Editing Spherical trigonometry (section)

==Identities==

===Supplemental cosine rules===
Applying the cosine rules to the polar triangle gives (Todhunter,<ref name=todhunter/> Art.47), ''i.e.'' replacing {{mvar|A}} by {{math|{{pi}} – ''a''}}, {{mvar|a}} by {{math|{{pi}} – ''A''}} etc.,
<math display=block>\begin{align}
\cos A &= -\cos B \, \cos C + \sin B \, \sin C \, \cos a, \\
\cos B &= -\cos C \, \cos A + \sin C \, \sin A \, \cos b, \\
\cos C &= -\cos A \, \cos B + \sin A \, \sin B \, \cos c.
\end{align}</math>

===Cotangent four-part formulae===
The six parts of a triangle may be written in cyclic order as ({{mvar|aCbAcB}}). The cotangent, or four-part, formulae relate two sides and two angles forming four ''consecutive'' parts around the triangle, for example ({{mvar|aCbA}}) or {{mvar|BaCb}}). In such a set there are inner and outer parts: for example in the set ({{mvar|BaCb}}) the inner angle is {{mvar|C}}, the inner side is {{mvar|a}}, the outer angle is {{mvar|B}}, the outer side is {{mvar|b}}. The cotangent rule may be written as (Todhunter,<ref name=todhunter/> Art.44)
<math display=block>
  \cos\!\Bigl(\begin{smallmatrix}\text{inner} \\ \text{side}\end{smallmatrix}\Bigr)
  \cos\!\Bigl(\begin{smallmatrix}\text{inner} \\ \text{angle}\end{smallmatrix}\Bigr) =
  \cot\!\Bigl(\begin{smallmatrix}\text{outer} \\ \text{side}\end{smallmatrix}\Bigr)
  \sin\!\Bigl(\begin{smallmatrix}\text{inner} \\ \text{side}\end{smallmatrix}\Bigr) -
  \cot\!\Bigl(\begin{smallmatrix}\text{outer} \\ \text{angle}\end{smallmatrix}\Bigr)
  \sin\!\Bigl(\begin{smallmatrix}\text{inner} \\ \text{angle}\end{smallmatrix}\Bigr),
</math>
and the six possible equations are (with the relevant set shown at right):
<math display=block>\begin{alignat}{5}
  \text{(CT1)}&& \qquad \cos b\,\cos C &= \cot a\,\sin b - \cot A \,\sin C \qquad&&(aCbA)\\[0ex]
  \text{(CT2)}&& \cos b\,\cos A &= \cot c\,\sin b - \cot C \,\sin A &&(CbAc)\\[0ex]
  \text{(CT3)}&& \cos c\,\cos A &= \cot b\,\sin c - \cot B \,\sin A &&(bAcB)\\[0ex]
  \text{(CT4)}&& \cos c\,\cos B &= \cot a\,\sin c - \cot A \,\sin B &&(AcBa)\\[0ex]
  \text{(CT5)}&& \cos a\,\cos B &= \cot c\,\sin a - \cot C \,\sin B &&(cBaC)\\[0ex]
  \text{(CT6)}&& \cos a\,\cos C &= \cot b\,\sin a - \cot B \,\sin C &&(BaCb)
\end{alignat}</math>
To prove the first formula start from the first cosine rule and on the right-hand side substitute for {{math|cos ''c''}} from the third cosine rule:
<math display=block>\begin{align}
  \cos a &= \cos b \cos c + \sin b \sin c \cos A \\
  &= \cos b\ (\cos a \cos b + \sin a \sin b \cos C) + \sin b \sin C \sin a \cot A \\
  \cos a \sin^2 b &= \cos b \sin a \sin b \cos C + \sin b \sin C \sin a \cot A.
\end{align}</math>
The result follows on dividing by {{math|sin ''a'' sin ''b''}}. Similar techniques
with the other two cosine rules give CT3 and CT5. The other three equations follow by applying rules 1, 3 and 5 to the polar triangle.

===Half-angle and half-side formulae===
With <math>2s=(a+b+c)</math> and <math>2S=(A+B+C),</math>
<math display="block">
\begin{alignat}{5}
  \sin{\tfrac{1}{2}}A &= \sqrt{\frac{\sin(s-b)\sin(s-c)}{\sin b\sin c}}
&\qquad\qquad
 \sin{\tfrac{1}{2}}a &= \sqrt{\frac{-\cos S\cos (S-A)}{\sin B\sin C}} \\[2ex]
 \cos{\tfrac{1}{2}}A &= \sqrt{\frac{\sin s\sin(s-a)}{\sin b\sin c}}
  & \cos{\tfrac{1}{2}}a &= \sqrt{\frac{\cos (S-B)\cos (S-C)}{\sin B\sin C}} \\[2ex]
  \tan{\tfrac{1}{2}}A &= \sqrt{\frac{\sin(s-b)\sin(s-c)}{\sin s\sin(s-a)}}
  & \tan{\tfrac{1}{2}}a &= \sqrt{\frac{-\cos S\cos (S-A)}{\cos (S-B)\cos(S-C)}}
 \end{alignat}
</math>

Another twelve identities follow by cyclic permutation.

