User:Jacobolus/HalfTan/Ext

This article is a draft, not yet ready for inclusion into the Wikipedia main namespace. This is a fork of certain sections of User:Jacobolus/HalfTan with trigonometry written in terms of exterior angles rather than interior angles.

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Planar trigonometry

Planar trigonometry (the metrical relations between angles and sides of a triangle in the Euclidean plane) can be described in terms of half-tangents instead of angle measures. Let ${\textstyle a,}$ ${\textstyle b,}$ and ${\textstyle c}$ be the lengths of the sides of a planar triangle. Let the respective exterior angles opposite each side have half-tangents ${\textstyle \alpha ,}$ ${\textstyle \beta ,}$ and ${\textstyle \gamma .}$ Then ${\textstyle 1/\alpha ,}$ ${\textstyle 1/\beta ,}$ and ${\textstyle 1/\gamma }$ are their supplements, the respective interior-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,

${\begin{aligned}{\frac {1}{\alpha }}\oplus {\frac {1}{\beta }}\oplus {\frac {1}{\gamma }}&=\infty ,&{\frac {1}{\alpha }}\oplus {\frac {1}{\beta }}&=\gamma ,\\[6mu]\alpha \oplus \beta \oplus \gamma &=0,&\alpha \oplus \beta &=-\gamma .\end{aligned}}$

Fully expanded in terms of ordinary addition and multiplication,

${\begin{aligned}{\frac {1}{\alpha }}{\frac {1}{\beta }}+{\frac {1}{\alpha }}{\frac {1}{\gamma }}+{\frac {1}{\beta }}{\frac {1}{\gamma }}&=1,\\[6mu]\alpha +\beta +\gamma &=\alpha \beta \gamma .\end{aligned}}$

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 two equivalent equations, the first of which is a simple modification of the law of cotangents and the second of which is the law of cosines written in terms of half-tangents, where $C(h)=(1-h^{2})/(1+h^{2})$ is the stereographic cosine.

${\begin{aligned}\gamma ^{2}&={\frac {(b+a)^{2}-c^{2}}{c^{2}-(b-a)^{2}}}={\frac {(a+b-c)(a+b+c)}{(-a+b+c)(a-b+c)}},\\[10mu]c^{2}&={\frac {\gamma ^{2}(b+a)^{2}+(b-a)^{2}}{\gamma ^{2}+1}}=a^{2}+b^{2}-2ab\,C(\gamma ).\end{aligned}}$

and likewise for ${\textstyle \beta }$ and ${\textstyle \alpha .}$ (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 ${\textstyle \{\alpha ,\beta ,\gamma \}}$ and ${\textstyle \{-\alpha ,-\beta ,-\gamma \}}$ 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,

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

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

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

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),

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

This simplifies to the law of sines,

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

where the common ratio ${\textstyle D}$ is the diameter of the circumcircle of the triangle.

This can alternately be written

${\frac {a+a}{\alpha \boxplus \alpha }}={\frac {b+b}{\beta \boxplus \beta }}={\frac {c+c}{\gamma \boxplus \gamma }}.$

Triangle area

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

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

$\varepsilon ^{2}={\tfrac {1}{4}}(-a+b+c)(a-b+c)(a+b-c)(a+b+c).$

As corollaries,

$\varepsilon \gamma ={\tfrac {1}{2}}(a+b-c)(a+b+c),\quad {\frac {\varepsilon }{\gamma }}={\tfrac {1}{2}}(-a+b+c)(a-b+c),$

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

$\alpha \beta \gamma \varepsilon ={\tfrac {1}{2}}(a+b+c)^{2}.$

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

Circumcircle, incircle, and excircles

The diameter ${\textstyle D}$ of the triangle's circumscribed circle (circumcircle) is

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

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,

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

The diameter ${\textstyle d}$ of the triangle's inscribed circle (incircle) is

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

The diameter ${\textstyle d_{a}}$ of the triangle's escribed circle (excircle) touching side ${\textstyle a}$ is

${\begin{aligned}d_{a}^{2}&={\frac {(a-b+c)(a+b-c)(a+b+c)}{-a+b+c}},\\[8mu]d_{a}&={\frac {2\varepsilon }{-a+b+c}}={\frac {a+b+c}{\alpha }},\end{aligned}}$

and likewise for the excircles touching sides ${\textstyle b}$ and ${\textstyle c}$.

