is sometimes called the half-tangent substitution, is sometimes misnamed the Weierstrass substitution, This substitution was used by Leonhard Euler to evaluate the integral $\int dx / (a + b\cos x)$ in his 1768 integral calculus textbook, was described as a general method by Adrien-Marie Legendre in 1817, and was in wide use by the middle of the 19th century. In 1966, William Eberlein misattributed it to Karl Weierstrass (1815–1897); two decades later, James Stewart did the same in his popular calculus textbook. Later authors, citing Stewart, have sometimes referred to this as the Weierstrass substitution. E342, Translation by Ian Bruce. in Russian is called the universal trigonometric substitution, or is sometimes just called $t$-substitution, but modern sources more often give it no name at all or use a descriptive compound modifier as in tangent half-angle substitution. A similar compound modifier is often used for the half-angle tangent formula in trigonometry. Number theorists call the half-tangent the rational parameter for a point on the unit circle. Sometimes, notably in directional statistics, it is called the stereographic projection. Other names include half-slope and tan-half-angle. The name half-tangent is also still used, especially in kinematics. Examples: This article will consistently adopt the name half-tangent for convenience.

The trigonometric half-tangent should not be confused with half-tangent meaning a ray tangent to a curve, half of a tangent line.

The hyperbolic half-tangent is rarely named, and is usually just used symbolically as e.g. $$\tanh\tfrac12\psi$$ or $$\tanh(\psi/2),$$ or sometimes descriptively called something like the hyperbolic tangent of half the angle. It has been specifically named the Lorentzian stereographic representation, or the menhir. In the context of higher-dimensional hyperbolic spaces, the (unsigned) half-tangent of a geodesic arc length is sometimes called the pseudo-chordal distance, the pseudo-hyperbolic distance, or the hyperbolic length. The substitution $$t = \tanh\tfrac12 x$$ has been called the hyperbolic tangent half-argument substitution, and the half-argument identity for the hyperbolic tangent has been called the  hyperbolic formula for the tangent half-angle and the half-tanh relation. This article will adopt the name (hyperbolic) half-tangent for convenience.

Tangent addition
When real numbers $$h_1$$ and $$h_2$$ are taken to be half-tangents representing circular or hyperbolic rotations, those rotations can be composed using the circular ($h_1 \oplus h_2$) or hyperbolic ($h_1 \boxplus h_2$) tangent addition operations, respectively.

Circular tangent addition


The composition of two planar rotations $$R_1$$ and $$R_2$$ is the new rotation $$R = R_1 \circ R_2$$ which results from rotating first by $$R_2$$ and then by $$R_1$$ (planar rotation is commutative, $$R_1 \circ R_2 = R_2 \circ R_1$$). When rotations are represented as complex numbers, the composition operation is complex multiplication, $$z = z_1z_2$$. When rotations are represented as angle measures, composition is addition $$\theta = \theta_1 + \theta_2$$. When rotations are represented as matrices, composition is matrix multiplication $$M = M_1M_2$$. When rotations are represented as half-tangents, composition is a circular tangent addition operation $$h = h_1 \oplus h_2$$ defined by


 * $$\begin{align}

h_1 \oplus h_2 &:= \frac{h_1 + h_2}{1 - h_1h_2} = \tan\tfrac12(\theta_1 + \theta_2) = i \frac{1 - z_1z_2}{1 + z_1z_2} \\[5mu] &\phantom{:}= \tan\left(\arctan h_1 + \arctan h_2\right). \end{align}$$

The tangent addition identity on which this operation is based was proved by Jakob Hermann in 1706 and independently by several other mathematicians shortly afterward.

If the two half-tangents are written as quotients,


 * $$\frac{p_1}{q_1} \oplus \frac{p_2}{q_2} = \frac{p_1q_2 + p_2q_1}{q_1q_2 - p_1p_2},$$

the relation to complex multiplication becomes clear:


 * $$(q_1 + ip_1)(q_2 + ip_2) = (q_1q_2 - p_1p_2) + i(p_1q_2 + p_2q_1).$$

As with multiplication and addition, this operation has an inverse $$\ominus$$, corresponding to division $$z_1/z_2$$ and subtraction $$\theta_1 - \theta_2\colon$$


 * $$\begin{align}

h_1 \ominus h_2 &:= \frac{h_1 - h_2}{1 + h_1h_2} = \tan\tfrac12(\theta_1 - \theta_2) = i \frac{z_2 - z_1}{z_2 + z_1} \\[5mu] &\phantom{:}= h_1 \oplus (-h_2) = \tan\left(\arctan h_1 - \arctan h_2\right). \end{align}$$

This tangent difference operation yields the half-tangent of $$h_1$$ after the circle has been rotated so that $$h_2$$ is at the origin.

Hyperbolic tangent addition
The operation $$\oplus$$ is the circular analog of a hyperbolic tangent addition operation $$\boxplus,$$



h_1 \boxplus h_2
 * = \frac{h_1 + h_2}{1 + h_1h_2}

= \tanh(\operatorname{artanh} h_1 + \operatorname{artanh} h_2), $$

with inverse operation $$\boxminus,$$



h_1 \boxminus h_2 := h_1 \boxplus (-h_2) = \frac{h_1 - h_2}{1 - h_1h_2}. $$

Because in special relativity the operation $$\boxplus$$ is the composition law for parallel velocities (in a coordinate system with natural units where the speed of light $c = 1$ ) it is sometimes called Einstein addition.

The circular and hyperbolic operations are related by


 * $$\begin{align}

(h_1 \oplus h_2)i &= (h_1i) \boxplus (h_2i), & (h_1 \ominus h_2)i &= (h_1i) \boxminus (h_2i), \\[5mu] (h_1 \boxplus h_2)i &= (h_1i) \oplus (h_2i), & (h_1 \boxminus h_2)i &= (h_1i) \ominus (h_2i), \end{align}$$

where $i$ is the imaginary unit.

Point at infinity
Half-tangents are naturally values on the projectively extended real line $\R\cup\{\infty\}$. A half-turn rotation is represented by the complex number $-1.$ As this is the center of the stereographic projection, it is projected to $$\infty = -\infty,$$ the point at infinity. If half-tangents are interpreted as ratios, this is the ratio $$1:0.$$

The tangent sum is well defined for $\infty\colon$


 * $$\begin{align}

\infty \oplus h &= \frac{h + \infty}{1 - h\infty} = -\frac{1}{h}, \\[10mu] \infty \boxplus h &= \frac{h + \infty}{1 + h\infty} = \!\!\phantom{-}\frac{1}{h}, \\[10mu] \infty \oplus \infty &= \infty \boxplus \infty = 0, \\[6mu] 0 \oplus \infty &= 0 \boxplus \infty = \infty. \end{align}$$

Singularities
Corresponding to the invariance of the speed of light, and analogous to ordinary multiplication by $$\infty$$ or $$0,$$ the values $\pm1$ absorb other arguments to the hyperbolic tangent sum $$\boxplus.$$ For any value $$h$$ such that $h^2 \neq 1$  (including $h = \infty$ ),


 * $$\begin{align}

h \boxplus 1 = h \boxminus (-1) &= \frac{h + 1}{1 + h} = 1, \\[6mu] h \boxminus 1 = h \boxplus (-1) &= \frac{h - 1}{1 - h} = -1. \end{align}$$

The exceptional sum $1 \boxplus (-1)$ and the exceptional differences $1 \boxminus 1$  and $(-1) \boxminus (-1)$  are undefined.

For complex-valued arguments, the circular tangent sum $$\oplus$$ has singularities at $$\pm i,$$ the imaginary units. For any value $$h$$ such that $h^2 \neq -1,$


 * $$\begin{align}

h \oplus i = h \ominus (-i) &= \frac{h + i}{1 - ih} = i, \\[6mu] h \oplus (-i) = h \ominus i &= \frac{h - i}{1 + ih} = -i. \end{align}$$

The sum $i \oplus (-i)$ and differences $i \ominus i$  and $(-i) \ominus (-i)$  are undefined.

Commutative group structure
The projectively extended real line $\widehat \R := \R \cup \{\infty \}$ is a commutative group under $$\oplus.$$ It is associative, $$(h_1 \oplus h_2) \oplus h_3 = h_1 \oplus (h_2 \oplus h_3) =: h_1 \oplus h_2 \oplus h_3.$$ It is commutative, $$h_1 \oplus h_2 = h_2 \oplus h_1.$$ It has identity element $0,$  $$h \oplus 0 = h.$$ Each element $h$  has an inverse $-h,$  $$h \oplus (-h) = 0.$$ This group is isomorphic to the circle group (the complex unit circle $$z\bar{z} = 1$$ under multiplication), the group of angle measures on the periodic interval $$(-\pi, \pi]$$ under addition modulo $$2\pi,$$ and the special orthogonal group $\operatorname{SO}(2)$  of planar Euclidean rotation matrices.

The interval $$(-1, 1)$$ is a commutative group under $$\boxplus.$$ It is associative, $$(h_1 \boxplus h_2) \boxplus h_3 = h_1 \boxplus (h_2 \boxplus h_3) =: h_1 \boxplus h_2 \boxplus h_3.$$ It is commutative, $$h_1 \boxplus h_2 = h_2 \boxplus h_1.$$ It has identity element $0,$ $$h \boxplus 0 = h.$$ Each element $h$  has an inverse $-h,$  $$h \boxplus (-h) = 0.$$  This group is isomorphic to the right branch of the split-complex unit hyperbola $$z\bar{z} = 1$$ under multiplication, to the group of hyperbolic-function arguments on the real line under addition, and to one connected component of the indefinite special orthogonal group $\operatorname{SO}^{+}(1,1)$  of rotation matrices in the pseudo-Euclidean plane with signature $(1, 1).$

$\widehat \R \smallsetminus \{\pm1\},$ the projectively extended real line punctured at the points $$\pm1,$$ is also a commutative group under $$\boxplus$$, isomorphic to multiplication on both branches of the split-complex unit hyperbola and to $\operatorname{SO}(1,1)$. This group has two connected components: the interval $$(-1, 1)$$ and the "exterior interval" $\widehat \R \smallsetminus [-1, 1].$

The projectively extended line restricted to rational numbers $\widehat \Q := \Q \cup \{\infty \},$ has analogous structure under $$\oplus$$: it is isomorphic to the group of rational points on the complex unit circle under multiplication. Likewise $\widehat \Q$ forms structures under $$\boxplus$$ isomorphic to the rational points on the split-complex unit hyperbola under multiplication.

Multiple sum
The tangent sum of several half-tangents is the ratio of sums of alternating elementary symmetric polynomials,


 * $$\begin{aligned}

&h_1 \oplus h_2 = \frac{h_1 + h_2}{1 - h_1h_2}, \qquad h_1 \oplus h_2 \oplus h_3 = \frac{h_1 + h_2 + h_3 - h_1h_2h_3}{1 - h_1h_2 - h_1h_3 - h_2h_3}, \quad \ldots \end{aligned}$$

The multiple hyperbolic tangent sum $$\boxplus$$ is the above with each minus sign replaced by a plus sign.

In general, letting $$e_m$$ be the $π$th elementary symmetric polynomial,
 * $$\begin{align}

h_1 \boxplus \cdots \boxplus h_k &= \frac {\sum\limits_{m \text{ odd}} e_m(h_1,..., h_k) } {\sum\limits_{m \text{ even}} e_m(h_1,..., h_k) }, \\[10mu]

h_1 \oplus \cdots \oplus h_k &= \frac1i \left(h_1i \boxplus \cdots \boxplus h_ki\right). \end{align}$$

Iterated tangent sum


Analogous to integer multiplication $$k\theta$$ of angle measures or exponentiation $$z^k$$ of complex numbers, we can define an iterated tangent sum operation,


 * $$h^{\oplus k} := \underbrace{h \oplus h \oplus \dots \oplus h}_{k \text{ times}}, \quad

h^{\boxplus k} := \underbrace{h \boxplus h \boxplus \dots \boxplus h}_{k \text{ times}}, $$

using superscript notation rather than an inline symbol such as $$\otimes$$ because unlike multiplication the operation is neither commutative nor associative but inherits properties from complex exponentiation.

Then,


 * $$\begin{align}

h^{\oplus 0} &= \dfrac{0}{1}, \quad h^{\oplus 1} = \dfrac{h}{1}, \quad h^{\oplus 2} = \dfrac{2h}{1 - h^2}, \quad h^{\oplus 3} = \dfrac{3h - h^3}{1 - 3h^2}, \\[10mu] h^{\oplus 4} &= \dfrac{4h - 4h^3}{1 - 6h^2 + h^4}, \quad h^{\oplus 5} = \dfrac{5h - 10h^3 + h^5}{1 - 10h^2 + 5h^4}, \quad \ldots \end{align}$$

The iterated hyperbolic tangent sum $$h^{\boxplus k}$$ is the above with each minus replaced by a plus.

For a general integer $k$, the coefficients of the polynomials in numerator and denominator are alternating binomial coefficients, the quotient


 * $$\begin{align}

h^{\boxplus k} &= \frac{(1 + h)^k - (1 + h)^{-k}}{(1 + h)^k + (1 + h)^{-k}} = \frac {\sum\limits_{m \text{ odd }} \!\Bigl(\begin{matrix}k \\[-6mu] ~\!\!m~\!\! \end{matrix}\Bigr) h^m} {\sum\limits_{m \text{ even}} \!\Bigl(\begin{matrix}k \\[-6mu] ~\!\!m~\!\! \end{matrix}\Bigr) h^m}, \\[5mu] h^{\oplus k} &= -i\frac{(1 + hi)^k - (1 + hi)^{-k}}{(1 + hi)^k + (1 + hi)^{-k}} = -i \frac {\sum\limits_{m \text{ odd }} \!\Bigl(\begin{matrix}k \\[-6mu] ~\!\!m~\!\! \end{matrix}\Bigr) (hi)^m} {\sum\limits_{m \text{ even}} \!\Bigl(\begin{matrix}k \\[-6mu] ~\!\!m~\!\! \end{matrix}\Bigr) (hi)^m} \\ &= -i (hi)^{\boxplus k}. \end{align}$$

The $m$th roots of these functions, values for which $h^{\oplus k} = 0,$ are the stereographic projection of the roots of unity.

The iterated tangent sum operation satisfies identities analogous to exponentiation of complex numbers:


 * $$\begin{align}

h_1^{\oplus k} \oplus h_2^{\oplus k} &= (h_1 \oplus h_2)^{\oplus k}, \\[10mu] h^{\oplus k_1} \oplus h^{\oplus k_2} &= h^{\oplus k_1 + k_2}, \\[10mu] \bigl(h^{\oplus k_1\!}\bigr)^{\oplus k_2} &= h^{\oplus k_1k_2}, \\[10mu] \end{align}$$

and likewise for the iterated hyperbolic tangent sum.

The iterated tangent sum can be generalized to an arbitrary (real or complex) "exponent" $w\colon$


 * $$\begin{align}

h^{\boxplus w} &:= \tanh(w \operatorname{artanh} h), \\[6mu] h^{\oplus w} &:= \tan(w \arctan h). \end{align}$$

Square root


In particular, an analog of the square root complex number $$\sqrt{z} = z^{1/2}$$ or the half angle measure $$\tfrac12\theta$$ is the "quarter-tangent" $$\sqrt[\oplus]{h} := h^{\oplus 1/2} = \tan\tfrac14\theta$$ (for $ satisfying $$\sqrt[\oplus]{h} \oplus \sqrt[\oplus]{h} = h\colon$$



\sqrt[\oplus]{h} = \frac{h}{1 + \sqrt{1 + h^2}} = \frac{-1 + \sqrt{1+h^2}}{h}. $$

The other branch of the square root represents the antipodal rotation,



\infty \oplus \sqrt[\oplus]{h} = -\frac1{\sqrt[\oplus]{h} } = \frac{h}{1 - \sqrt{1 + h^2}} = \frac{-1 - \sqrt{1+h^2}}{h}. $$

For the hyperbolic tangent sum, the analog of square root, $$\sqrt[\boxplus]{h}$$ satisfying $$\sqrt[\boxplus]{h} \boxplus \sqrt[\boxplus]{h} = h$$ is



\sqrt[\boxplus]{h} = \frac{h}{1+\sqrt{1-h^2}} = \frac{1-\sqrt{1-h^{2}}}{h}. $$

The other branch of the square root represents a point on the other branch of the hyperbola,



\infty \boxplus \sqrt[\boxplus]{h} = \frac1{\sqrt[\boxplus]{h} } = \frac{h}{1-\sqrt{1-h^2}} = \frac{1+\sqrt{1-h^{2}}}{h}. $$

Notice that when $$|h| > 1,$$ $\sqrt[\boxplus]{h}$ is not a real number.

Circular–hyperbolic identities
The three operations $\boxplus,$ $\oplus,$  and $+$  can be related to each-other via quotient identities,


 * $$\begin{align}

h_1 \oplus h_2\, = \frac{h_1 \boxplus h_2}{1 \ominus h_1h_2} &= \frac{h_1 + h_2}{1 - h_1h_2},

& h_1 \boxplus h_2\, = \ \ \,\,\frac{h_1 \oplus h_2}{1 \oplus h_1h_2} \ \ \ \, &= \frac{h_1 + h_2}{1 + h_1h_2}, \\[10mu]

\frac{h_1 \oplus h_2}{h_1 \boxplus h_2} = \ \frac{h_1 \boxminus h_2}{h_1 \ominus h_2}\ &= 1 \oplus h_1h_2,

& \frac{h_1 \boxplus h_2}{h_1 \boxminus h_2} = \ \frac{(h_1 \oplus h_2)^{\boxplus2}}{(h_1 \ominus h_2)^{\boxplus2}}\ &= 1 \oplus \frac{h_2^{\oplus2}}{h_1^{\oplus2}},\\[10mu]

\frac{h_1 \oplus h_2}{h_1 \boxminus h_2} = \ \frac{h_1 \boxplus h_2}{h_1 \ominus h_2}\ &= 1 \oplus \frac{h_2}{h_1},

& \frac{h_1 \oplus h_2}{h_1 \ominus h_2} = \ \frac{(h_1 \boxplus h_2)^{\oplus2}}{(h_1 \boxminus h_2)^{\oplus2}}\ &= 1 \oplus \frac{h_2^{\boxplus2}}{h_1^{\boxplus2}},

\end{align}$$

and product identities,


 * $$\begin{align}

(h_1 \oplus h_2)(h_1 \boxplus h_2) &= h_1^2\oplus h_2^2 + (h_1h_2)^{\oplus2},\\[6mu] (h_1 \ominus h_2)(h_1 \boxminus h_2) &= h_1^2\oplus h_2^2 - (h_1h_2)^{\oplus2},\\[6mu] (h_1 \oplus h_2)(h_1 \boxminus h_2) &= (h_1^2\boxminus h_2^2)(1 \oplus h_1h_2), \\[6mu] (h_1 \ominus h_2)(h_1 \boxplus h_2) &= (h_1^2\boxminus h_2^2)(1 \ominus h_1h_2), \\[6mu] (h_1 \oplus h_2)(h_1 \ominus h_2) &= (h_1 \boxplus h_2)(h_1 \boxminus h_2) = h_1^2\boxminus h_2^2. \end{align}$$

The last of these is the tangent-sum analog of the difference of two squares.