The proof (Todhunter,<ref name=todhunter/> Art.49) of the first formula starts from the identity <math>2\sin^2\!\tfrac{A}{2} = 1 - \cos A,</math> using the cosine rule to express {{mvar|A}} in terms of the sides and replacing the sum of two cosines by a product. (See [[List of trigonometric identities#Product-to-sum and sum-to-product identities|sum-to-product identities]].) The second formula starts from the identity <math>2\cos^2\!\tfrac{A}{2} = 1 + \cos A,</math> the third is a quotient and the remainder follow by applying the results to the polar triangle.

===Delambre analogies===
The Delambre analogies (also called Gauss analogies) were published independently by Delambre, Gauss, and Mollweide in 1807–1809.<ref>{{cite journal |last=Todhunter |first=Isaac |year=1873 |title=Note on the history of certain formulæ in spherical trigonometry |journal=The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science| volume=45 |issue=298 |pages=98–100|doi=10.1080/14786447308640820 }}</ref>

<math display="block">
\begin{align}
 \frac{\sin{\tfrac{1}{2}}(A+B)}
      {\cos{\tfrac{1}{2}}C}
=\frac{\cos{\tfrac{1}{2}}(a-b)}
      {\cos{\tfrac{1}{2}}c}
&\qquad\qquad
&
 \frac{\sin{\tfrac{1}{2}}(A-B)}
      {\cos{\tfrac{1}{2}}C}
=\frac{\sin{\tfrac{1}{2}}(a-b)}
      {\sin{\tfrac{1}{2}}c}
\\[2ex]
 \frac{\cos{\tfrac{1}{2}}(A+B)}
      {\sin{\tfrac{1}{2}}C}
=\frac{\cos{\tfrac{1}{2}}(a+b)}
      {\cos{\tfrac{1}{2}}c}
&\qquad
&
 \frac{\cos{\tfrac{1}{2}}(A-B)}
      {\sin{\tfrac{1}{2}}C}
=\frac{\sin{\tfrac{1}{2}}(a+b)}
      {\sin{\tfrac{1}{2}}c}
 \end{align}
</math>
Another eight identities follow by cyclic permutation.

Proved by expanding the numerators and using the half angle formulae. (Todhunter,<ref name=todhunter/> Art.54 and Delambre<ref>{{cite book
|last = Delambre
|first = J. B. J.
|author-link = Delambre
|title = Connaissance des Tems 1809
|year = 1807
|page = 445
|url = https://books.google.com/books?id=M8Mi6hU5tR0C&pg=PA445
|access-date = 2016-05-14
|archive-date = 2020-07-22
|archive-url = https://web.archive.org/web/20200722071420/https://books.google.com/books?id=M8Mi6hU5tR0C&pg=PA445
|url-status = live
}}</ref>)

===Napier's analogies===

<math display="block">\begin{align}
  \tan\tfrac{1}{2}(A+B) = \frac{\cos\tfrac{1}{2}(a-b)}{\cos\tfrac{1}{2}(a+b)} \cot\tfrac{1}{2}C
&\qquad&
  \tan\tfrac{1}{2}(a+b) = \frac{\cos\tfrac{1}{2}(A-B)}{\cos\tfrac{1}{2}(A+B)}\tan\tfrac{1}{2}c
\\[2ex]
  \tan\tfrac{1}{2}(A-B) = \frac{\sin\tfrac{1}{2}(a-b)}{\sin\tfrac{1}{2}(a+b)} \cot\tfrac{1}{2}C
&\qquad& 
  \tan\tfrac{1}{2}(a-b) =\frac{\sin\tfrac{1}{2}(A-B)}{\sin\tfrac{1}{2}(A+B)} \tan\tfrac{1}{2}c
\end{align}</math>

Another eight identities follow by cyclic permutation.