As a corollary,

$d_{a}\alpha =d_{b}\beta =d_{c}\gamma =d\alpha \beta \gamma =a+b+c.$

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,

$h_{a}^{2}={\frac {1}{4a^{2}}}(-a+b+c)(a-b+c)(a+b-c)(a+b+c).$

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

${\frac {1}{h_{a}}}+{\frac {1}{h_{b}}}+{\frac {1}{h_{c}}}={\frac {2}{d}}.$

Right triangles

The triangle is called a right triangle when one angle $\gamma =1$ is a right angle. The side ${\textstyle c}$ 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, ${\textstyle \alpha \oplus \beta =1,}$ and can be computed in terms of the sides as

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

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

$S(\alpha )=-C(\beta )={\frac {a}{c}},\quad -C(\alpha )=S(\beta )={\frac {b}{c}}.$

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 ${\textstyle a,}$ ${\textstyle b,}$ and ${\textstyle c}$ be the half-tangents of the central angles subtending sides of a spherical triangle (the "sides"). Let the exterior dihedral angles at the vertices opposite each side have respective half-tangents ${\textstyle \alpha ,}$ ${\textstyle \beta ,}$ and ${\textstyle \gamma }$ (the "exterior angles"). Then ${\textstyle 1/\alpha ,}$ ${\textstyle 1/\beta ,}$ and ${\textstyle 1/\gamma }$ are their supplements, the respective interior-dihedral-angle half-tangents (the "interior 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 ${\textstyle \varepsilon }$ be the half-tangent of the triangle's spherical excess.

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

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

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

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

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

Relations between dihedral and central angles

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

${\textstyle C(c)=C(a)C(b)-S(a)S(b)C(\gamma ),}$

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

${\begin{aligned}\gamma ^{2}&={\frac {(b+a)^{2}-c^{2}(1-ab)^{2}}{c^{2}(1+ab)^{2}-(b-a)^{2}}}={\frac {(a+b-c+abc)(a+b+c-abc)}{({-a}+b+c+abc)(a-b+c+abc)}},\\[10mu]c^{2}&={\frac {(b+a)^{2}+\gamma ^{2}(b-a)^{2}}{(1-ab)^{2}+\gamma ^{2}(1+ab)^{2}}}={\frac {a^{2}\boxplus b^{2}+S(ab)C(\gamma )}{1-S(ab)C(\gamma )}},\end{aligned}}$

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

As corollaries,

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

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 ${\textstyle c}$ results in the spherical law of tangents,

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

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

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

This simplifies to the spherical law of sines,

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

which can alternately be written

${\frac {a\boxplus a}{\alpha \boxplus \alpha }}={\frac {b\boxplus b}{\beta \boxplus \beta }}={\frac {c\boxplus c}{\gamma \boxplus \gamma }}.$

The spherical law of cosines for sides relates one side ${\textstyle c}$ to the three angles. Because of the duality between sides and exterior angles, every relation in spherical trigonometry still holds when the sides and exterior angles are interchanged. In terms of half-tangents,

${\textstyle C(\gamma )=C(\alpha )C(\beta )-S(\alpha )S(\beta )C(c),}$

When expanded as a rational equation then simplified this is

${\begin{aligned}c^{2}&={\frac {(\beta +\alpha )^{2}-\gamma ^{2}(1-\alpha \beta )^{2}}{\gamma ^{2}(1+\alpha \beta )^{2}-(\beta -\alpha )^{2}}}={\frac {(\alpha +\beta -\gamma +\alpha \beta \gamma )(\alpha +\beta +\gamma -\alpha \beta \gamma )}{({-\alpha }+\beta -\gamma +\alpha \beta \gamma )(\alpha -\beta -\gamma +\alpha \beta \gamma )}},\\[10mu]\gamma ^{2}&={\frac {(\beta +\alpha )^{2}+c^{2}(\beta -\alpha )^{2}}{(1-\alpha \beta )^{2}+c^{2}(1+\alpha \beta )^{2}}}={\frac {\alpha ^{2}\boxplus \beta ^{2}+S(\alpha \beta )C(c)}{1-S(\alpha \beta )C(c)}},\end{aligned}}$