Tangent addition series
Analogous to ordinary addition series, it is possible to add infinitely many quantities using the tangent addition operation. We can write


 * $$\bigoplus_{k=1}^\infty h_k = h_1 \oplus h_2 \oplus h_3 \oplus \cdots.$$

Where such a series converges, it can be written as an ordinary series of arctangents, equal up to some integer multiple of $$2\pi$$:


 * $$\begin{align}

\bigoplus_{k=1}^\infty h_k &= \tan\tfrac12\biggl(\sum_{k=1}^\infty 2\arctan h_k \biggr) \\[5mu] &= \tan\tfrac12(\,2\arctan h_1 + 2\arctan h_2 + 2\arctan h_3 + \cdots ). \end{align}$$

Antipodal, inverse, supplementary, and complementary rotations
In the figure, the original rotation $$R$$ and its representations as a point $z$ on the circle and a half-tangent $h$  are drawn in red.

Antipodes
Two rotations $$R_1$$ and $$R_2$$ are said to be diametrically opposite or antipodal if they are separated by a half-turn: as complex numbers $$z_1/z_2 = -1,$$ as angle measures $\theta_1 - \theta_2 \equiv \pi \pmod{2\pi},$ or as half-tangents $h_1 \ominus h_2 = \infty \iff h_1h_2 = -1.$

Given a rotation $$R,$$ the antipodal rotation can be represented as a complex number by $$-z,$$ as an angle measure by $$\pi + \theta,$$ or as a half-tangent by $$\infty \oplus h = -1/h.$$ (Green in the figure.)

Inverses
Two rotations $$R_1$$ and $$R_2$$ are said to be inverse if they compose to the identity rotation: as complex numbers $$z_1z_2 = 1,$$ as angle measures $\theta_1 + \theta_2 \equiv 0 \pmod{2\pi},$ or as half-tangents $h_1 \oplus h_2 = 0 \iff h_1 + h_2 = 0,$  i.e. inverse half-tangents are additive inverses

Given a rotation $$R,$$ the inverse rotation can be represented as a complex number by $$1/z,$$ as an angle measure by $$-\theta,$$ or as a half-tangent by $$0 \ominus h = -h.$$ (Orange in the figure.)

Supplements
Two rotations are said to be supplementary if they compose to make a half turn: as complex numbers, $$z_1z_2 = -1,$$ as angle measures $\theta_1 + \theta_2 \equiv \pi \pmod{2\pi},$ or as half-tangents $$h_1 \oplus h_2 = \infty \iff h_1h_2 = 1,$$ i.e. supplementary half-tangents are reciprocals.

Given a rotation $$R,$$ the supplementary rotation can be represented as a complex number by $$-1/z,$$ as an angle measure by $$\pi - \theta,$$ or as a half-tangent by $$\infty \ominus h = 1/h.$$ (Purple in the figure.)

For half-tangents $h_1$ and $h_2$  and their respective supplements $1/h_1$  and $1/h_2,$


 * $$\frac1{h_1} \oplus \frac1{h_2} = -(h_1 \oplus h_2), \quad \frac1{h_1} \boxplus \frac1{h_2} = h_1 \boxplus h_2.$$

Complements
Two rotations are said to be complementary if they compose to make a quarter turn: as complex numbers, $$z_1z_2 = i,$$ as angle measures $\theta_1 + \theta_2 \equiv \tfrac12\pi \pmod{2\pi},$ or as half-tangents $$h_1 \oplus h_2 = 1 \iff h_1 + h_2 + h_1h_2 = 1.$$

Given a rotation $$R,$$ the complementary rotation can be represented as a complex number by $$i/z,$$ as an angle measure by $$\tfrac12\pi - \theta,$$ or as a half-tangent by $$1 \ominus h = (1 - h)/(1 + h).$$ (Blue in the figure.)

Two rotations are each antipodal to the other's complement if they compose to make a negative quarter turn: as complex numbers, $$z_1z_2 = -i,$$ as angle measures $\theta_1 + \theta_2 \equiv -\tfrac12\pi \pmod{2\pi},$ or as half-tangents $$h_1 \oplus h_2 = -1 \iff h_1 + h_2 - h_1h_2 = -1.$$

Given a rotation $$R,$$ the complement of its antipodal rotation can be represented as a complex number by $$-i/z,$$ as an angle measure by $$-\tfrac12\pi - \theta,$$ or as a half-tangent by $$\ominus1 \ominus h = (h + 1)/(h - 1).$$

Quarter-turned rotations
Let $$Q$$ be the quarter turn represented as the complex number $$i,$$ the angle measure $$\tfrac12\pi,$$ or the half-tangent $$1.$$

Given a rotation $$R$$ the supplement of its complement is the quarter-turned rotation $$Q \circ R$$ represented as complex number by $$zi,$$ as an angle measure by $$\theta + \tfrac12\pi,$$ or as a half-tangent by $$h \oplus 1 = (1 + h)/(1 - h).$$

The complement of its supplement is the quarter-turned rotation $$Q^{-1} \circ R$$ represented as complex number by $$-zi,$$ as an angle measure by $$\theta - \tfrac12\pi,$$ or as a half-tangent by $$h \ominus 1 = (h - 1)/(h + 1).$$

Circular distances


Between two rotations or two points on a circle, there are several related concepts of distance or separation. In the following, $$h_1 = \tan\tfrac12\theta_1,$$ $$z_1 = \exp \theta_1 i,$$ and likewise for $$h_2,$$ $$\theta_2,$$ $$z_2.$$

Intrinsic distance
The distance intrinsic to the circle is proportional to arc length and is called angular distance, circular distance, or angle measure. For points represented as angle measures this is the ordinary difference. For complex numbers it is the absolute value of the argument of the quotient. For half-tangents this is the absolute value of twice the arctangent of the stereographic difference. Here we measure angle in radians.


 * $$\begin{align}d(h_1, h_2)

= 2\arctan|h_1 \ominus h_2| = \left|\operatorname{rem}(\theta_1 - \theta_2, 2\pi)\right| = \left|\arg\frac{z_1}{z_2}\right|, \end{align}$$

where $\operatorname{rem}(x, q) = x - q \bigl\lfloor\tfrac12 + x/q\bigr\rfloor$ is a rounding modulo operation.

Stereographic distance
The half-tangent or stereographic distance between two points on the circle is proportional to the distance after the circle has been stereographically projected through the point antipodal to one of them. This can also be thought of as the half-tangent of one point after the circle has been rotated so the other is at the origin.


 * $$\begin{align}d_s(h_1, h_2)

= |h_1 \ominus h_2| = \left|\frac{h_1 - h_2}{1 + h_1h_2}\right| = \left|\tan\tfrac12(\theta_1 - \theta_2)\right| = \left|\frac{z_1 - z_2}{z_1 + z_2}\right|. \end{align}$$

This distance function is not a metric under the conventional definition because it does not satisfy the triangle inequality under addition (making it a semimetric). However, any three points do satisfy a triangle inequality under tangent sum,


 * $$|h_1 \ominus h_2| \leq |h_1 \ominus h_3| \oplus |h_2 \ominus h_3|,$$

with equality whenever $$h_3$$ lies on the shorter arc between $$h_1$$ and $$h_2.$$

Chordal distance
The chordal distance between two rotations or points on the circle is proportional to the length of a chord, the extrinsic Euclidean distance when the circle is embedded in the Euclidean plane (or complex plane). Here we normalize these distances to a circle of unit diameter (sometimes chordal distances are doubled, representing distances for a unit-radius circle).


 * $$\begin{align}d_c(h_1, h_2)

= \frac{|h_1 - h_2|}{\sqrt{1 + h_1^2}\sqrt{1 + h_2^2}} = \left|\sin\tfrac12(\theta_1 - \theta_2)\right| = \tfrac12\left|z_1 - z_2\right|. \end{align}$$

Also see § Half-angle identities below.

Normal distance
The normal distance between two points on the circle, historically called the versed sine (versine) and corresponding to the sagitta [arrow] of twice the arc, is the distance between the projections of the two points onto the diameter through one of them. Here we normalize it to a unit-diameter circle (haversine).


 * $$\begin{align}d_n(h_1, h_2)

= \frac{(h_1 - h_2)^2}{\bigl(1 + h_1^2\bigr)\bigl(1 + h_2^2\bigr)} = \tfrac12 \bigl(1 - \cos(\theta_1 - \theta_2) \bigr) = \tfrac14\left|z_1 - z_2\right|^2. \end{align}$$

Relation between distances
Letting $$\kappa = d(h_1, h_2),$$ $$\chi = d_c(h_1, h_2),$$ $$\sigma = d_s(h_1, h_2),$$ $$\nu = d_n(h_1, h_2),$$


 * $$\begin{align}

\kappa &= 2\arctan \sigma = 2\arcsin \chi = 2\arccos(1 - \nu) \\[4mu] \sigma &= \tan\tfrac12 \kappa = \frac{\chi}{\sqrt{1 - \chi^2}} = \sqrt{\frac{\nu}{1 - \nu}}, \\[8mu] \chi &= \sin\tfrac12 \kappa = \frac{\sigma}{\sqrt{1 + \sigma^2}} = \sqrt{\nu}, \\[8mu] \nu &= \tfrac12(1 - \cos\kappa) = \frac{\sigma^2}{1 + \sigma^2} = \chi^2. \end{align}$$

Differential geometry


The projectively extended real line is a model for the circle under the differential relation


 * $$d\theta = \frac{2\,dh}{1 + h^2},$$

where $\theta$ is angle measure on the circle and $h = \tan \tfrac12\theta.$

To differentiate an arbitrary function of half-tangent $f(h)$ uniformly with respect to the circle,


 * $$\frac{d}{d\theta}f(h) = \frac{dh}{d\theta}\frac{d}{dh}f(h) = \tfrac12(1 + h^2)\frac{d}{dh}f(h).$$

To integrate an arbitrary function of half-tangent $f(h)$ uniformly with respect to the circle,


 * $$\int f(h)\, d\theta = \int \frac{2f(h)}{1 + h^2}dh.$$

In particular, the derivative of the identity function $f(h)=h$ is


 * $$\frac{d}{d\theta} h = \tfrac12(1 + h^2),$$

and its antiderivative is


 * $$\int h\,d\theta = \int \frac{2h}{1 + h^2}dh = \log(1 + h^2) + c,$$

where $$\log$$ is the natural logarithm.

The signed angle measure (along the shortest arc) between two half-tangents $h_1 = \tan\tfrac12\theta_1$ and $h_2 = \tan\tfrac12\theta_2$  is the integral of the constant function $f(h)=1$ ,


 * $$\begin{align}

\int\limits_{0}^{h_2 \ominus h_1} \!\!\! \frac{2\,dh}{1 + h^2} &= 2\arctan(h_2 \ominus h_1) = 2\arctan\left(\frac{h_2 - h_1}{1 + h_1h_2}\right) \\[6mu] &\equiv \theta_2 - \theta_1 \pmod{2\pi}. \end{align}$$

The circular distance between two points on the circle is thus the (unsigned)


 * $$\begin{align}

d(h_1, h_2) = 2\arctan|h_2 \ominus h_1|. \end{align}$$

Cayley transform
The Cayley transform $E$ is the half-tangent analog of the exponential function $\exp$ :


 * $$E(h) := \frac{1 + h}{1 - h} = 1 \oplus h, \qquad E(hi) = 1 \oplus hi = \frac{i \boxminus h}i.$$

When $h = \tan\tfrac12\theta$ is a real-valued circular half-tangent, the transform $E(h) = 1 \oplus h$  is a quarter-turn rotation. The transform of $hi$ is $z = E(hi) = \exp\theta,$  a unit-magnitude complex number. The transform $E$ cyclically permutes


 * $$\begin{alignat}{5}

&{\phantom-0}    &&\to     && 1              &&\to    && {\phantom-\infty} &&\to    && {-1}                     &&\to    && {\phantom{-}0}, \\[5mu] &{\phantom-h}    &&\to     && 1 \oplus h     &&\to    && {-1/h}\ \         &&\to    && {\phantom{-}h \ominus 1} &&\to    && {\phantom{-}h}, \\[5mu] &{-h}            &&\to     && 1 \ominus h\ \ &&\to    && {\phantom-1/h}    &&\to    && {-(1 \oplus h)}\ \       &&\to    && {-h},\\[5mu] &{\phantom-hi}\ \ &&\to    && z              &&\to    && {\phantom-i/h}    &&\to    && {-1/z}                   &&\to    && {\phantom{-}hi},\\[5mu] &{-hi}           &&\to\quad && 1/z            &&\to\ \ && {-i/h}\           &&\to\ \ && {-z}                     &&\to\ \ && {-hi}. \end{alignat}$$

The transform $$E$$ fixes the imaginary unit: $$E(i) = i,$$ $$E(-i) = -i.$$

The reciprocal transform $$h \mapsto 1/E(h)$$ applied to half-tangents takes the complement,


 * $$\frac1{E(h)} = \frac{1 - h}{1 + h} = 1 \ominus h = E(-h).$$

This is an involution, $$1 \big/ E\bigl(1 / E(h)\bigr) = 1 \ominus (1 \ominus h) = h$$, exchanging


 * $$\begin{align}

0 \ &\leftrightarrow \ 1, & \infty \ &\leftrightarrow \ {-1}, & i\ &\leftrightarrow \ {-i}, \\[5mu] h \ &\leftrightarrow \ 1 \ominus h, &{-h} \ &\leftrightarrow \ 1 \oplus h, & {1/h} \ &\leftrightarrow \ h \ominus 1, & {-1/h} \ &\leftrightarrow \ {-(1 \oplus h)}, \\[5mu] hi \ &\leftrightarrow \ 1/z, & {-hi} \ &\leftrightarrow \ z, & 1/hi \ &\leftrightarrow \ {-1/z}, & {-1/hi} \ &\leftrightarrow \ {-z}, \end{align}$$

with fixed points $\sqrt[\oplus]{1} = \sqrt2 - 1$ and $\sqrt[\oplus]{1} \oplus \infty = -\sqrt2 - 1.$

The inverse transform $E^{-1}$ (analogous to the natural logarithm) and its reciprocal are


 * $$\begin{alignat}{4}

E^{-1}(h) &= \frac{h - 1}{1 + h} &&= \ \ \ \, h \ominus 1 &&= -\frac1{E(h)} &&= E(-1/h), \\[8mu] \frac1{E^{-1}(h)} &= \frac{1 + h}{h - 1} &&= - (1 \oplus h) &&= \,{-E(h)} &&= E(1/h) = E^{-1}(-h). \end{alignat}$$

$E^{-1}$ permutes $ 0 \to {-1} \to \infty \to 1 \to 0,$  while its reciprocal reflects $ 0 \leftrightarrow -1,$  $\infty \leftrightarrow 1.$

Analogous to the exponential function, the transform $$E$$ converts tangent addition of the arguments to multiplication. Unlike the exponential function, $$E$$ also converts multiplication of the arguments to tangent addition.


 * $$\begin{align}

E(h_1 \boxplus h_2) &= E(h_1)\,E(h_2), & E(h_1 \boxminus h_2) &= E(h_1) \big/ E(h_2), \\[6mu] E\bigl((h_1 \oplus h_2)i\bigr) &= E(h_1i)\,E(h_2i), & E\bigl((h_1 \ominus h_2)i\bigr) &= E(h_1i) \big/ E(h_2i), \\[6mu] E(h_1h_2) &= E(h_1) \boxplus E(h_2), & E(h_1/h_2) &= E(h_1) \boxminus E(h_2), \\[6mu] E(h_1h_2)i &= E(h_1)i \oplus E(h_2)i, & E(h_1/h_2)i &= E(h_1)i \ominus E(h_2)i. \end{align}$$

These identities above also hold if $$E$$ is replaced by $$E^{-1}.$$

Similar identities can be written in terms of the tangent addition operations, without explicitly naming $E$ :

Quarter-turned and complement product identities
From a half-tangent $$h = \tan\tfrac12\theta,$$ the complement half-tangent $$E(-h) = 1 \ominus h = \sec\theta - \tan\theta$$ and quarter-turned half-tangent $$E(h) = 1 \oplus h = \sec\theta + \tan\theta$$ appear often. The two are supplements,


 * $$(1 \ominus h)(1 \oplus h) = 1.$$

The product (or quotient) of quarter-turned or complement half-tangents can be rewritten as a quarter-turned or complement hyperbolic tangent sum (or difference):


 * $$\begin{align}

(1 \oplus h_1)(1 \oplus h_2) &= 1 \oplus (h_1 \boxplus h_2), \\[8mu] (1 \ominus h_1)(1 \ominus h_2) &= 1 \ominus (h_1 \boxplus h_2), \\[5mu] (1 \oplus h_1)(1 \ominus h_2) &= \frac{1 \oplus h_1}{1 \oplus h_2} = 1 \oplus (h_1 \boxminus h_2) \\[5mu] &= \frac{1 \ominus h_2}{1 \ominus h_1} = 1 \ominus (h_2 \boxminus h_1). \end{align}$$

The above identity can be applied recursively to a quotient of arbitrary factors,


 * $$\begin{align}

&(1 \oplus p_1)(1 \oplus p_2) \cdots (1 \oplus p_k)(1 \ominus q_1)(1 \ominus q_2) \cdots (1 \ominus q_m) \\[5mu] &\quad =\frac{(1 \oplus p_1)(1 \oplus p_2) \cdots (1 \oplus p_k)} {(1 \oplus q_1)(1 \oplus q_2) \cdots (1 \oplus q_m)} = 1 \oplus (p_1 \boxplus \cdots \boxplus p_k \boxminus q_1 \boxminus \cdots \boxminus q_m), \\[5mu] &\quad = \frac{(1 \ominus q_1)(1 \ominus q_2) \cdots (1 \ominus q_m)} {(1 \ominus p_1)(1 \ominus p_2) \cdots (1 \ominus p_k)} = 1 \ominus (q_1 \boxplus \cdots \boxplus q_m \boxminus p_1 \boxminus \cdots \boxminus p_k). \end{align}$$

The complement of a product or quotient can be factored as the hyperbolic tangent sum or difference of complements:


 * $$\begin{align}

1 \ominus h_1h_2 &= (1 \ominus h_1) \boxplus (1 \ominus h_2), \\[8mu] 1 \ominus \frac{h_1}{h_2} &= (1 \ominus h_1) \boxminus (1 \ominus h_2), \\[8mu] \end{align}$$

This identity can also be applied recursively to a quotient of arbitrary factors,


 * $$\begin{align}

1 \ominus \frac{p_1p_2\cdots p_k}{q_1q_2\cdots q_m} &= (1 \ominus p_1) \boxplus (1 \ominus p_2) \boxplus \cdots \boxplus (1 \ominus p_k) \\ &\qquad \boxminus (1 \ominus q_1) \boxminus (1 \ominus q_2) \boxminus \cdots \boxminus (1 \ominus q_m). \end{align}$$

In particular, if the half-tangents are repeated this turns the complement of a power into an iterated hyperbolic sum of complements,


 * $$\begin{align}

1 \ominus h^k &= (1 \ominus h)^{\boxplus k}. \end{align}$$

Circular functions


The circular functions (a.k.a. trigonometric functions) of angle measure $$\theta$$ can alternately be written as rational functions of the half-tangent $$h = \tan \tfrac12\theta.$$ calls these functions the stereographic sine, stereographic cosine, and stereographic tangent, which we will denote $S,$  $C,$  and $T$  respectively, these are:


 * $$\begin{align}

\sin\theta &= \frac{2h}{1+h^2}  =: S(h), & \csc\theta &= \frac{1+h^2}{2h} = \frac1{S(h)},    \\[10mu] \cos\theta &= \frac{1-h^2}{1+h^2} =: C(h), & \sec\theta &= \frac{1+h^2}{1-h^2} = \frac1{C(h)}, \\[10mu] \tan\theta &= \frac{2h}{1-h^2} =: T(h), & \cot\theta &= \frac{1-h^2}{2h} = \frac1{T(h)}. \end{align}$$

The tangent and sine of a half-tangent are also respectively its circular and hyperbolic tangent squares,


 * $$\begin{align}

\tan \theta = T(h) &= h \oplus h = \, h^{\oplus2}, \\[6mu] \sin \theta = S(h) &= h \boxplus h = \, h^{\boxplus2}, \\[6mu] \cos \theta = C(h) &= \frac{S(h)}{T(h)} = \frac{h^{\boxplus 2}}{h^{\oplus 2}} = 1 \ominus h^2. \end{align}$$

The unit complex number $z = \exp \theta i = 1 \oplus hi = E(hi)$ is also a rational function of $$h,$$


 * $$\begin{align}

z = \exp\theta i &= \frac{1-h^2}{1+h^2} + \frac{2h}{1+h^2}i = \frac{1 + hi}{1 - hi} = 1 \oplus hi \\[8mu] &= C(h) + S(h)i = E(hi). \end{align}$$

In terms of $z,$ the sine and cosine are related to the Joukowsky transform:


 * $$\begin{alignat}{5}

S(h) &= \ \frac{z - z^{-1}}{2i} &&= i\frac{1 - z^2}{2z} &&= \frac{i}{T(z)} &&= \frac{i}{z^{\oplus2}}, \\[8mu] C(h) &= \ \frac{z + z^{-1}}2  &&= \frac{1 + z^2}{2z} &&= \frac{1}{S(z)} &&= \frac{1}{z^{\boxplus2}}, \\[8mu] T(h) &= \frac1i\, \frac{z - z^{-1}}{z + z^{-1}}     &&= i\frac{1 - z^2}{1 + z^2} &&= i C(z) &&= i(1 \ominus z^2). \end{alignat}$$

Less common circular functions chord ($t = tan(x/2)$), versine ($sin x$), vercosine ($cos x$), and exsecant ($y = x^{⊕k}$) can also be written in terms of the half-tangent:


 * $$\begin{align}

\operatorname{crd}\theta  &:= 2\sin\tfrac12\theta = \frac{2h}{\sqrt{1+h^2}} = 2S\bigl(\sqrt[\oplus]{h}\bigr), \quad |\theta| < \pi    \\[10mu] \operatorname{vers}\theta &:= 1 - \cos\theta = \frac{2h^2}{1+h^2} =  h\,S(h), \\[10mu] \operatorname{vercos}\theta &:= 1 + \cos\theta = \frac{2}{1+h^2} =  \frac{S(h)}h = \frac{d\theta}{dh}, \\[10mu] \operatorname{exsec}\theta &:= \sec\theta - 1 = \frac{2h^2}{1-h^2} = h\,T(h). \end{align}$$

Derivatives and antiderivatives
Just as for the circular functions,


 * $$\begin{align}

\frac{d}{d\theta}S(h) &= C(h), & \int S(h)\,d\theta &= -C(h) + c, \\[10mu] \frac{d}{d\theta}C(h) &= -S(h), & \int C(h)\,d\theta &= S(h) + c, \\[10mu] \frac{d}{d\theta}T(h) &= \frac1{C(h)^2}, & \int T(h)\,d\theta &= -\log|C(h)| + c, \\[10mu] \frac{d}{d\theta}\frac1{S(h)} &= -\frac{C(h)}{S(h)^2}, & \int \frac{1}{S(h)}d\theta &= \log|h| + c, \\[10mu] \frac{d}{d\theta}\frac1{C(h)} &= \frac{S(h)}{C(h)^2}, & \int \frac{1}{C(h)}d\theta &= \log|1 \oplus h| + c, \\[10mu] \frac{d}{d\theta}\frac1{T(h)} &= -\frac{1}{S(h)^2}, & \int \frac{1}{T(h)}d\theta &= \log |S(h)| + c. \end{align}$$

where $d\theta = 2\,dh / \bigl(1 + h^2\bigr)$ ; see above.

Identities
For each trigonometric identity relating the circular functions of angle measure, an analogous identity relates these stereographic circular functions of half-tangent.

Pythagorean identity
The Pythagorean identity does not depend on the parametrization of the circle,


 * $$\begin{align}

C(h)^2 + S(h)^2 &= 1. \end{align}$$

Reflections
As with sine and cosine, the function $S$ is odd while $C$  is even so taking the inverse half-tangent ($0 \ominus h = -h$) flips the sign of $S$  but not $C$, while taking the supplementary half-tangent ($\infty \ominus h = 1/h$) flips the sign of $C$  but not $S$ , and taking the complementary half-tangent ($1 \ominus h$) swaps $S$  and $C,$


 * $$\begin{align}

S(-h) &= -S(h), & S(1/h) &= \phantom{-}S(h), & S(1 \ominus h) &= C(h), \\[6mu] C(-h) &= \phantom{-}C(h), & C(1/h) &= -C(h), & C(1 \ominus h) &= S(h), \\[6mu] T(-h) &= -T(h), & T(1/h) &= -T(h), & T(1 \ominus h) &= 1 / T(h). \end{align}$$

A half-tangent's complement is the supplement of the quarter-turned half-tangent, and also its negatively quarter-turned supplement,



1 \ominus h = \frac1{1 \oplus h} = \frac1h \ominus 1. $$

Quarter and half turns
Shifts via the tangent sum operation correspond to shifts of sine and cosine via angle addition,


 * $$\begin{align}

S(h \oplus 1) &= \phantom-C(h), & S(h \oplus \infty) &= -S(h), & S(h \ominus 1) &= -C(h), \\[6mu] C(h \oplus 1) &= -S(h), & C(h \oplus \infty) &= -C(h), & C(h \ominus 1) &= \phantom-S(h), \\[6mu] T(h \oplus 1) &= -1 / T(h), & T(h \oplus \infty) &= \phantom-T(h), & T(h \ominus 1) &= -1 / T(h). \end{align}$$

The tangent and secant are also related to the complement and quarter-turned half-tangents,


 * $$\begin{alignat}{2}

T(h) &= \ h^{\oplus2} &&= \tfrac12\bigl((1 \oplus h) - (1 \ominus h)\bigr) = \frac{1}{T(1 \ominus h)} = \frac{-1}{T(1 \oplus h)}, \\[6mu] \frac1{C(h)} &= \frac{h^{\oplus2}}{h^{\boxplus2}} &&= \tfrac12\bigl((1 \oplus h) + (1 \ominus h)\bigr) = \frac{1}{S(1 \ominus h)} = \frac{1}{S(1 \oplus h)}. \end{alignat}$$

Combining both sides above,


 * $$\begin{align}

1 \oplus h &= \frac1{C(h)} + T(h), \\[6mu] 1 \ominus h &= \frac1{C(h)} - T(h). \end{align}$$

Tangent sum identities
Analogous to trigonometric angle sum identities,


 * $$\begin{align}

h_1 \oplus h_2 &= \frac{h_1 + h_2}{1 - h_1h_2} = \frac{S(h_1) + S(h_2)}{C(h_1) + C(h_2)} = \frac{C(h_2) - C(h_1)}{S(h_1) - S(h_2)}, \\[10mu]

h_1 \ominus h_2 &= \frac{h_1 - h_2}{1 + h_1h_2} = \frac{S(h_1) - S(h_2)}{C(h_1) + C(h_2)} = \frac{C(h_2) - C(h_1)}{S(h_1) + S(h_2)}, \\[10mu]

(h_1 \ominus h_2)(h_1 \oplus h_2) &= \frac{C(h_2) - C(h_1)}{C(h_1) + C(h_2)} = 1 \ominus \frac{C(h_1)}{C(h_2)} = h_1^2\boxminus h_2^2, \\[10mu]

\frac{h_1 \ominus h_2}{h_1 \oplus h_2} &= \frac{S(h_1) - S(h_2)}{S(h_1) + S(h_2)} = 1 \ominus \frac{S(h_2)}{S(h_1)} = \frac{T(h_1 \boxminus h_2)}{T(h_1 \boxplus h_2)}, \\[10mu] \end{align}$$

Taking sines or cosines of tangent sums:


 * $$\begin{align}

S(h_1 \oplus h_2) &= \ \ \, \frac{2(h_1 + h_2)(1 - h_1h_2)}{(1 + h_1^2)(1 + h_2^2)} \quad = S(h_1)C(h_2) + C(h_1)S(h_2), \\[10mu]

C(h_1 \oplus h_2) &= \frac{(1 - h_1h_2)^2 - (h_1 + h_2)^2}{(1 + h_1^2)(1 + h_2^2)} = C(h_1)C(h_2) - S(h_1)S(h_2), \\[10mu]

\frac{S(h_1 \ominus h_2)}{S(h_1 \oplus h_2)} &= \frac{T(h_1) - T(h_2)}{T(h_1) + T(h_2)} = 1 \ominus \frac{T(h_2)}{T(h_1)} = \frac{h_1 \boxminus h_2}{h_1 \boxplus h_2}, \\[10mu]

\frac{C(h_1 \ominus h_2)}{C(h_1 \oplus h_2)} &= \frac{1 - T(h_1)T(h_2)}{1 + T(h_1)T(h_2)} = 1 \oplus T(h_1)T(h_2)

\end{align}$$

Product identities
By rearranging the tangent sum identities above, we obtain identities for the product of sines and cosines:


 * $$\begin{align}

S(h_1)S(h_2) &= \tfrac12\bigl(C(h_1 \ominus h_2) - C(h_1 \oplus h_2)\bigr), \\[5mu] C(h_1)C(h_2) &= \tfrac12\bigl(C(h_1 \ominus h_2) + C(h_1 \oplus h_2)\bigr), \\[5mu] S(h_1)C(h_2) &= \tfrac12\bigl(S(h_1 \ominus h_2) + S(h_1 \oplus h_2)\bigr), \\[5mu] C(h_1)S(h_2) &= \tfrac12\bigl(S(h_2 \ominus h_1) + S(h_1 \oplus h_2)\bigr). \\[5mu] \end{align}$$

Combining these, the products or quotients of tangents are


 * $$\begin{align}

T(h_1)T(h_2) &= \frac {C(h_1 \ominus h_2) - C(h_1 \oplus h_2)} {C(h_1 \ominus h_2) + C(h_1 \oplus h_2)}, \\[8mu] \frac{T(h_1)}{T(h_2)} &= \frac {S(h_1 \ominus h_2) + S(h_1 \oplus h_2)} {S(h_2 \ominus h_1) + S(h_1 \oplus h_2)}. \end{align}$$

The sines and tangents of a product or quotient are


 * $$\begin{align}

S(h_1 h_2) &= (h_1h_2)^{\boxplus2} = \frac{2h_1h_2}{1 + h_1^2h_2^2}, & S(h_1/h_2) &= (h_1/h_2)^{\boxplus2} = \frac{2h_1h_2}{h_1^2 + h_2^2}, \\[10mu] T(h_1 h_2) &= (h_1h_2)^{\oplus2} = \frac{2h_1h_2}{1 - h_1^2h_2^2}, & T(h_1/h_2) &= (h_1/h_2)^{\oplus2} = \frac{2h_1h_2}{h_2^2 - h_1^2}, \\[10mu] \frac{S(h_1 h_2)}{S(h_1/h_2)} &= h_1^2 \boxplus h_2^2, & \frac{T(h_1 h_2)}{T(h_1/h_2)} &= h_2^2 \boxminus h_1^2, \\[10mu] \frac{T(h_1 h_2)}{S(h_1/h_2)} &= h_1^2 \oplus h_2^2, & \frac{S(h_1 h_2)}{T(h_1/h_2)} &= h_2^2 \ominus h_1^2, \end{align}$$

The cosecants and cotangents of products therefore satisfy


 * $$\begin{align}

\frac1{S(h_1 h_2)} &= \frac{1}{S(h_1)\,S(h_2)} + \frac{1}{T(h_1)\,T(h_2)}, \\[10mu] \frac1{S(h_1/h_2)} &= \frac{1}{S(h_1)\,S(h_2)} - \frac{1}{T(h_1)\,T(h_2)} = \frac1{S(h_2/h_1)}, \\[10mu] \frac1{T(h_1 h_2)} &= \frac{1}{T(h_1)\,S(h_2)} + \frac{1}{S(h_1)\,T(h_2)}, \\[10mu] \frac1{T(h_1/h_2)} &= \frac{1}{T(h_1)\,S(h_2)} - \frac{1}{S(h_1)\,T(h_2)} = \frac{-1}{T(h_2/h_1)}, \\[10mu] \end{align}$$

and the products of cosecants and cotangents satisfy


 * $$\begin{align}

\frac{1}{S(h_1)\,S(h_2)} &= \frac12\left(\frac1{S(h_1 h_2)} + \frac1{S(h_1/h_2)}\right), \\[6mu] \frac{1}{T(h_1)\,T(h_2)} &= \frac12\left(\frac1{S(h_1 h_2)} - \frac1{S(h_1/h_2)}\right), \\[8mu] \frac{1}{T(h_1)\,S(h_2)} &= \frac12\left(\frac1{T(h_1 h_2)} + \frac1{T(h_1/h_2)}\right) = \frac12\left(\frac1{T(h_1 h_2)} - \frac1{T(h_2/h_1)}\right). \\[8mu] \end{align}$$

The products or quotients of cosines can be written as shifted hyperbolic tangent sums or differences of squares. See above.


 * $$\begin{align}

C(h_1)\,C(h_2) &= \frac{S(h_1/h_2) - S(h_1 h_2)}{S(h_1/h_2) + S(h_1 h_2)} = \left(1 \ominus h_1^2\right)\left(1 \ominus h_2^2\right) = 1 \ominus \bigl(h_1^2 \boxplus h_2^2 \bigr), \\[8mu] \frac{C(h_1)}{C(h_2)} &= \frac{T(h_1/h_2) + T(h_1 h_2)}{T(h_1/h_2) - T(h_1 h_2)} = 1 \ominus \bigl(h_1^2 \boxminus h_2^2 \bigr) = 1 \oplus \bigl(h_2^2 \boxminus h_1^2 \bigr). \end{align}$$

These identities extend naturally to the product or quotient of arbitrary cosines,


 * $$\begin{align}

&\frac{C(p_1)\,C(p_2)\cdots C(p_k)}{C(q_1)\,C(q_2)\cdots C(q_m)} \\[5mu] &\qquad= \left(1 \ominus p_1^2\right)\left(1 \ominus p_2^2\right)\cdots \left(1 \ominus p_k^2\right) \left(1 \oplus q_1^2\right)\left(1 \oplus q_2^2\right)\cdots \left(1 \oplus q_m^2\right) \\[5mu] &\qquad= 1 \ominus \bigl(p_1^2 \boxplus p_2^2 \boxplus \cdots \boxplus p_k^2 \boxminus q_1 \boxminus q_2 \boxminus \cdots \boxminus q_m \bigr). \end{align}$$

The cosine of a product or quotient can be separated as a hyperbolic tangent sum or difference of cosines


 * $$\begin{align}

C(h_1 h_2) &= C(h_1) \boxplus C(h_2), \\[6mu] C(h_1/h_2) &= C(h_1) \boxminus C(h_2), \\[6mu] C{\left(\frac{p_1 p_2 \cdots p_k}{q_1 q_2 \cdots q_m}\right)} &= C(p_1) \boxplus C(p_2) \boxplus \cdots \boxplus C(p_k) \\ &\qquad \boxminus C(q_1) \boxminus C(q_2) \boxminus \cdots \boxminus C(q_m). \end{align}$$

Inverse functions
The inverse stereographic circular functions (analogous to arcsine, arccosine, arctangent) taking a sine $$y$$, cosine $$x$$, or tangent $$t$$ to a half-tangent $$h$$ are


 * $$\begin{align}

C^{-1}(x) &= \sqrt{1\ominus x\vphantom l} = \frac{\sqrt{1 - x^2}}{1 + x} = \frac{1 - x}{\sqrt{1 - x^2}}, \\[8mu] S^{-1}(y) &= \sqrt[\boxplus]{y\vphantom t} = \frac{y}{1 + \sqrt{1 - y^2}} = \frac{1 - \sqrt{1 - y^2}}{y} = 1 \ominus \sqrt{1\ominus y\vphantom l}, \\ T^{-1}(t) &= \sqrt[\oplus]{t \vphantom y} = \frac{t}{1 + \sqrt{1 + t^2}} = \frac{1 - \sqrt{1 + t^2}}{t}.

\end{align}$$

The other branch of each square root also returns a half-tangent $h$ satisfying $$C(h) = x,$$ $$S(h) = y,$$ or $T(h) = t$:


 * $$\begin{alignat}{2}

0 \ominus C^{-1}(x) &= -C^{-1}(x) &&= \frac{-\sqrt{1 - x^2}}{1 + x} = \frac{1 - x}{-\sqrt{1 - x^2}} = -\sqrt{1\ominus x\vphantom l}, \\[8mu] \infty \ominus S^{-1}(y) &= \ \frac1{S^{-1}(y)} &&= \frac{y}{1 - \sqrt{1 - y^2}} = \frac{1 + \sqrt{1 - y^2}}{y} = 1 \oplus \sqrt{1\ominus y\vphantom l}, \\[8mu] \infty \oplus T^{-1}(t) &= \ \frac{-1}{T^{-1}(t)} &&= \frac{t}{1 - \sqrt{1 + t^2}} = \frac{1 + \sqrt{1 + t^2}}{t}.