and likewise for ${\textstyle a}$ and ${\textstyle b.}$

As corollaries,

${\begin{aligned}ab&={\frac {\alpha +\beta +\gamma -\alpha \beta \gamma }{\alpha +\beta -\gamma +\alpha \beta \gamma }}={\frac {\alpha \oplus \beta }{\gamma }}\oplus 1,&ab\ominus 1&={\frac {\alpha \oplus \beta }{\gamma }},\\[10mu]{\frac {a}{b}}&=\,\,{\frac {{-\alpha }+\beta +\gamma +\alpha \beta \gamma }{\alpha -\beta +\gamma +\alpha \beta \gamma }}\ \,=1\ominus {\frac {\alpha \ominus \beta }{\gamma }},&1\ominus {\frac {a}{b}}&={\frac {\alpha \ominus \beta }{\gamma }}.\end{aligned}}$

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,

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

One last set of relations between all six parts:

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

This can alternately be rewritten in any of sixteen total ways because:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle \begin{alignat}{3} \frac{a + b + c - abc}{\alpha\beta\gamma} &= \ \frac{-a + b + c - abc}{\alpha} &&= \ \frac{a - b + c + abc}{\beta} &&= \frac{a + b - c + abc}{\gamma} \\[5mu] \frac{\alpha + \beta + \gamma - \alpha\beta\gamma}{abc} &= \frac{-\alpha + \beta + \gamma + \alpha\beta\gamma}{a} &&= \frac{\alpha - \beta + \gamma + \alpha\beta\gamma}{b} &&= \frac{\alpha + \beta - \gamma + \alpha\beta\gamma}{c} \end{alignat}}

Spherical excess

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

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

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

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

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:

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

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

${\begin{aligned}{\frac {\varepsilon }{\gamma }}&={\frac {1}{f}}(-a+b+c+abc)(a-b+c+abc),\\[10mu]\varepsilon \gamma &={\frac {1}{f}}(a+b-c+abc)(a+b+c-abc)\end{aligned}}$

and likewise for ${\textstyle \alpha }$ and ${\textstyle \beta }$. Furthermore,

$\varepsilon \alpha \beta \gamma ={\frac {1}{f}}(a+b+c-abc)^{2}.$

Spherical triangles where the half-tangents of central angles ${\textstyle a,}$ ${\textstyle b,}$ ${\textstyle c,}$ and the half-tangent of excess ${\textstyle \varepsilon }$ are all rational numbers are called spherical Heron 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 ${\textstyle D}$. (The half-tangent of the radius is ${\sqrt[{\oplus }]{D}}$.)

For clarity, define

${\begin{aligned}f&=2+a^{2}+b^{2}+c^{2}-a^{2}b^{2}c^{2},\\[6mu]g^{2}&={\tfrac {1}{4}}f^{2}(1+\varepsilon ^{2})=(1+a^{2})(1+b^{2})(1+c^{2}).\end{aligned}}$

Then the half-tangent ${\textstyle D}$ of the diameter of the circumcircle is

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

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 ${\textstyle d,}$

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

The half-tangent of the diameter ${\textstyle d_{a}}$ of the triangle's escribed circle (excircle) touching side ${\textstyle a}$ is

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

and likewise for the excircles touching sides ${\textstyle b}$ and ${\textstyle c}$.

As a corollary,

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

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

$\varepsilon =ab=-(\alpha \oplus \beta \oplus 1).$

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:

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

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

$S(\alpha )={\frac {S(a)}{S(c)}},\quad S(\beta )={\frac {S(b)}{S(c)}}.$

References

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