\end{alignat}$$

The inverse of $E,$ a modified Cayley transform analogous to the natural logarithm taking a unit complex number to a half-tangent times the imaginary unit, is



E^{-1}(z) = \frac{z - 1}{z + 1} = z \ominus 1 = hi. $$

Multiple-angle identities
The half-tangent analogs of circular functions of multiple angles are the functions
 * $$\begin{align}

C_k(h) &:= C(h^{\oplus k}) = \cos k\theta, \\[6mu] S_k(h) &:= S(h^{\oplus k}) = \sin k\theta, \\[6mu] E_k(h) &:= E(h^{\boxplus k}), \\[6mu] E_k(hi) \!&\phantom{:}= E(h^{\oplus k}i) = C_k(h) + iS_k(h) = \exp k\theta i. \end{align}$$

The first few are:


 * $$\begin{align}

C_0(h) &= 1, & S_0(h) &= 0, \\[6mu]

C_1(h) &= \dfrac{1 - h^2}{1 + h^2}, & S_1(h) &= \dfrac{2h}{1 + h^2}, \\[10mu]

C_2(h) &= \dfrac{1 - 6h^2 + h^4}{(1 + h^2)^2}, & S_2(h) &= \dfrac{4h - 4h^3}{(1 + h^2)^2}, \\[10mu]

C_3(h) &= \dfrac{1 - 15h^2 + 15h^4 - h^6}{(1 + h^2)^3}, & S_3(h) &= \dfrac{6h - 20h^3 + 6h^5}{(1 + h^2)^3}, \\[10mu]

& \ldots & & \ldots

\end{align}$$

or in general,


 * $$\begin{align}

E_k(hi) &= \left(\frac{1 + hi}{1 - hi}\right)^{\!k} = \frac{(1 + hi)^{2k}}{(1 + h^2)^k}, \\[10mu] E_{-k}(hi) &= \left(\frac{1 - hi}{1 + hi}\right)^{\!k} = \frac{(1 - hi)^{2k}}{(1 + h^2)^k}, \\[10mu] C_k(h) &= \frac{E_k(hi) + E_{-k}(hi)}{2} = \dfrac{1}{(1 + h^2)^k}\sum_{m} \binom{2k}{2m}\bigl({-h^2}\bigr)^m, \\[10mu] S_k(h) &= \frac{E_k(hi) - E_{-k}(hi)}{2i} = \dfrac{h}{(1 + h^2)^k}\sum_{m} \binom{2k}{2m + 1}\bigl({-h^2}\bigr)^m. \end{align}$$

Half-angle identities
The stereographic circular functions of the "quarter-tangent" $$\sqrt[\oplus]h$$ (see above) are:


 * $$\begin{align}

C_{1/2}(h) := C\bigl(\sqrt[\oplus]h\bigr) &= \frac{1}{\sqrt{1 + h^2}}, \\[6mu] S_{1/2}(h) := S\bigl(\sqrt[\oplus]h\bigr) &= \frac{h}{\sqrt{1 + h^2}} = h\,C_{1/2}(h), \\[6mu] E_{1/2}(hi) := E\bigl(i\sqrt[\oplus]h\bigr) &= \frac{1 + hi}{\sqrt{1 + h^2}} = \sqrt{\frac{1 + hi\vphantom{)}}{1 - hi}} = \sqrt{E(hi)}. \end{align}$$

The sine of a half-angle $$S_{1/2}(h) = \sin\tfrac12\theta$$ is noteworthy as the chord length in a unit-diameter circle, see above.

Taking the supplement (reciprocal) of the argument exchanges half-angle sine with half-angle cosine:


 * $$S_{1/2}(p/q) = \frac{p/q}{\sqrt{1 + p^2/q^2}} = \frac{p}{\sqrt{q^2 + p^2}} = \frac{1}{\sqrt{1 + q^2/p^2}} = C_{1/2}(q/p)$$

The stereographic circular functions can be described in terms of half-angle functions:


 * $$\begin{aligned}

S(h) &= 2C_{1/2}(h)S_{1/2}(h), \\[6mu] C(h) &= 1 - 2S_{1/2}(h)^2 = 2C_{1/2}(h)^2 - 1. \end{aligned}$$

Sines and cosines of half-angle sums and differences are found in spherical trigonometry, and can be translated to half-tangent form using the identities


 * $$\begin{align}

C_{1/2}(h_1 \oplus h_2) &= \frac{1 - h_1h_2}{\sqrt{1 + h_1^2}\sqrt{1 + h_2^2}}, & S_{1/2}(h_1 \oplus h_2) &= \frac{h_1 + h_2}{\sqrt{1 + h_1^2}\sqrt{1 + h_2^2}}, \\[6mu]

\frac{S_{1/2}(h_1 \oplus h_2)}{C_{1/2}(h_1 \oplus h_2)} &= \frac{h_1 + h_2}{1 - h_1h_2} = h_1 \oplus h_2, & \frac{S_{1/2}(h_1 \ominus h_2)}{C_{1/2}(h_1 \ominus h_2)} &= \frac{h_1 - h_2}{1 + h_1h_2} = h_1 \ominus h_2, \\[10mu]

\frac{S_{1/2}(h_1 \oplus h_2)}{C_{1/2}(h_1 \ominus h_2)} &= \frac{h_1 + h_2}{1 + h_1h_2} = h_1 \boxplus h_2, & \frac{S_{1/2}(h_1 \ominus h_2)}{C_{1/2}(h_1 \oplus h_2)} &= \frac{h_1 - h_2}{1 - h_1h_2} = h_1 \boxminus h_2, \\[10mu]

\frac{C_{1/2}(h_1 \ominus h_2)}{C_{1/2}(h_1 \oplus h_2)} &= \frac{1 + h_1h_2}{1 - h_1h_2} = h_1h_2 \oplus 1, & \frac{S_{1/2}(h_1 \ominus h_2)}{S_{1/2}(h_1 \oplus h_2)} &= \frac{h_1 - h_2}{h_1 + h_2} = \frac{h_1}{h_2} \ominus 1, \\[10mu]

E_{1/2}\bigl((h_1 \oplus h_2)i\bigr) &= \frac{1 + h_1i + h_2i - h_1h_2}{\sqrt{1 + h_1^2}\sqrt{1 + h_2^2}}.

\end{align}$$

For three arguments (also found in spherical trigonometry),


 * $$\begin{align}

C_{1/2}\left(h_1 \oplus h_2 \oplus h_3\right) &= \frac {1 - h_1h_2 - h_1h_3 - h_2h_3} {\sqrt{1 + h_1^2}\sqrt{1 + h_2^2}\sqrt{1 + h_3^2} }\,,\\[10mu]

S_{1/2}\left(h_1 \oplus h_2 \oplus h_3\right) &= \frac {h_1 + h_2 + h_3 - h_1h_2h_3} {\sqrt{1 + h_1^2}\sqrt{1 + h_2^2}\sqrt{1 + h_3^2} }\,,\\[10mu]

E_{1/2}\bigl((h_1 \oplus h_2 \oplus h_3)i\bigr) &= C_{1/2}\left(h_1 \oplus h_2 \oplus h_3\right) + S_{1/2}\left(h_1 \oplus h_2 \oplus h_3\right)i. \end{align}$$

For any number of arguments,


 * $$\begin{alignat}{3}

E_{1/2}\bigl((h_1 \oplus \cdots \oplus h_k)i\bigr) &= \,\prod_m E_{1/2}(h_mi) \ = \,\prod_m \frac{1 + h_mi}{\sqrt{1 + h_m^2 }} \ &&= \frac {\displaystyle \sum_m e_m{\left(h_1i,..., h_ki\right)}} {\displaystyle \prod_m \sqrt{1 + h_m^2} }\,,\\[10mu]

C_{1/2}\left(h_1 \oplus \cdots \oplus h_k\right) &= \frac{\displaystyle \prod_m E_{1/2}(h_mi) + \prod_m E_{1/2}(-h_mi)}{2} &&= \frac {\displaystyle \sum_{m \text{ even}} e_m{\left(h_1i,..., h_ki\right)}} {\displaystyle \prod_m \sqrt{1 + h_m^2} }\,, \\[10mu]

S_{1/2}\left(h_1 \oplus \cdots \oplus h_k\right) &= \frac{\displaystyle \prod_m E_{1/2}(h_mi) - \prod_m E_{1/2}(-h_mi) }{2i} &&= \frac {\displaystyle -i \sum_{m \text{ odd}} e_m{\left(h_1i,..., h_ki\right)}} {\displaystyle \prod_m \sqrt{1 + h_m^2} }\,. \end{alignat}$$

where $$m \in \{0, 1, 2, ..., k\}$$ and $$e_m$$ is the $h$th elementary symmetric polynomial. The denominators above come from a product of cosines:



\prod_m C_{1/2}(h_m) = \frac1{\displaystyle \prod_m \sqrt{1 + h_m^2}} = \frac1 \sqrt{\displaystyle \,\sum_m e_m{\left(h_1^2,..., h_k^2\right)\vphantom{\Big|} }} \,. $$

Ptolemy's theorem that the sum of products of lengths of opposite sides of a convex cyclic quadrilateral is equal to the product of the lengths of the diagonals can be rewritten as an algebraic relationship of four arbitrary half-tangents representing the vertices:


 * $$\begin{align}

0 = &\phantom{{}+{}} S_{1/2}(h_1\ominus h_2)S_{1/2}(h_4\ominus h_3) \\ &{} + S_{1/2}(h_1\ominus h_3)S_{1/2}(h_2\ominus h_4) \\ &{} +S_{1/2}(h_1\ominus h_4)S_{1/2}(h_3\ominus h_2). \end{align}$$

This can be proven by expanding it in terms of the previous identities:



0 = \frac{(h_1 - h_2)(h_4 - h_3) + (h_1 - h_3)(h_2 - h_4) + (h_1 - h_4)(h_3 - h_2)} {\sqrt{1 + h_1^2}\sqrt{1 + h_2^2}\sqrt{1 + h_3^2}\sqrt{1 + h_4^2}}.$$

After expanding the products of binomials in the numerator, every term cancels.

Stereographic polynomials
Analogous to Laurent polynomials of unit-magnitude complex numbers or trigonometric polynomials of angle measures, stereographic polynomials can be defined for half-tangents. These are rational functions of the form $s(h) = p(h)\big/\bigl(1 + h^2\bigr)\vphantom{)}^n$ where $p$  is a polynomial of degree at most $$2n$$.

If the polynomial in the numerator is $p(h) = \sum_{k=0}^{2n} p_k h^k,$ with coefficients $$\{p_k\},$$ then the stereographic polynomial can be written in terms of powers of half-angle sines and cosines as:



s(h) = \sum_{k=0}^{2n}\frac{p_k h^k}{\bigl(1 + h^2\bigr)\vphantom{)}^n} = \sum_{k=0}^{2n} p_kS_{1/2}(h)^{k}C_{1/2}(h)^{2n-k}, $$

because $$S_{1/2}(h)^{k}C_{1/2}(h)^{2n-k} = h^k \big/ \bigl(1 + h^2\bigr)\vphantom{)}^n.$$

As a complex function, a stereographic polynomial has all of its poles at $$\pm i$$ (and none at $$\infty$$), compared to a Laurent polynomial with poles at $$\{0, \infty\}$$ or an ordinary polynomial with poles only at $$\infty.$$

Just as a trigonometric polynomial $$t(\theta)$$ can be written in terms of a basis of cosines and sines or complex exponentials of multiple angles,


 * $$t(\theta) = \tfrac12a_0 + \sum_{k=1}^n a_k \cos k\theta + b_k \sin k\theta = \sum_{k=-n}^{n} c_k \exp k\theta i,$$

or under the change of variables $$z = \exp \theta i$$ the resulting Laurent polynomial $$\ell(z) = t(\theta)$$ can be broken into even and odd parts or written in monomial basis,


 * $$\ell(z) = \tfrac12a_0 + \sum_{k=1}^n a_k \frac{z^k + z^{-k}}{2} + b_k \frac{z^k - z^{-k}}{2i} = \sum_{k=-n}^{n} c_k z^k,$$

under the change of variables $$h = \tan\tfrac12\theta$$ this is the stereographic polynomial $$s(h) = t(\theta),$$ and can be written in either of the bases,


 * $$s(h) = \tfrac12a_0 + \sum_{k=1}^n a_k C_k(h) + b_k S_k(h) = \sum_{k=-n}^{n} c_k E_k(hi).$$

In all three of the corresponding polynomials above, the coefficients $$a_k,$$ $$b_k,$$ and $$c_k$$ are the same.

Hyperbolic functions
The hyperbolic functions of argument $\psi$ can alternately be written as rational functions of the hyperbolic half-tangent $h = \tanh \tfrac12\psi.$  As functions of $h$  these turn out to be equivalent to circular functions we had above, but with tangent and sine exchanged. We will continue to use the letters $S,$ $C,$  and $T$  to refer to the circular functions of a half-tangent.


 * $$\begin{align}

\sinh \psi = \tan \theta &= \frac{2h}{1-h^2}   = T(h), & \operatorname{csch}\psi = \cot \theta &= \frac{1-h^2}{2h} = \frac1{T(h)}, \\[10mu] \cosh \psi = \sec \theta &= \frac{1+h^2}{1-h^2} = \frac1{C(h)}, & \operatorname{sech}\psi = \cos \theta &= \frac{1-h^2}{1+h^2} = C(h), \\[10mu] \tanh \psi = \sin \theta &= \frac{2h}{1+h^2}   = S(h), & \coth\psi = \csc \theta &= \frac{1+h^2}{2h} = \frac1{S(h)}. \end{align}$$

The circular angle measure $\theta$ is the gudermannian of the hyperbolic angle measure $\psi,$  with common half-tangent $h = \tan\tfrac12\theta = \tanh\tfrac12\psi,$  which can be defined by


 * $$\theta = \operatorname{gd}\psi := 2\arctan \tanh \tfrac12 \psi.$$

The hyperbolic tangent and sine of a half-tangent are also respectively its hyperbolic and circular tangent squares,


 * $$\begin{align}

\tanh \psi = \,S(h)\ &= h \boxplus h = \, h^{\boxplus2}, \\[6mu] \sinh \psi = \,T(h)\ &= h \oplus h = \, h^{\oplus2}, \\[6mu] \cosh \psi = \frac1{C(h)} &= \frac{T(h)}{S(h)} = \frac{h^{\oplus 2}}{h^{\boxplus 2}} = 1 \oplus h^2. \end{align}$$

There are two common geometric interpretations of a hyperbolic angle measure $\psi.$ The first is as the logarithm of a multiplicative scaling by $w = \exp \psi,$  which can be combined using complex numbers with a circular rotation $\exp(\alpha + \beta i)$  to scale and rotate complex numbers or vectors in the Euclidean plane by multiplication.

Under this interpretation, hyperbolic functions are the even and odd parts of the exponential function of a real (or perhaps complex) argument,


 * $$\begin{align}

\sinh \psi = \frac{e^\psi - e^{-\psi}}{2}, \quad \cosh \psi = \frac{e^\psi + e^{-\psi}}{2}, \quad \tanh \psi = \frac{e^\psi - e^{-\psi}}{e^\psi + e^{-\psi}}. \end{align}$$

The second is as the logarithm of a hyperbolic rotation (Lorentz boost) in pseudo-Euclidean space using split-complex numbers of the form $$\zeta = \exp \psi j = \xi + \eta j$$ with an imaginary unit $$j^2 = +1,$$ analogous to a circular rotation in Euclidean space expressed via the complex number $$z = \exp \theta i = x + iy,$$ with imaginary unit $$i^2 = -1.$$

Under this interpretation, hyperbolic functions are the even and odd parts of the exponential function of a split-complex valued argument, either pure-imaginary or general,


 * $$\begin{align}

\sinh \psi = \frac{e^{\psi j} - e^{-\psi j}}{2j}, \quad \cosh \psi = \frac{e^{\psi j} + e^{-\psi j}}{2}, \quad \tanh \psi = j \frac{e^{\psi j} - e^{-\psi j}}{e^{\psi j} + e^{-\psi j}}. \end{align}$$

Compare to the circular functions:


 * $$\begin{align}

\sin \theta = \frac{e^{\theta i} - e^{-\theta i}}{2i}, \quad \cos \theta = \frac{e^{\theta i} + e^{-\theta i}}{2}, \quad \tan \theta = -i \frac{e^{\theta i} - e^{-\theta i}}{e^{\theta i} + e^{-\theta i}}. \end{align}$$

When using the hyperbolic half-tangent $$h$$ instead of the hyperbolic angle measure $$\psi,$$ it is possible to represent points on both branches of the unit hyperbola $$\zeta\bar\zeta = 1,$$ instead of only the right branch.

Möbius transformations
For any four half-tangents $$h_1, h_2, h_3, h_4 \in \widehat \R$$ their cross-ratio is the quantity,


 * $$(h_1, h_2; h_3, h_4) := \frac{h_1 - h_3}{h_1 - h_4} \bigg/ \frac{h_2 - h_3}{h_2 - h_4} = \frac{(h_1-h_3)(h_2-h_4)}{(h_1-h_4)(h_2-h_3)}.$$

When one of the half-tangents is the half-turn $$\infty,$$ this remains well defined, reducing to e.g.


 * $$(h_1, \infty; h_3, h_4) = \frac{h_1-h_3}{h_1-h_4}.$$

This quantity is the same between the corresponding points $$z_k = E(h_ki) = (1 + h_ki)\big/(1 - h_ki)$$ on the complex unit circle,


 * $$(h_1, h_2; h_3, h_4) = (z_1, z_2; z_3, z_4) = \frac{(z_1-z_3)(z_2-z_4)}{(z_1-z_4)(z_2-z_3)}.$$

This gives another way to express the definition of the map $$z \mapsto h,$$ as a cross ratio:


 * $$h = (h, 1; 0, \infty) = (z, i; 1, -1) = \frac{z - 1}{z + 1} \bigg/ \frac{i - 1}{i + 1} = -i \frac{z - 1}{z + 1}.$$

Any transformation $$M$$ of the projectively extended real line which preserve the cross-ratio,
 * $$\bigl(M(h_1), M(h_2); M(h_3), M(h_4)\bigr) = (h_1, h_2; h_3, h_4),$$

is called a linear fractional transformation or Möbius transformation, is a homography, and is an element of the projective special linear group $$\operatorname{PSL}_{2}(\R)$$. It is a function of the form


 * $$M(h) = \frac{ah + b}{ch + d},$$

and can be written as the matrix


 * $$\begin{pmatrix} a & b \\ c & d \end{pmatrix},$$

where $$a, b, c, d \in \widehat \R,$$ and $$ad - bc \neq 0.$$ Any uniform scaling of $$a, b, c, d$$ represents the same transformation. Composition of Möbius transformations corresponds to matrix multiplication.

The general transformation can have zero, one, or two (real) fixed points, which can be found by solving


 * $$\begin{align}

&h = \frac{ah + b}{ch + d} \implies ch^2 - (a-d)h - b = 0 \\[5mu] &\implies h = \frac{a-d \pm \sqrt{(a-d)^2 + 4bc}}{2c} \end{align}$$

When the real line is considered as the set of ideal points of the hyperbolic plane (cf. Poincaré half-plane model), the group of Möbius transformations with real coefficients which preserve orientation ($ad > bc$) is isomorphic to the group of isometries of a the hyperbolic plane. The orientation-reversing transformations ($ad < bc$) correspond to isometries of paired hyperbolic planes which exchange their points (in the half-plane model, exchanging upper and lower half-planes; in the hyperboloid model exchanging two sheets of the hyperboloid).

Types of transformations
The only homographies of half-tangents preserving distance on the circle and orientation are rotations $h \mapsto a \oplus h$ using the tangent addition operation with one fixed argument. Rotations have no fixed points, except for the zero rotation which fixes every point. Three basic rotations are the the half-turn $$h \mapsto \infty \oplus h$$ and the quarter turns $$h \mapsto 1 \oplus h$$ and $$h \mapsto (-1) \oplus h = h \ominus 1.$$


 * $$\begin{align}

a \oplus h &\quad \text{has matrix} \quad \begin{pmatrix} 1 & a \\ -a & 1 \end{pmatrix}, \\[8mu] 0 \oplus h &\quad \text{has matrix} \quad \begin{pmatrix} 1 & 0 \\ 0 & 1  \end{pmatrix}, & \infty \oplus h &\quad \text{has matrix} \quad \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \\[8mu] 1 \oplus h &\quad \text{has matrix} \quad \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}, & \ominus1 \oplus h &\quad \text{has matrix} \quad \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}. \end{align}$$

Also preserving distance but reversing orientation are the reflections $h \mapsto a \ominus h$ using the tangent subtraction operation with one fixed argument. Reflections fix a pair of antipodal points $$\sqrt[\oplus]a$$ and $$\sqrt[\oplus]a \oplus \infty,$$ and exchange $$0 \leftrightarrow a.$$ The most basic reflections are $$h \mapsto -h = 0 \ominus h$$ with $$\sqrt[\oplus]a = 0$$ and $$h \mapsto 1/h = \infty \ominus h$$ with $$\sqrt[\oplus]a = 1.$$ The reflections $$h \mapsto 1 \ominus h$$ and $$h \mapsto \ominus1 \ominus h$$ exchange the real and imaginary axes when transplanted to the complex unit circle.


 * $$\begin{align}

a \ominus h &\quad \text{has matrix} \quad \begin{pmatrix} -1 & a \\ a & 1 \end{pmatrix}, \\[8mu] 0 \ominus h &\quad \text{has matrix} \quad \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}, & \infty \ominus h &\quad \text{has matrix} \quad \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \\[8mu] 1 \ominus h &\quad \text{has matrix} \quad \begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix}, & \ominus1 \ominus h &\quad \text{has matrix} \quad \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}. \end{align}$$

Another kind of transformation is the origin-centered dilation $$h \mapsto sh$$ with fixed points $$0$$ and $$\infty.$$ More generally a dilation can be centered at some other point, so that $$h \oplus a \mapsto sh \oplus a,$$ with antipodal fixed points $$a$$ and $$a \oplus \infty.$$ When $$s > 0$$ these transformations can be interpreted as the apparent movement of the "celestial circle" in a $2π$-dimensional spacetime when the observer changes relativistic velocity, a lorentz boost.


 * $$\begin{align}

&sh \quad \text{has matrix} \quad \begin{pmatrix} s & 0 \\ 0 & 1 \end{pmatrix}, \\[12mu] &s(h \ominus a) \oplus a \quad \text{has matrix ...} \\[8mu] &\qquad \begin{pmatrix} 1 & a \\ -a & 1 \end{pmatrix} \begin{pmatrix} s & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & -a \\ a & 1 \end{pmatrix} = \begin{pmatrix} s + a^2 & (1-s)a \\ (1-s)a & 1 + sa^2 \end{pmatrix}. \end{align}$$

The only homographies of hyperbolic half-tangents preserving hyperbolic distance are hyperbolic rotations using the hyperbolic tangent addition operation and reflections using the hyperbolic subtraction operation with one fixed argument. Any hyperbolic rotation fixes $$1$$ and $-1$, while a hyperbolic reflection exchanges them.



a \boxplus h \quad \text{has matrix} \quad \begin{pmatrix} 1 & a \\ a & 1 \end{pmatrix}, \qquad a \boxminus h \quad \text{has matrix} \quad \begin{pmatrix} -1 & a \\ -a & 1 \end{pmatrix}. $$

When $$h$$ is treated as a circular half-tangent, $$a \boxplus h,$$ is a dilation by $$1 \ominus a$$ around the equatorial point $$1,$$ or by $$1 \oplus a$$ around $$-1.$$


 * $$\begin{align}

a \boxplus h &= (1 \oplus a)(h \oplus 1) \ominus 1 \\[5mu] &= (1 \ominus a)(h \ominus 1) \oplus 1. \end{align}$$

This leads to the identities from above.

Planar trigonometry


Planar trigonometry (the metrical relations between angles and sides of a triangle in the Euclidean plane) is conventionally written down in terms of side lengths symbolized by $$a, b, c$$ and angle measures (in degrees or radians) symbolized by $$\alpha, \beta, \gamma,$$ following the convention established by Euler, who built on the ancient tradition of labeling the vertices $$A, B, C.$$ When set up this way, several conventional trigonometry identities involve the half-tangents $$\tan\tfrac12\alpha, \tan\tfrac12\beta, \tan\tfrac12\gamma,$$ alongside the trigonometric sines, cosines, and tangents of the angles.

Occasionally, however, the half-tangent of each angle is instead treated as the basic quantity and directly given a symbol, whereupon the transcendental trigonometric functions of angle measure become rational functions of half-tangent, and all of the traditional trigonometric identities can be written as strictly rational relationships. This is the approach we will adopt here:

Let $a,$ $b,$  and $c$  be the lengths of the sides of a planar triangle. Let the respective (interior) angles opposite each side have half-tangents $\alpha,$ $\beta,$  and $\gamma.$  Then $1/\alpha,$  $1/\beta,$  and $1/\gamma$  are their supplements, the respective exterior-angle half-tangents.

Relations among angles
In any triangle, the interior angle measures sum to a half turn or equivalently the exterior angle measures sum to a full turn. In terms of half-tangents this relation can be written as any of,


 * $$\begin{align}

\alpha \oplus \beta \oplus \gamma &= \infty, & \alpha \oplus \beta &= \frac1\gamma,\\[6mu] \frac1\alpha \oplus \frac1\beta \oplus \frac1\gamma &= 0, & \frac1\alpha \oplus \frac1\beta &= - \frac1\gamma. \end{align}$$

Fully expanded in terms of ordinary addition and multiplication,


 * $$\begin{align}

\alpha\beta + \alpha\gamma + \beta\gamma &= 1, \\[6mu] \frac1\alpha + \frac1\beta + \frac1\gamma &= \frac1\alpha \frac1\beta \frac1\gamma. \end{align}$$

Expressed in terms of angle measure, these identities are sometimes called the "triple tangent identity" or "triple cotangent identity".

Relations between sides and angles
Angle $$\gamma$$ can be related to the side lengths by the equivalent equations below, the first of which is a simple modification of the law of cotangents and the last of which is the law of cosines written in terms of half-tangents, where $$C(h) = 1 \ominus h^2$$ is the stereographic cosine.


 * $$\begin{align}

\gamma^2 &= \frac{c^2 - (b - a)^2}{(b + a)^2 - c^2} = \frac{(-a + b + c)(a - b + c)}{(a + b - c)(a + b + c)},\\[10mu]

C(\gamma) &= \frac {a - c}{b} \boxplus \frac{b}{a + c} = \frac {b - c}{a} \boxplus \frac{a}{b + c} \\[10mu]

c^2 &=\frac{\gamma^2(b + a)^2 + (b - a)^2}{\gamma^2 + 1} = a^2 + b^2 - 2ab\,C(\gamma). \end{align}$$

and likewise for $\beta$ and $\alpha.$  (The squares on the left hand side arise because two different triangle shapes can be found with the given side lengths, with angular half-tangents (or angle measures) of opposite signs $\{\alpha,\beta,\gamma\}$  and $\{-\alpha,-\beta,-\gamma\}$  indicating anticlockwise and clockwise turns, respectively. These two triangles are congruent under reflection.)

The half-tangent expressions of Mollweide's formulas (first published by Isaac Newton in 1707) are corollaries,


 * $$\begin{align}

\alpha\beta &= \,\, \frac{a + b - c}{a + b + c} \ \, = \frac{a + b}{c} \ominus 1, & \alpha\beta \oplus 1 = \frac{1 + \alpha\beta}{1 - \alpha\beta} &= \frac{a + b}{c}, \\[10mu] \frac\alpha\beta &= \frac{a - b + c}{-a + b + c} = \frac{a - b}{c} \oplus 1, & \frac\alpha\beta \ominus 1 = \,\,\frac{\alpha - \beta}{\alpha + \beta}\, &= \frac{a - b}{c}. \end{align}$$

and likewise for other pairs of angles. Taking the quotient of these to eliminate $c$ results in the law of tangents,



\frac{\alpha \ominus \beta}{\alpha \oplus \beta} = \frac{a - b}{a + b}. $$

The left side of the law of tangents can be written in terms of $$S(h) = 2h/(1 + h^2),$$ the stereographic sine (see § Circular functions › Tangent sum identities above),



1 \ominus \frac{S(\beta)}{S(\alpha)} = 1 \ominus \frac{b}{a}, \qquad \frac{S(\alpha)}{S(\beta)} = \frac ab. $$

This is the law of sines,


 * $$\frac{a}{S(\alpha)} = \frac{b}{S(\beta)} = \frac{c}{S(\gamma)} = \pm D,$$

where the common ratio $D$ is the diameter of the circumcircle of the triangle.

Unlike in spherical and hyperbolic geometry, in Euclidean geometry the dual of the law of cosines degenerates: in the infinitessimal limit a squared side of a spherical triangle vanishes $$c^2 = 0$$ and $$C(0) = 1.$$ So the result is merely a rearrangement of the angle relationship $$\alpha \oplus \beta \oplus \gamma = \infty$$ or $\alpha\beta + \alpha\gamma + \beta\gamma = 1,$ demonstrating the tangent-sum identity for stereographic cosine,


 * $C(\gamma) = -C(\alpha)C(\beta) + S(\alpha)S(\beta).$

Compare that to below about the spherical versions.

But we can salvage the rational expression for one side in terms of the three angles by dividing by the spherical excess. In the infinitesimal limit the ratio $c^2/\varepsilon$ of squared side to excess of a spherical triangle degenerates to the ratio of squared side to twice the area of a planar triangle, so for notational consistency we will use the symbol $$\varepsilon$$ to mean twice the area of a planar triangle (see  below):


 * $$\begin{align}

\frac{c^2}{\varepsilon} &= \frac {({-\alpha\beta} + \alpha\gamma + \beta\gamma + 1) (\alpha + \beta + \gamma - \alpha\beta\gamma)} {(\alpha\beta - \alpha\gamma + \beta\gamma + 1) (\alpha\beta + \alpha\gamma - \beta\gamma + 1)} = \frac {S(\gamma)}{S(\alpha)S(\beta)}. \end{align}$$

As corollaries,


 * $$\begin{align}

\frac{ab}{\varepsilon} &= \frac {\alpha + \beta + \gamma - \alpha\beta\gamma} {-\alpha\beta + \alpha\gamma + \beta\gamma + 1} = \frac{1}{S(\gamma)}, \\[10mu]

\frac ab &= \,\,\frac {\alpha\beta + \alpha\gamma - \beta\gamma + 1} {\alpha\beta - \alpha\gamma + \beta\gamma + 1} \ \, = \bigl((\alpha \ominus \beta)\gamma\bigr) \oplus 1, & \frac ab \ominus 1 &= (\alpha \ominus \beta)\gamma. \end{align}$$

and likewise for other pairs of sides. In the latter equation, the areas cancel and the ratio of stereographic side lengths does not vanish in the planar limit, and we are left with a proper dual to one of Mollweide's formulas – one of Napier's analogies transplanted directly to the plane. However, it is more commonly written as


 * $$\begin{align}

\frac ab = \frac {\alpha\beta + \alpha\gamma - \beta\gamma + 1} {\alpha\beta - \alpha\gamma + \beta\gamma + 1} = \frac {S(\alpha)} {S(\beta)}, \end{align}$$

an expression of the law of sines.

Triangle area
Let $$\varepsilon$$ be twice the (signed) area of the triangle; for a triangle with base $b$ and altitude $h$, $\varepsilon = bh.$

In terms of two sides and the included angle, the area is


 * $$\varepsilon = ab\,S(\gamma).$$

In terms of the three sides, Heron's formula is


 * $$\varepsilon^2 = \tfrac14(-a + b + c)(a - b + c)(a + b - c)(a + b + c).$$

As corollaries,



\varepsilon\gamma = \tfrac12(-a + b + c)(a - b + c), \quad \frac\varepsilon\gamma = \tfrac12(a + b - c)(a + b + c), $$

and likewise for $$\alpha$$ and $$\beta.$$ Furthermore,



\frac\varepsilon{\alpha\beta\gamma} = \tfrac12(a + b + c)^2. $$

Triangles where $a,$ $b,$  $c,$  and $\varepsilon$  are all rational numbers are called Heronian triangles; in such triangles, the half-tangents $$\alpha,$$ $$\beta,$$ and $$\gamma$$ are also rational numbers.

Circumcircle, incircle, and excircles
The diameter $D$ of the triangle's circumscribed circle (circumcircle) is


 * $$\begin{align}

D^2 &= \frac{4a^2b^2c^2}{(-a + b + c)(a - b + c)(a + b - c)(a + b + c)}, \\[8mu] D &= \frac{abc}{\varepsilon}. \end{align}$$

As a corollary, if the triangle is scaled so that the diameter of the circumcircle is $$1,$$ then twice the area is the product of the sines. For a general triangle,


 * $$\frac{\varepsilon}{D^2} = \frac{\varepsilon^3}{a^2b^2c^2} = S(\alpha)S(\beta)S(\gamma).$$

The diameter $d$ of the triangle's inscribed circle (incircle) is


 * $$\begin{align}

d^2 &= \frac{(-a + b + c)(a - b + c)(a + b - c)}{a + b + c}, \\[8mu] d &= \frac{2\varepsilon}{a + b + c} = \alpha\beta\gamma(a + b + c) = \gamma(a + b - c), \end{align}$$

and likewise for $$a$$ and $$b.$$

The diameter $d_a$ of the triangle's escribed circle (excircle) touching side $a$  is


 * $$\begin{align}

d_a^2 &= \frac{(a - b + c)(a + b - c)(a + b + c)}{-a + b + c}, \\[8mu] d_a &= \frac{2\varepsilon}{-a + b + c} = \alpha(a + b + c) = \frac\alpha{\beta\gamma}(-a + b + c) = \frac1\beta(a + b - c), \end{align}$$

and likewise for the excircles touching sides $b$ and $c$.

As a corollary,


 * $$\begin{alignat}{4}

\frac{d_a}{\alpha} &= \ \frac{d_b}{\beta} &&= \quad \ \frac{d_c}{\gamma} &&= \frac{d}{\alpha\beta\gamma} &{}= a + b + c, \\[8mu]

\frac{d_a}{\beta^{-1}} &= \frac{d_b}{\alpha^{-1}} &&= \frac{d_c}{\alpha^{-1}\beta^{-1}\gamma} &&= \ \ \frac{d}{\gamma} &{}= a + b - c. \end{alignat}$$

Altitudes
An altitude $$h_a$$ is the signed distance from the "base" side $$a$$ to the opposite vertex $$\alpha.$$ It can be computed by dividing the double area $$\varepsilon$$ by the base side, among other ways,



h_a = \frac\varepsilon a = bS(\gamma) = cS(\beta) = \frac{bc}D, $$

and likewise for $$h_b$$ and $$h_c.$$

Applying the relation between $$\varepsilon$$ and the three sides,



h_a^2 = \frac1{4a^2}(-a + b + c)(a - b + c)(a + b - c)(a + b + c). $$

The sum of the reciprocal altitudes is the reciprocal inradius (the inradius is half the diameter of the incircle),


 * $$\frac1{h_a} + \frac1{h_b} + \frac1{h_c} = \frac{2}{d} = \frac{a + b + c}{\varepsilon}.$$

Right triangles
The triangle is called a right triangle when one angle $$\gamma = 1$$ is a right angle. The side $c$ is called the hypotenuse and the other two sides are called legs.

Twice the area of the triangle, $$\varepsilon,$$ is the product of the legs,


 * $$\varepsilon = ab.$$

The other two angles are complements, $\alpha \oplus \beta = 1,$ and can be computed in terms of the sides as



\alpha = \frac{a}{c + b} = \frac{c - b}{a},\quad \beta = \frac{b}{a + c} = \frac{c - a}{b}. $$

For the right angle, $$C(\gamma) = 0$$ and $$S(\gamma) = 1,$$ while for the other two angles sines and cosines are the side ratios,



S(\alpha) = C(\beta) = \frac ac,\quad C(\alpha) = S(\beta) = \frac bc. $$

The Pythagorean identity is obtained from the law of cosines,


 * $$c^2 = a^2 + b^2.$$

When all three sides are integers, the triangle is called a Pythagorean triangle. For such a triangle, the half-tangents $$\alpha$$ and $$\beta$$ are rational numbers. Conversely, whenever $$\gamma = 1$$ and $$\alpha$$ or $$\beta$$ is rational the triangle can be uniformly scaled into a Pythagorean triangle.

Spherical trigonometry
Spherical trigonometry (the metrical relations between dihedral angles and central angles of a spherical triangle) can also be described in terms of half-tangents instead of angle measures. Let $a,$ $b,$  and $c$  be the half-tangents of the central angles subtending sides of a spherical triangle (the "sides"). Let the (interior) dihedral angles at the vertices opposite each side have respective half-tangents $\alpha,$ $\beta,$  and $\gamma$  (the "interior angles"). Then $1/\alpha,$ $1/\beta,$  and $1/\gamma$  are their supplements, the respective exterior-dihedral-angle half-tangents (the "exterior angles").

Relation between dihedral angles and spherical excess
In the Euclidean plane, the three interior angles of a triangle always compose to a half turn, but on a sphere the composition of the three interior dihedral angles of a triangle always exceeds a half turn, by an angular quantity called the triangle's spherical excess. For a sphere of unit radius, the measure of a triangle's spherical excess (also called solid angle) is equal to the spherical surface area enclosed by the triangle (this identity is Girard's theorem).

Here, let $\varepsilon$ be the half-tangent of the triangle's spherical excess.



\alpha \oplus \beta \oplus \gamma = \infty \oplus \varepsilon. $$

The three exterior angles of a spherical triangle and the excess $\varepsilon$ sum to a full turn,



\frac1\alpha \oplus \frac1\beta \oplus \frac1\gamma \oplus \varepsilon = 0. $$

Rearranging the above, the excess can be written in terms of angles as



\varepsilon = \alpha \oplus \beta \oplus \gamma \ominus \infty = \frac{-1}{\alpha \oplus \beta \oplus \gamma} = - {\left(\frac1\alpha \oplus \frac1\beta \oplus \frac1\gamma\right)}. $$

Relations between dihedral and central angles
The spherical law of cosines for angles relates one dihedral angle ("angle") $\gamma$ to the three central angles ("sides"). In terms of half-tangents,


 * $C(c) = C(a)C(b) + S(a)S(b)C(\gamma),$

where $$C(h) = (1 - h^2)/(1 + h^2)$$ is the stereographic cosine and $$S(h) = 2h/(1 + h^2)$$ is the stereographic sine. When expanded as a rational equation then simplified this is


 * $$\begin{align}

\gamma^2 &= \frac {c^2(1 + ab)^2 - (b - a)^2} {(b + a)^2 - c^2(1 - ab)^2} = \frac {({-a} + b + c + abc)(a - b + c + abc)} {(a + b - c + abc)(a + b + c - abc)}, \\[10mu]

C(\gamma) &= \frac {a \ominus c}{b} \boxplus \frac{b}{a \oplus c} = \frac {b \ominus c}{a} \boxplus \frac{a}{b \oplus c}, \\[10mu]

c^2 &= \frac {\gamma^2(b + a)^2 + (b - a)^2} {\gamma^2(1 - ab)^2 + (1 + ab)^2} = \frac{a^2 \boxplus b^2 - S(ab)C(\gamma)}{1 + S(ab)C(\gamma)}, \end{align}$$

and likewise for $\alpha$ and $\beta.$  In the small-triangle limit with $$ab \ll 1$$, this reduces to the planar law of cosines.

As corollaries,


 * $$\begin{align}

\alpha\beta &= \,\, \frac{a + b - c + abc}{a + b + c - abc} \ \, = \frac{a \oplus b}{c} \ominus 1, & \alpha\beta \oplus 1 &= \frac{a \oplus b}c, \\[10mu]

\frac\alpha\beta &= \frac{a - b + c + abc}{-a + b + c + abc} = \frac{a \ominus b}{c} \oplus 1, & \frac\alpha\beta \ominus 1 &= \frac{a \ominus b}c. \end{align}$$

and likewise for other pairs of angles. The two identities above on the right are the half-tangent expressions for two of Napier's analogies (the spherical analog of Mollweide's formulas for a planar triangle). Taking their quotient to eliminate $c$ results in the spherical law of tangents,



\frac{\alpha \ominus \beta}{\alpha \oplus \beta} = \frac{a \ominus b}{a \oplus b}. $$

The two sides of the law of tangents can be written in terms of sines,


 * $$\frac{S(\alpha) - S(\beta)}{S(\alpha) + S(\beta)} = \frac{S(a) - S(b)}{S(a) + S(b)}.$$

This simplifies to the spherical law of sines,


 * $$\frac{S(a)}{S(\alpha)} = \frac{S(b)}{S(\beta)} = \frac{S(c)}{S(\gamma)}.$$

The spherical law of cosines for sides relates one side $c$ to the three angles. In terms of half-tangents,


 * $C(\gamma) = -C(\alpha)C(\beta) + S(\alpha)S(\beta)C(c),$

When expanded as a rational equation then simplified this is


 * $$\begin{align}

c^2 &= \frac {\gamma^2(\beta + \alpha)^2 - (1 - \alpha\beta)^2} {(1 + \alpha\beta)^2 - \gamma^2(\beta - \alpha)^2} = \frac {({-\alpha\beta} + \alpha\gamma + \beta\gamma + 1) (\alpha\beta + \alpha\gamma + \beta\gamma - 1)} {(\alpha\beta - \alpha\gamma + \beta\gamma + 1) (\alpha\beta + \alpha\gamma - \beta\gamma + 1)}, \\[10mu]

\gamma^2 &= \frac {c^2(1 + \alpha\beta)^2 + (1 - \alpha\beta)^2} {c^2(\beta - \alpha)^2 + (\beta + \alpha)^2} = \frac{1 - S(\alpha\beta)C(c)}{\alpha^2 \boxplus \beta^2 + S(\alpha\beta)C(c)}, \end{align}$$

and likewise for $a$ and $b.$

As corollaries,


 * $$\begin{align}

ab &= \frac {\alpha\beta + \alpha\gamma + \beta\gamma - 1} {-\alpha\beta + \alpha\gamma + \beta\gamma + 1} = \bigl((\alpha \oplus \beta)\gamma\bigr) \ominus 1, & ab \oplus 1 &= (\alpha \oplus \beta)\gamma, \\[10mu]

\frac ab &= \,\,\frac {\alpha\beta + \alpha\gamma - \beta\gamma + 1} {\alpha\beta - \alpha\gamma + \beta\gamma + 1} \ \, = \bigl((\alpha \ominus \beta)\gamma\bigr) \oplus 1, & \frac ab \ominus 1 &= (\alpha \ominus \beta)\gamma. \end{align}$$

and likewise for other pairs of sides. The two above on the right are the rest of Napier's analogies.

Combining the two laws of cosines we obtain four more corollaries,


 * $$\begin{align}

c^2 &= \frac {(a \boxminus b)(a \boxplus b)(\alpha \boxplus \beta)} {\alpha \boxminus \beta}, & c\gamma &= \frac {a \boxminus b} {\alpha \boxminus \beta}, \\[10mu]

\gamma^2 &= \frac {a \boxminus b} {(a \boxplus b)(\alpha \boxplus \beta)(\alpha \boxminus \beta)}, & \frac c\gamma &= (a \boxplus b)(\alpha \boxplus \beta). \end{align}$$

One last set of relations between all six parts:


 * $$\begin{align}

\frac{a + b + c - abc}{1}\, \frac{\beta\gamma + \alpha\gamma + \alpha\beta - 1}{abc} = 4 \end{align}$$

This can alternately be rewritten in any of sixteen total ways because:
 * $$\begin{alignat}{3}

\frac{a + b + c - abc}{1} &= \ \ \, \frac{-a + b + c - abc}{\beta\gamma} &&= \ \ \frac{a - b + c + abc}{\alpha\gamma} &&= \frac{a + b - c + abc}{\alpha\beta} \\[15mu] \frac{\beta\gamma + \alpha\gamma + \alpha\beta - 1}{abc} &= \frac{-\beta\gamma + \alpha\gamma +\alpha\beta + 1}{a} &&= \frac{\beta\gamma - \alpha\gamma + \alpha\beta + 1}{b} &&= \frac{\beta\gamma + \alpha\gamma - \alpha\beta + 1}{c} \end{alignat}$$

Spherical excess
As mentioned previously, the half-tangent $$\varepsilon$$ of spherical excess can be described in terms of angles,



\varepsilon = \alpha \oplus \beta \oplus \gamma \ominus \infty = \frac{-1}{\alpha \oplus \beta \oplus \gamma} = \frac {\alpha\beta + \alpha\gamma + \beta\gamma - 1} {\alpha + \beta + \gamma - \alpha\beta\gamma}. $$

It can also be described in terms of two sides and their included angle,



\varepsilon = \frac{a b \, S(\gamma)}{1 + a b \, C(\gamma)} = \frac{2ab\gamma}{(1 + ab) + \gamma^2(1 - ab)}. $$

L'Huilier's formula is somewhat similar to Heron's formula, and describes the quarter-tangent of spherical excess in terms of the quarter-tangents of the three sides. To use the notation of this article,



\left(\sqrt[\oplus]\varepsilon\right)^2 = \sqrt[\oplus]{\ominus a \oplus b \oplus c}\, \sqrt[\oplus]{a \ominus b \oplus c}\, \sqrt[\oplus]{a \oplus b \ominus c}\, \sqrt[\oplus]{a \oplus b \oplus c}. $$

Another way to write this relationship is Cagnoli's formula,



S_{1/2}(\varepsilon)^2 = \frac{ S_{1/2}(\ominus a \oplus b \oplus c)\, S_{1/2}(a \ominus b \oplus c)\, S_{1/2}(a \oplus b \ominus c)\, S_{1/2}(a \oplus b \oplus c)} {4\,C_{1/2}(a)^2\,C_{1/2}(b)^2\,C_{1/2}(c)^2}. $$

A third way, expressing the half-tangent of spherical excess in terms of the cosines of the three sides, was known to Euler and Lagrange in the 1770s. After being expanded in half-tangents and simplified, this is quite similar to the planar Heron's formula, to which it reduces in the small-triangle limit:


 * $$\begin{align}

\varepsilon^2 &= \frac {1 - C(a)^2 - C(b)^2 - C(c)^2 + C(a)C(b)C(c)} {\bigl(1 + C(a) + C(b) + C(c)\bigr)\vphantom{)}^2} \\[8mu] &= \frac {(-a + b + c + abc)(a - b + c + abc)(a + b - c + abc)(a + b + c - abc)}  {(2 + a^2 + b^2 + c^2 - a^2b^2c^2)^2}. \end{align}$$

For clarity in the following, define $f = 2 + a^2 + b^2 + c^2 - a^2b^2c^2.$ Then as corollaries,


 * $$\begin{align}

\varepsilon\gamma &= \frac1f(- a + b + c + abc)(a - b + c + abc), \\[10mu] \frac\varepsilon\gamma &= \frac1f(a + b - c + abc)(a + b + c - abc) \end{align}$$

and likewise for $\alpha$ and $\beta$. Furthermore,



\frac\varepsilon{\alpha\beta\gamma} = \frac1f(a + b + c - abc)^2. $$

Spherical triangles where the half-tangents of central angles $a,$ $b,$  $c,$  and the half-tangent of excess $\varepsilon$  are all rational numbers are called Heronian spherical triangles. (In such triangles, all three dihedral angle half-tangents $$\alpha,$$ $$\beta,$$ and $$\gamma$$ are also rational numbers.)

Circumscribed and inscribed small circles
A small circle circumscribed about a spherical triangle (the circumcircle) is the small circle passing through all three vertices of the triangle. When the sphere is embedded in 3-dimensional Euclidean space, this is the intersection of the sphere and the plane passing through the three vertices. Traditional spherical trigonometry books give formulas for the tangent of the central angle radius of this circle, but this is the half-tangent of the central angle diameter of the circle, which we will denote $D$. (The half-tangent of the radius is $\sqrt[\oplus]D$.)

For clarity, define


 * $$\begin{align}

f &= 2 + a^2 + b^2 + c^2 - a^2b^2c^2 \\[2mu] &= (1 + a^2) + (1 + b^2) + (1 + c^2) - (1 + a^2b^2c^2), \\[6mu] g^2 &= \tfrac14f^2(1 + \varepsilon^2) = (1 + a^2)(1 + b^2)(1 + c^2). \end{align}$$

Then the half-tangent $D$ of the diameter of the circumcircle is


 * $$\begin{align}

D^2 &= \frac {4a^2b^2c^2(1 + a^2)(1 + b^2)(1 + c^2)} {(-a + b + c + abc)(a - b + c + abc)(a + b - c + abc)(a + b + c - abc)}, \\[10mu] D &= \frac{2gabc}{f\varepsilon} = \frac{abc}{S_{1/2}(\varepsilon)}. \end{align}$$

A small circle inscribed in a spherical triangle (the incircle) is the small circle tangent to all three sides (great-circle arcs passing through the vertices). Again, traditional spherical trigonometry sources give formulas for the tangent of the incircle's radius, equal to the half-tangent of its diameter which we will call $d,$


 * $$\begin{align}

d^2 &= \frac {(-a + b + c + abc)(a - b + c + abc)(a + b - c + abc)} {g^2(a + b + c - abc)}, \\[8mu] d &= \frac {f\varepsilon} {g(a + b + c - abc)} = \frac1g\alpha\beta\gamma(a + b + c - abc) = \alpha\beta\gamma\,S_{1/2}(a \oplus b \oplus c) \\[4mu] &\phantom{{}=\frac {f\varepsilon} {g(a + b + c - abc)}} = \frac1g\gamma(a + b - c + abc) = \gamma\,S_{1/2}(a \oplus b \ominus c), \end{align}$$ and likewise for $$a$$ and $$b.$$

The half-tangent of the diameter $d_a$ of the triangle's escribed circle (excircle) touching side $a$  is


 * $$\begin{align}

d_a^2 &= \frac {(a + b + c + abc)(a - b + c + abc)(a + b - c + abc)} {g^2(-a + b + c - abc)}, \\[8mu] d_a &= \frac {f\varepsilon} {g(-a + b + c + abc)} = \frac1g\alpha(a + b + c - abc) = \alpha\,S_{1/2}(a \oplus b \oplus c). \end{align}$$

and likewise for the excircles touching sides $b$ and $c$.

As a corollary,



\frac{d_a}{\alpha} = \frac{d_b}{\beta} = \frac{d_c}{\gamma} = \frac{d}{\alpha\beta\gamma} = \frac1g(a + b + c - abc) = S_{1/2}(a \oplus b \oplus c). $$

Right-angled triangles
For a spherical triangle with $$\gamma = 1$$ a right angle, the half-tangent of spherical excess (analogous to the area of a planar triangle) is


 * $$\varepsilon = ab = \alpha \oplus \beta \ominus 1.$$

The spherical Pythagorean identity is the law of cosines for a right-angled triangle, conventionally formulated as $$C(c) = C(a)C(b).$$ In terms of half-tangents it appears more similar planar Pythagorean identity:


 * $$c^2 = a^2 \boxplus b^2 = \frac{a^2 + b^2}{1 + a^2b^2}.$$

Given any pair of other data, all of the sides and angles can be determined by the identities,


 * $$\begin{align}

C(c) &= C(a)C(b) = \frac1{T(\alpha)T(\beta)}, \quad S(c) = \frac{S(a)}{S(\alpha)}, \\[6mu] C(\alpha) &= \frac{T(b)}{T(c)}, \quad S(\alpha) = \frac{C(\beta)}{C(b)} = \frac{S(a)}{S(c)}, \\[10mu] C(a) &= \frac{C(\alpha)}{S(\beta)} = \frac{C(c)}{C(b)}, \quad S(a) = \frac{T(b)}{T(\beta)}, \end{align}$$

and likewise exchanging $$a \leftrightarrow b, \alpha \leftrightarrow \beta.$$

In practical computation, when one or both legs of the triangle is very small, taking cosines can result in loss of significance. It can improve precision to take the complement of each side ($h \mapsto 1 \ominus h$) and then algebraically manipulate to obtain,


 * $$\begin{align}

c^2 &= a^2 \boxplus b^2 = - \frac{C(\alpha \oplus \beta)}{C(\alpha \ominus \beta)}, \quad (1 \ominus c)^2 = \frac{\alpha \ominus a}{\alpha \oplus a}, \\[6mu] \alpha^2 &= \frac{S(c \ominus b)}{S(c \oplus b)},\quad (1 \ominus \alpha)^2 = (\beta \ominus b)(\beta \oplus b) = \frac{c \ominus a}{c \oplus a}, \\[6mu] a^2 &= (\alpha \oplus \beta \ominus 1)(\alpha \ominus \beta \oplus 1) = (c \oplus b)(c \ominus b), \quad (1 \ominus a)^2 = \frac{S(\beta \ominus b)}{S(\beta \oplus b)}. \\[6mu]

\end{align}$$

Inversive and Laguerre geometry
By the intersecting secants theorem and intersecting chords theorem from Euclid's Elements, given a point and a circle in the Euclidean plane, for any line through the point and intersecting the circle, the product of the segment length from the point to the two intersections of the line with the circle is a constant. Jakob Steiner called this value the power of the point with respect to the circle, and its study led to the concept of the radical axis of two circles and the radical center of three circles. An inversion of the plane with respect to a circle exchanges points in the plane such that the product of their distances to a common center is a given constant. More generally the group of Möbius transformations is generated by such circle inversions.

Edmond Laguerre found a dual concept: given a directed line and a directed circle in the Euclidean plane, for any point on the line, the product of the half-tangents of the angles between the given line and the two tangent lines to the circle passing through the point is a constant, the power of the line with respect to the circle. What Laguerre called a transformation by reciprocal semi-lines exchanges directed lines which intersect along a central axis, the product of whose respective half-tangents with that axis is a given constant.

The spherical analog of the intersecting secants and chords theorems replaces planar distances with stereographic distances (half-tangents of central angles), and was proven by Anders Lexell in 1786. Thus, analogously to the Euclidean case, the power of a point on the sphere with respect to a small circle is the product of the stereographic distances from the point to the two intersections of the circle and any great circle through the point which intersects the circle. An inversion of the sphere with respect to a small circle exchanges points such that the product of their stereographic distances is a given constant. The power of a directed great circle with respect to a directed small circle is the product of of the half-tangents of the angles between the given great circle and the two great-circle tangents to the small circle passing through any point along the line. A transformation by reciprocal directed great circles exchanges directed great circles which intersect along a central axis, the product of whose respective half-tangents with that axis is a given constant.



Cross ratio:
 * Foote, Robert L., and Xidian Sun. "An Intrinsic Formula for the Cross Ratio in Spherical and Hyperbolic Geometries." The College Mathematics Journal 46, no. 3 (2015): 182-188. https://doi.org/10.4169/college.math.j.46.3.182
 * Lu, Deng-Maw, Chi-Feng Chang, and Wen-Miin Hwang. "Cross ratio in sphere geometry and its application to mechanism design." Journal of the Franklin Institute 332, no. 2 (1995): 219-226. https://doi.org/10.1016/0016-0032(95)00042-1
 * Coxeter, H. S. M. "Inversive distance." Annali di Matematica Pura ed Applicata 71 (1966): 73-83. https://link.springer.com/content/pdf/10.1007/BF02413734.pdf

Conic sections


Given two fixed points $$F_1, F_2$$ in the Euclidean plane, an ellipse with foci $$F_1, F_2$$ is commonly defined to be a locus of points $$P$$ such that the sum $$a + b$$ of the distances $$a = \|P - F_1\|$$ and $$b = \|P - F_1\|$$ is some constant $$k_e.$$ Likewise, one branch of a hyperbola with foci $$F_1, F_2$$ is commonly defined to be a locus of points $$P$$ such that the difference of distances $$b - a$$ is some other constant $k_h$; the other branch of the hyperbola is the locus of points with difference $$b - a = -k_h.$$

If we construct the triangle $$F_1F_2P$$, then due to Mollweide's formula (see for the half-tangent expression), this sum or difference of side lengths is a function of the product or quotient of half-tangents of the opposite angles. Thus we can instead characterize an ellipse as the locus of points such that the product $$\alpha\beta$$ of half-tangents of angles $$\alpha = \tan\tfrac12\angle F_1F_2P$$ and $$\beta = \tan\tfrac12\angle PF_1F_2$$ is a constant $$\kappa_e$$. Likewise, one branch of a hyperbola is the locus of points such that the quotient $$\beta/\alpha$$ is a constant $\kappa_h$; the other branch is the locus of points with quotient $$\beta/\alpha = 1/\kappa_h.$$

Related feature of confocal parabolas: https://archive.org/details/elementaryconic00smituoft/page/110/mode/1up

The situation is analogous for conics on the sphere and hyperbolic plane, except that distances on the Euclidean plane sum using $$+,$$ stereographic distances on the sphere sum using $$\oplus$$ and stereographic distances on the hyperbolic plane sum using $$\boxplus.$$ (See above.)

On the sphere and the hyperbolic plane, the dual statement is closely related: if $$x_1, x_2$$ are two fixed focal geodesics, then the envelopes of the geodesics $$\ell$$ forming triangles (trilaterals) $$x_1x_2\ell$$ with constant product $$ab$$ or quotient $$b/a$$ (where $$a$$ is the stereographic length of the side along geodesic $$x_1$$ and $$b$$ is the stereographic length of the side along geodesic $x_2$) are confocal dual conics. On the Euclidean plane the metrical duality between points and lines is less exact. One of the duals results in the tangent–asymptotes triangle of a hyperbola: given two intersecting lines $$x_1, x_2$$ in the plane, any hyperbola with those lines as its asymptotes is the envelope of tangent lines $$\ell$$ such that the triangle formed by lines $$x_1x_2\ell$$ has constant area (equivalently, $$ab$$ is constant, where $$a$$ is the length of the side along geodesic $$x_1$$ and $$b$$ is the length of the side along geodesic $x_2$). The envelopes of lines $$\ell$$ forming triangles with $$b/a$$ constant is a degenerate conic: a pair of parallel lines.

Half-tangent function
The half-tangent function is the circular function $$\theta \mapsto h = \tan\tfrac12\theta,$$ the tangent of half of the argument.

Formal definition
The half-tangent function can be formally defined as the ratio of the power series for $$\sin \tfrac12 \theta$$ and $$\cos \tfrac12 \theta,$$ both of which converge throughout the complex plane.


 * $$\begin{align}

\tan \tfrac12 \theta = \frac{\sin \tfrac12 \theta}{\cos \tfrac12 \theta} =\frac{\displaystyle -i \sum_{n\text{ odd}} \frac{(\theta i)^n}{2^nn!} }{\displaystyle \phantom{-i}\sum_{n\text{ even}} \frac{(\theta i)^n}{2^nn!} }

= \frac{\displaystyle \frac \theta2 - \frac{\theta^3}{2^33!} + \frac{\theta^5}{2^55!} - \frac{\theta^7}{2^77!} + \cdots }{\displaystyle 1 - \frac{\theta^2}{2^22!} + \frac{\theta^4}{2^44!} - \frac{\theta^6}{2^66!} + \cdots }. \end{align}$$

In terms of the complex exponential function, it can be defined as


 * $$\begin{align}

\tan \tfrac12 \theta = -i\, \frac{\exp{\theta i} - 1}{\exp{\theta i} + 1}. \end{align}$$

Alternately, it can be defined for the interval $$|\theta| < \pi$$ as the solution to an initial value problem



\tan\tfrac12 0 = 0,\quad \frac{d}{d\theta}\tan\tfrac12 \theta = \tfrac12{\bigl(1 + \tan^2 \tfrac12 \theta\bigr)}, $$

and then analytically continued throughout the complex plane.

Relation to other circular functions
For real-valued $$\theta,$$ the half-tangent can be written in terms of other circular functions in a wide variety of ways,


 * $$\begin{aligned}

h = \tan \tfrac12\theta &= -i \frac{\exp \theta i - 1}{\exp \theta i + 1} = \frac{i - i\cos \theta + \sin\theta}{1 + \cos \theta + i\sin\theta} \\[10mu] &= \frac{\sin \theta}{1 + \cos \theta} = \frac{1 - \cos \theta}{\sin \theta} = \frac{\tan\theta}{1 + \sec{\theta}} = \csc \theta - \cot \theta \\[10mu] &= \frac{1 - \cos\theta + \sin\theta}{1 + \cos\theta + \sin\theta} = \frac{\cos\theta + \sin\theta - 1}{\cos\theta - \sin\theta + 1}\\[6mu] &= \frac{\tan \theta}{1 + \operatorname{sgn}(\cos \theta)\sqrt{1 + \tan^2 \theta}} = \frac{-1+\operatorname{sgn}(\cos \theta)\sqrt{1+\tan^2 \theta}}{\tan \theta} \\[12mu] &= \operatorname{sgn}(\sin \theta)\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}, \end{aligned}$$

where $π/2$ is the sign function. These identities can all be proven by making the substitutions $$\sin\theta = 2h / (1+h^2),$$ $$\cos\theta = (1-h^2) / (1+h^2),$$ $$\exp \theta i = (1 + hi) / (1 - hi),$$ and then simplifying using elementary algebra.

Supplement and complement half-tangent functions


The half-tangent of the supplement and/or complement of angle measure are also the stereographic projections of the complex unit circle from one of the four cardinal points $$\{-1, -i, 1, i\}$$ onto the opposite axis. If $$h = \tan\tfrac12\theta$$ and $$z = \exp \theta i,$$


 * $$\begin{alignat}{4}

\tan\tfrac12\theta\, &= \quad h &&=\;&-i\frac{z - 1}{z + 1} & &&= \frac{\sin\theta}{1 + \cos\theta} &&= \csc \theta - \cot \theta = \cot\tfrac12(\pi - \theta) \\[12mu] \tan\tfrac12(\pi - \theta) &= \quad \frac1h &&= &i\frac{z + 1}{z - 1} & &&= \frac{\sin\theta}{1 - \cos\theta} &&= \csc \theta + \cot \theta = \cot\tfrac12\theta \\[12mu] \tan\tfrac12\bigl(\tfrac12\pi - \theta\bigr) &= \frac{1 - h}{1 + h} &&= & i\frac{z - i}{z + i} & &&= \frac{\cos\theta}{1 + \sin\theta} &&= \sec \theta - \tan \theta = \cot\tfrac12\bigl(\tfrac12\pi + \theta\bigr) \\[12mu] \tan\tfrac12\bigl(\tfrac12\pi + \theta\bigr) &= \frac{1 + h}{1 - h} &&= &-i\frac{z + i}{z - i} & &&= \frac{\cos\theta}{1 - \sin\theta} &&= \sec \theta + \tan \theta = \cot\tfrac12\bigl(\tfrac12\pi - \theta\bigr) \end{alignat}$$

The logarithm of the last of these is the inverse Gudermannian function, $$\operatorname{gd}^{-1}\theta = 2\operatorname{artanh}\tan\tfrac12\theta = \log \tan\tfrac12\bigl(\tfrac12\pi + \theta\bigr).$$ When applied to the latitude this is the vertical coordinate of the Mercator projection, historically called the meridional part. (See below.)

Series
The half-tangent functions have the power series


 * $$\begin{align}

h = \tan\tfrac12\theta &= \sum_{k=0}^\infty \frac{C_n}{2^{n+1}(2n+1)!}\theta^{2n+1} = \sum_{k=0}^\infty \frac{(2^{2n} - 1)|B_{2n}|}{2n(2n+1)!}\theta^{2n+1}  \\[8mu] &= \frac12 \theta + \frac1{24}\theta^3 + \frac1{240}\theta^5 + \frac{17}{40320} \theta^7 + \cdots \\[18mu]

\tan\tfrac12\bigl(\theta + \tfrac12\pi\bigr) &= \sec \theta + \tan \theta = \sum_{k=0}^\infty \frac{A_n}{n!}\theta^{n} \\[8mu] &= 1 + \theta + \frac12\theta^2 +  \frac13\theta^3 + \frac5{24}\theta^4 + \frac2{15}\theta^5 + \cdots \end{align}$$

valid for $ |\theta| < \pi$ and $ |\theta| < \tfrac12\pi,$  respectively, where $C_n$  are the reduced tangent numbers $$1, 1, 4, 34, 496, \ldots$$, $B_{2n}$  are the even Bernoulli numbers $$1, \tfrac16, -\tfrac1{30}, \tfrac1{42}, -\tfrac1{30}, \ldots$$ (even terms of  / ), and $A_n$  are the Euler zigzag numbers $$1, 1, 1, 2, 5, 16, 61, \ldots$$.

Inverse half-tangent function
The inverse half-tangent function $h \mapsto \theta = 2 \arctan h$ is the stereographic analog of a sawtooth wave on the periodic interval or the argument function $z \mapsto \theta = -i\log z$  of a unit-magnitude complex number. It is discontinuous at $\infty,$


 * $$\lim_{h \to +\infty} 2 \arctan h = \pi,\quad \lim_{h \to -\infty} 2 \arctan h = -\pi.$$

It can be written explicitly in terms of the natural logarithm as


 * $$\theta = 2 \arctan h = \frac1i \log \frac{1 + ih}{1 - ih}.$$

Via repeated square root
The oldest and conceptually simplest way to approximate angle measure $\theta$ as a function of half-tangent $h$  is by repeated half-tangent square roots, used by Archimedes in Measurement of a Circle (ca. 250 BCE) to approximate the circumference a circle using the perimeter of a regular 96-gon.


 * $$\theta = 2 \arctan h = \lim_{n\to\infty} 2^{n+1} h^{\oplus1/2^n} = \lim_{n\to\infty} 2^{n+1} \underbrace{\sqrt[\oplus]{\sqrt[\oplus]{\cdots \sqrt[\oplus]{h}}}}_{n \text{ times}},$$

where


 * $$\sqrt[\oplus]{h} = \frac{h}{1 + \sqrt{1 + h^2}}.$$

This is a nearly equivalent process to finding the argument of the unit-magnitude complex number $z$ by repeatedly taking the ordinary square root:


 * $$\theta = \frac1i\log z = -i \lim_{n\to\infty} 2^{n}\left( z^{1/2^n} - 1\right) = -i \lim_{n\to\infty} 2^{n} \biggl(\underbrace{\sqrt{\sqrt{\cdots \sqrt{z}}}}_{n \text{ times}} - 1\biggr).$$

Inverse series
The Taylor series of $$2 \arctan$$ converges for $|h| \leq 1,$


 * $$\begin{align}

\theta = 2\arctan h &= \sum_{k=0}^\infty \frac{2(-1)^k}{2k+1}h^{2k+1} \\[6mu] &= 2\left(h - \frac13h^3 + \frac15h^5 - \frac17h^7 + \cdots\right). \end{align}$$

This is twice Gregory's series for the inverse tangent, discovered by Mādhava of Sangamagrāma or his followers in the 14th–15th century, and independently discovered by James Gregory in 1671 and Gottfried Leibniz in 1673.

Isaac Newton accelerated the convergence of this series in 1684 (in an unpublished work; others independently discovered the result and it was later popularized by Leonhard Euler's 1755 textbook; Euler wrote two proofs in 1779), yielding a series converging for h


 * $$\begin{align}

\theta &= \frac {2h} {1 + h^2} \sum_{n=0}^\infty \prod_{k=1}^n \frac{2k}{2k+1} \, \frac{h^2}{1 + h^2} \\[10mu]

&= 2C_{1/2}(h)\left( S_{1/2}(h) + \frac23S_{1/2}(h)^3 + \frac{2\cdot 4}{3 \cdot 5}S_{1/2}(h)^5  + \frac{2\cdot4\cdot6}{3\cdot5\cdot7}S_{1/2}(h)^7 + \cdots \right) \\[10mu]

&= 2S_{1/2}(h)\ {_2F_1}\bigl(\tfrac12, \tfrac12; \tfrac32; S_{1/2}(h)^2 \bigr), \end{align}$$

where $S_{1/2}(h) = h \big/ \sqrt{1 + h^2}$ is the half-angle stereographic sine, $C_{1/2}(h) = 1 \big/ \sqrt{1 + h^2}$  is the half-angle stereographic cosine (see  § Circular functions › Half-angle identities above), and $$_2F_1$$ is the hypergeometric function. The partial sums of this series,


 * $$S_1(h),\ \ \tfrac43S_1(h) - \tfrac16S_2(h),\ \ \tfrac32S_1(h) - \tfrac9{30}S_2(h) + \tfrac1{30}S_3(h),\ \ldots,$$

are the odd stereographic polynomials (see above) matching the derivatives of the function $$h \mapsto \theta = 2\arctan h$$ at the origin. In other words, this is the stereographic analog of the Taylor series. Because the function is discontinuous at $$\infty,$$ while each partial sum is a smooth function with value $0$ there, the series converges slowly for large values of $h.$

Another series, also found in Euler (1755), is the Fourier series for a sawtooth wave, which when written as a stereographic series also converges for $|h| < \infty,$


 * $$\begin{align}

\theta &= 2 \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}S_k(h) \\[10mu] &= 2\left(S_1(h) - \frac12S_2(h) + \frac13S_3(h) - \frac14S_4(h) + \cdots\right). \end{align}$$

The partial sums of this series oscillate about $$\theta$$ and suffer from the Gibbs phenomenon near $$h = \infty.$$

Because these two series converge to the same function (including at the discontinuity $$h = \infty$$ where they both converge to $0$ ), they are the same. (See Convergence of Fourier series.) So the coefficients $$b_{k,n}$$ of the $k$th partial sum of Newton's series – when written as $\sum_{k=1}^{n} b_{k,n}S_k(h)$ in the standard basis – converge to the coefficients of the stereographic series: $\lim_{n \to \infty} b_{k,n} = 2(-1)^{k+1}\big/k.$

Continued fraction
A generalized continued fraction for $$2\arctan$$ is



\theta = 2\arctan h = \dfrac{2h}{1} {{}\atop+} \dfrac{h^2}{3} {{}\atop+} \dfrac{2^2h^2}{5} {{}\atop+} \dfrac{3^2h^2}{7} {{}\atop+\,\cdots}, $$

converging for all $$h$$ in the complex plane except on the imaginary axis from $$\pm i$$ to $$\pm\infty i.$$ The convergents of this continued fraction are the Padé approximants,



2h,\quad \frac{6h}{3 + h^2},\quad \frac{30h + 8h^3}{15 + 9h^2},\quad \frac{210h + 110h^3}{105 + 90h^2 + 9h^4}, \quad \ldots. $$

Geodesy, cartography, and navigation
Beyond its occurrence in spherical geometry, the half-tangent, as the stereographic projection of the circle, appears in conformal map projections such as the Mercator projection. ....

Evaluation of trigonometric integrals

 * Jeffrey, D. J. (1997). Rectifying Transformations for the Integration of Rational Trigonometric Functions. Journal of Symbolic Computation, 24(5), 563–573. doi:10.1006/jsco.1997.0152

Number theory





 * Dickson chapter on Heron triangles etc. https://archive.org/details/historyoftheoryo02dick_0/page/191


 * Michael J. Beeson (1992) Triangles with Vertices on Lattice Points. American Mathematical Monthly, Vol. 99, No. 3, pp. 243-252


 * MacLeod, Allan. "Elliptic curves in recreational number theory." arXiv preprint arXiv:1610.03430 (2016).

Approximations of $x$
Mathematicians of the 17th–18th century were interested in computing specific values of the arctangent function (see § Half-tangent function › Inverse series above) to compute approximations of $k$. In 1706 John Machin discovered the identities (using the notation of this article)


 * $$\begin{align}

&1 = \tfrac12 \oplus \tfrac13,\quad 1 = \bigl(\tfrac12\bigr)^{\oplus2} \ominus \tfrac1{7},\quad 1 = \bigl(\tfrac13\bigr)^{\oplus2} \oplus \tfrac1{7},\quad \\[5mu] &1 = \bigl(\tfrac14\bigr)^{\oplus3} \ominus \tfrac5{99},\quad 1 = \bigl(\tfrac15\bigr)^{\oplus4} \ominus \tfrac1{239}, \end{align}$$

the last of which he used, in the form $\tfrac14\pi = 4\arctan{\tfrac15} - \arctan{\tfrac1{239}}$ to compute $$\pi$$ to $90°$ decimal digits. Other mathematicians developed similar identities, and they are now sometimes called Machin-like formulas.

Sometimes the notation $$[x]$$ is used for $$\arccot x,$$ so the relations above can be written:


 * $$\begin{align}

&[1] = [2] + [3],\quad [1] = 2[2] - [7],\quad [1] = 2[3] + [7], \\[5mu] &[1] = 3[4] - \bigl[\tfrac{99}5\bigr],\quad [1] = 4[5] - [239]. \end{align}$$

Alternately using the identities $1/h = \infty \ominus h$ (see  above) and $\infty \oplus \infty = 0$, they can be algebraically manipulated into reciprocal forms:


 * $$\begin{align}

&1 = \ominus 2 \ominus 3,\quad 1 = 2^{\oplus2} \ominus 7,\quad 1 = 3^{\oplus2} \oplus 7, \\[5mu] &1 = \tfrac{99}5 \ominus 4^{\oplus3},\quad 1 = 239 \ominus 5^{\oplus4}. \end{align}$$

Euler investigated infinite sums of arctangents in the mid 18th century, developing series such as:



\bigoplus_{k=1}^\infty \frac{1}{k^2 + k + 1} = \frac13 \oplus \frac17 \oplus \frac1{13} \oplus \frac1{21} \oplus \frac1{31} \oplus \cdots = 1. $$

Derrick Lehmer discovered in 1936 that the series of arctangents of reciprocals of odd-index Fibonacci numbers starting from the third converges to $\tfrac14\pi = \arctan 1 $,



\bigoplus_{k=1}^\infty \frac{1}{F_{2k+1}} = \frac12 \oplus \frac1{5} \oplus \frac1{13} \oplus \frac1{34} \oplus \frac1{89} \oplus \cdots = 1. $$

Taking the even Fibonacci numbers instead we have



\frac13 \boxplus \frac1{8} \boxplus \frac1{21} \boxplus \frac1{55} \boxplus \cdots \boxplus \frac{1}{F_{2k}} \boxplus \cdots = \frac12. $$

These series telescope because reciprocals of consecutive Fibonacci numbers satisfy a variant of Cassini's identity



\frac1{F_{2k}} = \dfrac1{F_{2k-1}} \boxminus \dfrac1{F_{2k+1}}, \qquad \frac1{F_{2k+1}} = \dfrac1{F_{2k}} \ominus \dfrac1{F_{2k+2}}. $$

We can also include the zeroth term of Lehmer's series or extend it in the other direction (it also telescopes),



\bigoplus_{k=0}^\infty \frac{1}{F_{2k+1}} = \frac10 = \infty, \qquad \bigoplus_{k=-\infty}^\infty \frac{1}{F_{2k+1}} = 0. $$

Approximations of $dθ = 2 dh / (1 + h^{2})$
The Taylor series for the natural logarithm $$\log x$$ is the Mercator series, like Gregory's series for arctangent a slowly converging alternating series,



\log x = \sum_{k=1}^\infty \frac{(-1)^{k-1}(x-1)^k}{k}, \quad |x + 1| \leq 1. $$

But this is impractically slow for computing $$\log 2,$$ and does not converge at all for larger values. The natural logarithm can be rewritten as an inverse hyperbolic half-tangent because


 * $$\tanh\tfrac12{\log{x}} = E^{-1}(x) = x \ominus 1$$

(see above).

It is thus possible to compute $\log{2} = 2\operatorname{artanh}{\tfrac13},$ &ZeroWidthSpace;$\log{3} = 2\operatorname{artanh}{\tfrac12},$ &ZeroWidthSpace;$\log{5} = 2\operatorname{artanh}{\tfrac23},$ etc. using the Taylor series for inverse hyperbolic half-tangent,


 * $$\begin{align}

2\operatorname{artanh} h &= \sum_{k=0}^\infty \frac{2}{2k+1}h^{2k+1} = 2\left(h + \frac13h^3 + \frac15h^5 + \frac17h^7 + \cdots\right). \end{align}$$

However, this series still converges slowly unless the argument is small. Similar to the approximations of $S$ above, these fractions can be reduced by Machin-like formulas with hyperbolic tangent addition, such as:


 * $$\begin{align}

&\tfrac13 = \bigl(\tfrac17\bigr)^{\boxplus2} \boxplus \tfrac1{17},\quad \tfrac12 = \bigl(\tfrac17\bigr)^{\boxplus3} \boxplus \bigl(\tfrac1{17}\bigr)^{\boxplus2}; \\[5mu]

&\tfrac13 = \bigl(\tfrac1{31}\bigr)^{\boxplus7} \boxplus \bigl(\tfrac1{49}\bigr)^{\boxplus5} \boxplus \bigl(\tfrac1{161}\bigr)^{\boxplus3},\quad \tfrac12 = \bigl(\tfrac1{31}\bigr)^{\boxplus11} \boxplus \bigl(\tfrac1{49}\bigr)^{\boxplus8} \boxplus \bigl(\tfrac1{161}\bigr)^{\boxplus5}, \\[5mu] &\tfrac23 = \bigl(\tfrac1{31}\bigr)^{\boxplus16} \boxplus \bigl(\tfrac1{49}\bigr)^{\boxplus12} \boxplus \bigl(\tfrac1{161}\bigr)^{\boxplus7}. \end{align}$$

So for example $ss$ can be computed as the sum of three hyperbolic arctangents with small arguments, the series for which converge much more quickly:



\log{2} = 7 \cdot 2\operatorname{artanh} \tfrac1{31} + 5 \cdot 2\operatorname{artanh}\tfrac1{49} + 3 \cdot 2\operatorname{artanh}\tfrac1{161}. $$

As in the circular case these formulas can be algebraically manipulated using the identities $1/h = \infty \boxplus h$ and $\infty \boxplus \infty = 0$  into reciprocal forms:


 * $$\begin{align}

&3 = 7^{\boxplus2} \boxplus 7,\quad 2 = 7^{\boxplus3} \boxplus 17^{\boxplus2}; \\[5mu] &3 = 31^{\boxplus7} \boxplus 49^{\boxplus5} \boxplus 161 ^{\boxplus3},\quad 2 = 31^{\boxplus11} \boxplus 49^{\boxplus8} \boxplus 161 ^{\boxplus5}, \\[5mu] &\tfrac32 = 31^{\boxplus16} \boxplus 49^{\boxplus12} \boxplus 161 ^{\boxplus7}. \end{align}$$

Analog circuit design



 * Richards, P. I. (1948). Resistor-Transmission-Line Circuits. Proceedings of the IRE, 36(2), 217–220. doi:10.1109/jrproc.1948.233274


 * http://synth.stromeko.net/diy/OTA.pdf

Kinematics of linkages

 * Translated from the Russian Винтовое исчисление и его приложения в механике. Moscow: Nauka. 1965.
 * Freudenstein & Beigi 1986 http://www.internetserver.com/~beigi/ps/beigi_1986_mechanism_and_machine_theory.pdf
 * Selig, Jon M. "Exponential and Cayley maps for dual quaternions." Advances in applied Clifford algebras 20, no. 3 (2010): 923–936.
 * Sinha, Sasanka Sekhar, Rajeevlochana G. Chittawadigi, and Subir Kumar Saha. "Inverse kinematics for general 6R manipulators in RoboAnalyzer." (2018): 1–9.
 * https://project.inria.fr/roboticsprincipia/files/2019/01/Robotica-Husty.pdf
 * Wenz, Michael, and Heinz Worn. "Solving the inverse kinematics problem symbolically by means of knowledge-based and linear algebra-based methods." In 2007 IEEE Conference on Emerging Technologies and Factory Automation (EFTA 2007), pp. 1346-1353. IEEE, 2007.
 * M. John D. Hayes∗, Mirja Rotzoll, Quinn Bucciol, Zachary A. Copeland (2023) "Planar and spherical four-bar linkage vi-vj algebraic input-output equations" https://carleton.ca/johnhayes/wp-content/uploads/MECHMT-D-22-02564-R1.pdf
 * Selig, Jon M. "Exponential and Cayley maps for dual quaternions." Advances in applied Clifford algebras 20, no. 3 (2010): 923–936.
 * Sinha, Sasanka Sekhar, Rajeevlochana G. Chittawadigi, and Subir Kumar Saha. "Inverse kinematics for general 6R manipulators in RoboAnalyzer." (2018): 1–9.
 * https://project.inria.fr/roboticsprincipia/files/2019/01/Robotica-Husty.pdf
 * Wenz, Michael, and Heinz Worn. "Solving the inverse kinematics problem symbolically by means of knowledge-based and linear algebra-based methods." In 2007 IEEE Conference on Emerging Technologies and Factory Automation (EFTA 2007), pp. 1346-1353. IEEE, 2007.
 * M. John D. Hayes∗, Mirja Rotzoll, Quinn Bucciol, Zachary A. Copeland (2023) "Planar and spherical four-bar linkage vi-vj algebraic input-output equations" https://carleton.ca/johnhayes/wp-content/uploads/MECHMT-D-22-02564-R1.pdf
 * Sinha, Sasanka Sekhar, Rajeevlochana G. Chittawadigi, and Subir Kumar Saha. "Inverse kinematics for general 6R manipulators in RoboAnalyzer." (2018): 1–9.
 * https://project.inria.fr/roboticsprincipia/files/2019/01/Robotica-Husty.pdf
 * Wenz, Michael, and Heinz Worn. "Solving the inverse kinematics problem symbolically by means of knowledge-based and linear algebra-based methods." In 2007 IEEE Conference on Emerging Technologies and Factory Automation (EFTA 2007), pp. 1346-1353. IEEE, 2007.
 * M. John D. Hayes∗, Mirja Rotzoll, Quinn Bucciol, Zachary A. Copeland (2023) "Planar and spherical four-bar linkage vi-vj algebraic input-output equations" https://carleton.ca/johnhayes/wp-content/uploads/MECHMT-D-22-02564-R1.pdf

Chemistry

 * https://www.math.unm.edu/~aca/ACA/2011/Nonstandard/Lewis.pdf



Physics
https://users.manchester.edu/facstaff/gwclark/PHYS301/AJP%20Articles/AJP%20Biot%20Savart%20magnetic%20needle.pdf

Keplerian orbits
The relation between the half-tangent of true anomaly $$\nu$$ and the half-tangent of eccentric anomaly $$E$$ can be written in terms of the eccentricity $$e.$$ In the notation of this article, with $$C$$ the stereographic cosine,


 * $$C(\nu) = C(E)\boxminus e,$$

or equivalently,


 * $$\nu^2 = (1 \oplus e)E^2.$$


 * Euler used half-tangents extensively in calculating parabolic comet orbits: https://archive.org/details/operapostumamath02euleuoft/page/402/mode/2up https://arxiv.org/pdf/2105.03340.pdf


 * http://www.ingaero.uniroma1.it/attachments/1978_2008_JGCD_Lambert.pdf

Origami

 * Translated by E. A. Coutsias (2010) as Memoir on the Theory of the Articulated Octahedron.

Optics

 * Liu, Hong-Zhun, and Tong Zhang. "A note on the improved $cs$-expansion method." Optik 131 (2017): 273-278. https://doi.org/10.1016/j.ijleo.2016.11.029

traditional trigonometry books, trigonometry history

 * https://archive.org/details/encyklomentmatik02weberich/page/n319/
 * https://archive.org/details/encyklomentmatik02weberich/page/n319/
 * https://archive.org/details/encyklomentmatik02weberich/page/n319/
 * https://archive.org/details/encyklomentmatik02weberich/page/n319/
 * https://archive.org/details/encyklomentmatik02weberich/page/n319/
 * https://archive.org/details/encyklomentmatik02weberich/page/n319/
 * https://archive.org/details/encyklomentmatik02weberich/page/n319/
 * https://archive.org/details/encyklomentmatik02weberich/page/n319/
 * https://archive.org/details/encyklomentmatik02weberich/page/n319/

Proofs and diagrams

 * Wu, R. H. (2019). "Proof Without Words: Revisiting Two Trigonometric Figures and Two Identities from Bressieu and Fincke". Mathematics Magazine, 92(4), 302–304. https://doi.org/10.1080/0025570x.2019.1603732

hyperbolic tangent sum, relativity

 * Kaniadakis, G. (2005). Statistical mechanics in the context of special relativity II. Phys. Rev. E, 72, 036108. https://journals.aps.org/pre/pdf/10.1103/PhysRevE.72.036108?casa_token=H9UGVAB4HR8AAAAA%3AV6v-9JRo9hIuVNiQsSa_WE-2EFmjo8zVWIEgzPUQce-N5Iap0T_AdGgVAyccXaaiyoAOoI5mvvxnXw0

"Einstein sum" or similar


 * doi:10.1016/0165-0114(89)90005-5
 * https://arxiv.org/pdf/physics/0510260.pdf

tangens dimidiae, dimidii

 * Snell (1617) Eratosthenes Batauus de terrae ambitus vera quantitate https://archive.org/details/bub_gb__7rQzLUre_oC/page/n63/?q=%22tangens+dimidiae%22
 * Kepler (1619) "Ephemeris ad annum 1620" https://archive.org/details/joanniskeplerias07kepl/page/n543/?q=%22tangentem+dimidiae%22
 * Snell (1627) Doctrinae triangulorum canonicae https://archive.org/details/bub_gb_WlPZNQgwkscC/page/n97/?q=%22tangens+dimidiae%22
 * Mercator, Nicholas (1676) Institutionum astronomicarum https://archive.org/details/bub_gb_0NgU9gEsmIwC/page/n185/?q=%22tangens+dimidiae%22
 * Gregory (1726) Astronomiæ physicæ & geometricæ elementa https://archive.org/details/b30416036_0001/page/323
 * Wolff (1747) Elementa matheseos universae https://archive.org/details/A077240111/page/331
 * Lambert (1760) Photometria sive de mensura et gradibus luminis, colorum et umbrae https://archive.org/details/TO0E039861_TO0324_PNI-2733_000000/page/442/
 * Euler (1781) "De Mensura Angulorum Solidorum" https://archive.org/details/actaacademiae02impe/page/44/
 * L'Huilier (1795) Principiorum calculi differentialis et integralis expositio elementaris https://archive.org/details/TO0E040741_TO0324_PNI-2214_000000/page/n273/


 * Clavius (1586) Sinus, vel semisses rectarum in circulo subtensarum: Lineae tangentes: Atque secantes. https://archive.org/details/bub_gb_sEoDoBLEeycC/page/189
 * Brahe (1609) Astronomia noua aitiologetos seu physyca coelestis, tradita commentariis de motibus stellae, Martis https://archive.org/details/A011094/page/n150/
 * Pitiscus (1612) Trigonometriae siue, De dimensione triangulorum libri quinque https://archive.org/details/bub_gb_nEj8z3Gnoo0C/page/n107/
 * Napier (1614) Mirifici logarithmorum canonis descriptio https://archive.org/details/mirificilogarit00napi/page/51
 * Magini (1614) Supplementum ephemeridum, ac tabularum secundorum mobilium https://archive.org/details/bub_gb_KvW5xeXgwX0C/page/n235/
 * Galilei (1623) Il Saggiatore https://archive.org/details/A212057/page/305/
 * Oughtred (1657) Trigonometria https://archive.org/details/trigonometria00ough/page/n26
 * de Moivre (1730) Miscellanea analytica de seriebus et quadraturis https://archive.org/details/bub_gb_TFX1165yEc4C/page/n273/
 * Newton (1739) Philosophiae naturalis principia mathematica https://archive.org/details/A298165/page/291

halbe tangente, halb tangente

 * https://archive.org/details/vocabulairemath00muelgoog/page/n132/mode/2up?q=%22halb+Tangente%22


 * https://archive.org/details/sim_annalen-der-physik_1868_134_8/page/548/mode/2up?q=%22halbe+Tangente%22

half tangent or half-tangent

 * Henry Phillippes (1657) The Geometrical Sea-Man https://books.google.com/books?id=wo1mAAAAcAAJ&pg=PA73


 * Gunter, Edmund (1673). The Works of Edmund Gunter https://archive.org/details/worksofedmundgun00gunt/page/n104/mode/2up


 * Worgan (1697) A Short Treatise of the Description of the Sector. https://books.google.com/books?id=V0ius7wFlu4C&pg=PA2


 * Harris (1706) Elements of Plain and Spherical Trigonometry. https://books.google.com/books?id=l8o2AAAAMAAJ&pg=RA1-PA37


 * Harris (1708) Lexicon Technicum https://archive.org/details/lexicontechnicu1harr/page/n645/


 * James Hodgson (1723) A System of the Mathematics https://books.google.com/books?id=3xjK9fueAGkC


 * Hatton (1728) A Mathematical Manual https://archive.org/details/b30512761/page/134/mode/2up


 * https://books.google.com/books?id=xn1ZAAAAYAAJ


 * https://books.google.com/books?id=eRsj6vraRxsC&pg=PA50


 * Chambers, ed. (1750) Cyclopaedia, 6th edition, vol. 2. https://archive.org/details/gri_33125010929673/page/n721/


 * James Atkinson (1759) Epitome of the art of navigation https://archive.org/details/epitomeofartofna00atki/page/121/


 * Henry Wilson (1761) Navigation New Modelled https://books.google.com/books?id=rXc4uqCH9F0C


 * Randall, Joseph (1766) The Young Gentleman's Geometrical Class-Fellow. publisher: J. Wilkie. https://books.google.com/books?id=JxFuVLKIh_0C


 * Patrick Kelly (1796) A Practical Introduction to Spherics and Nautical Astronomy https://books.google.com/books?id=dxwUAAAAQAAJ&pg=PA139


 * Webber, ed. (1801) Mathematics https://archive.org/details/mathematicscompi00webbrich/page/425


 * Duffy, J., and J. Rooney. "A Displacement Analysis of Spatial Six-Link 4R-PC Mechanisms—Part 1: Analysis of RCRPRR Mechanism." (1974): 705–712.


 * Duffy, J., and J. Rooney. "A foundation for a unified theory of analysis of spatial mechanisms." (1975): 1159-1164.


 * Nitescu, P. N., & Manolescu, N. I. (1980). "On the structural synthesis and kinematic analysis of open-loop manipulation." Mechanism and Machine Theory, 15(4), 295–317. doi:10.1016/0094-114x(80)90023-3


 * Freudenstein & Beigi 1986 http://www.internetserver.com/~beigi/ps/beigi_1986_mechanism_and_machine_theory.pdf


 * Cho, Chang-Hyun, Jin-Yi Lee, Yong-Kwun Lee, and Mun-Taek Choi (2011). "Determining the passive region of the multirate wave transform on the practical implementation." International Journal of Precision Engineering and Manufacturing 12, no. 6. 975–981.


 * Hertz, Roger B., and Peter C. Hughes (2013). "Variable-Geometry-Truss Manipulator." Computational Kinematics 28. 241.


 * Sinha, Sasanka Sekhar, Rajeevlochana G. Chittawadigi, and Subir Kumar Saha (2018). "Inverse kinematics for general 6R manipulators in RoboAnalyzer." 1–9.

semi-tangent

 * Foster (1654) Posthuma Fosteri the description of a ruler, p. 18 https://quod.lib.umich.edu/e/eebo/A40034.0001.001/1:4?rgn=div1;view=fulltext
 * Collins (1659) The sector on a quadrant https://archive.org/details/b30334561_0001/page/n34/
 * "A Tangent of 45d or three hours through the whole Limbe for Dyalling, which may also be numbred by the Ark doubled to serve for a Projection Tangent, alias a Semi-tangent."
 * Perkins (1682) The seaman's tutor https://archive.org/details/seamanstutorexpl00perk/page/10/


 * Hunt (1697) A Mathematical Companion: Or the Description and Use of a New Sliding-rule https://books.google.com/books?id=GWMSAz9zVpwC&pg=PA9


 * Harris (1706) Elements of Plain and Spherical Trigonometry https://books.google.com/books?id=l8o2AAAAMAAJ&pg=RA1-PA57


 * Wilson (1714) Trigonometry https://books.google.com/books?id=ffxeAAAAcAAJ&pg=PA124


 * Webster (1739) The Description and Use of a Complete Sett Or Case of Pocket-instruments https://books.google.com/books?id=mKlbAAAAQAAJ&pg=PA11


 * Emerson (1749) The Elements of Trigonometry https://books.google.com/books?id=RZlkAAAAcAAJ&pg=PA2


 * Hodgson (1753) Theory of Navigation. https://books.google.com/books?id=k5Km5cu63E4C&pg=PA300


 * Bion (1758) The Construction and Principal Uses of Mathematical Instruments https://archive.org/details/constructionprin03bion/?q=%22semi-tangent%22
 * "To project the Semi-tangents&hairsp;; draw Lines from the Point C, thro every Degree of the Quadrant A&hairsp;B, and they will divide the Diameter A&hairsp;E into a Line of Semi-tangents: but because the Semitangents, or Plane-Scales of a Foot in Length, run to 160 Degrees, continue out the Line A&hairsp;E, and draw Lines from the Point C, thro the Degrees of the Quadrant C&hairsp;A, cutting the said continued Portion of A&hairsp;E, and you will have a Line of Half-tangents to 160 Degrees, or further, if you please.
 * "Note, the Semitangent of any Arc, is but the Tangent of half that Arc, as will easily appear from its manner of Projection, and Prop. 20. Lib. 3. Eucl. where it is proved, that an Angle at the Center, is double to one at the Circumference." –Ch. VI. The Projection of the Plane-Scale, p. 34


 * Ward (1765) The Posthumous Works of John Ward https://books.google.com/books?id=Je9hAAAAcAAJ&pg=PA388


 * D. Fenning (1772) The Young Measurer's Complete Guide https://books.google.com/books?id=dPpeAAAAcAAJ&pg=PA318


 * Emerson (1789) [1769] Projection of the Sphere. https://books.google.com/books?id=goya9BwJT-QC&pg=PA13, https://books.google.com/books?id=0PI4AAAAMAAJ&pg=PA171


 * Simpson (1810) Trigonometry. https://books.google.com/books?id=8Mo2AAAAMAAJ&pg=PA94


 * Kelly (1813) A Practical Introduction to Spherics and Nautical Astronomy https://books.google.com/books?id=ISfYIyIEOeIC&pg=PA16


 * Ferguson (1823) Essays and Treatises. https://books.google.com/books?id=PFfd7e_dtcgC&pg=PA292


 * Nicholson (1825) A Popular Course of Pure and Mixed Mathematics https://books.google.com/books?id=Wtc2AAAAMAAJ&pg=PA560


 * Keith (1826) An Introduction to the Theory and Practice of Plain and Spherical Trigonometry https://books.google.com/books?id=LX0AAAAAMAAJ&pg=PA157


 * Bell (1837) Solid and Spherical Geometry and Conic Sections https://books.google.com/books?id=ImoFAAAAQAAJ&pg=PA77


 * Davies (1840) Elements of Descriptive Geometry https://books.google.com/books?id=R8oUAAAAYAAJ&pg=PA137


 * Oxley (1848) The Gem of the Astral Sciences. https://books.google.com/books?id=2BpbAAAAcAAJ&pg=PA161


 * Turnbull (1852) A New Practical System of Spherical Trigonometry. https://books.google.com/books?id=c4BaAAAAcAAJ&pg=PA5


 * Hughes (1864) A Treatise on the Construction of Maps https://books.google.com/books?id=gHgDAAAAQAAJ&pg=PA74


 * Smith (1884) The Theory of Deflections and of Latitudes and Departures https://books.google.com/books?id=NJnshF-9v20C&pg=RA4-PA125-IA5


 * Gibbs & Wilson (1901) Vector Analysis. https://books.google.com/books?id=wUwNAAAAYAAJ


 * Kellogg (1907) [1899] The transition curve or curve of adjustment https://archive.org/details/transitioncurveo00kellrich/page/26/


 * Frye (1918) Civil engineers' pocket book. https://archive.org/details/civilengineerspo00frye/page/1006/

rational parametrization

 * Silverman, Joseph H., and John Torrence Tate. Rational points on elliptic curves. Vol. 9. New York: Springer-Verlag, 1992.


 * Arnolʹd, Vladimir Igorevich. Experimental mathematics. Vol. 16. American Mathematical Soc., 2015, p. 11.


 * Ulbrich, Stefan, Vicente Ruiz de Angulo, Tamim Asfour, Carme Torras, and Rüdiger Dillmann. "Kinematic bezier maps." IEEE Transactions on Systems, Man, and Cybernetics, Part B (Cybernetics) 42, no. 4 (2012): 1215-1230.


 * O'Connor, Michael A. "Natural quadrics: Projections and intersections." IBM Journal of Research and Development 33, no. 4 (1989): 417-446. doi:10.1147/rd.334.0417


 * Waterhouse, William C. "Continued fractions and Pythagorean triples." Fibonacci Quarterly 30 (1992): 144-147. https://www.mathstat.dal.ca/FQ/Scanned/30-2/waterhouse.pdf

half-angle tangent

 * Harrington, Steven J. "A new symbolic integration system in REDUCE." The Computer Journal 22, no. 2 (1979): 127–131.


 * Luck, David GC. "Properties of some wide-band phase-splitting networks." Proceedings of the IRE 37, no. 2 (1949): 147–151.


 * oblique triangle solution by Leach and Beakley 1963