These identities follow by division of the Delambre formulae. (Todhunter,<ref name=todhunter/> Art.52)

Taking quotients of these yields the [[Law of tangents#Spherical version|law of tangents]], first stated by [[Mathematics in medieval Islam|Persian mathematician]] [[Nasir al-Din al-Tusi]] (1201–1274),

<math display=block>
\frac{\tan\tfrac12(A-B)}{\tan\tfrac12(A+B)}
= \frac{\tan\tfrac12(a-b)}{\tan\tfrac12(a+b)} 
</math>

===Napier's rules for right spherical triangles===
[[File:Spherical trigonometry Napier right-angled.svg|center|thumb|300px]]

When one of the angles, say {{math|C}}, of a spherical triangle is equal to {{pi}}/2 the various identities given above are considerably simplified. There are ten identities relating three elements chosen from the set {{mvar|a}}, {{mvar|b}}, {{mvar|c}}, {{mvar|A}}, and {{mvar|B}}.

[[John Napier|Napier]]<ref>{{cite book
|last = Napier
|first = J
|author-link = John Napier
|title = Mirifici Logarithmorum Canonis Constructio
|year = 1614
|page = 50
|url = https://books.google.com/books?id=VukHAQAAIAAJ
|access-date = 2016-05-14
|archive-date = 2013-04-30
|archive-url = https://web.archive.org/web/20130430105056/http://books.google.com/books?id=VukHAQAAIAAJ
|url-status = live
}}
An 1889 translation ''The Construction of the Wonderful Canon of Logarithms'' is available as en e-book from [https://www.abebooks.co.uk/servlet/SearchResults?tn=Construction+Wonderful+Canon+Logarithms Abe Books] {{Webarchive|url=https://web.archive.org/web/20200303190642/https://www.abebooks.co.uk/servlet/SearchResults%3Ftn%3DConstruction%2BWonderful%2BCanon%2BLogarithms |date=2020-03-03 }}</ref> provided an elegant [[mnemonic|mnemonic aid]] for the ten independent equations: the mnemonic is called Napier's circle or Napier's pentagon (when the circle in the above figure, right, is replaced by a pentagon).

First, write the six parts of the triangle (three vertex angles, three arc angles for the sides) in the order they occur around any circuit of the triangle: for the triangle shown above left, going clockwise starting with {{mvar|a}} gives {{mvar|aCbAcB}}. Next replace the parts that are not adjacent to {{mvar|C}} (that is {{mvar|A}}, {{mvar|c}}, and {{mvar|B}}) by their complements and then delete the angle {{mvar|C}} from the list. The remaining parts can then be drawn as five ordered, equal slices of a pentagram, or circle, as shown in the above figure (right). For any choice of three contiguous parts, one (the ''middle'' part) will be adjacent to two parts and opposite the other two parts. The ten Napier's Rules are given by
*sine of the middle part = the product of the tangents of the adjacent parts
*sine of the middle part = the product of the cosines of the opposite parts
The key for remembering which trigonometric function goes with which part is to look at the first vowel of the kind of part: middle parts take the sine, adjacent parts take the tangent, and opposite parts take the cosine.
For an example, starting with the sector containing {{mvar|a}} we have:
<math display=block>\begin{align}
  \sin a &= \tan(\tfrac{\pi}{2} - B)\,\tan b  \\[2pt]
  &= \cos(\tfrac{\pi}{2} - c)\, \cos(\tfrac{\pi}{2} - A) \\[2pt]
  &= \cot B\,\tan b \\[4pt]
  &= \sin c\,\sin A. 
\end{align}</math>
The full set of rules for the right spherical triangle is (Todhunter,<ref name=todhunter/> Art.62)
<math display="block">\begin{alignat}{4}
  &\text{(R1)}&\qquad \cos c&=\cos a\,\cos b,
&\qquad\qquad
  &\text{(R6)}&\qquad \tan b&=\cos A\,\tan c,\\
  &\text{(R2)}& \sin a &= \sin A\,\sin c,
&&\text{(R7)}& \tan a &= \cos B\,\tan c,\\
  &\text{(R3)}& \sin b &= \sin B\,\sin c,
&&\text{(R8)}& \cos A &= \sin B\,\cos a,\\
  &\text{(R4)}& \tan a &= \tan A\,\sin b,
&&\text{(R9)}& \cos B &= \sin A\,\cos b,\\
  &\text{(R5)}& \tan b &= \tan B\,\sin a,
&&\text{(R10)}& \cos c &= \cot A\,\cot B.
\end{alignat}</math>

===Napier's rules for quadrantal triangles===
[[File:Spherical trigonometry Napier quadrantal 01.svg|center|thumb|300px|A quadrantal spherical triangle together with Napier's circle for use in his mnemonics]]

A quadrantal spherical triangle is defined  to be a spherical triangle in which one of the sides subtends an angle of {{pi}}/2 radians at the centre of the sphere: on the unit sphere the side has length {{pi}}/2. In the case that the side  {{mvar|c}} has length {{pi}}/2 on the unit sphere the equations governing the remaining sides and angles may be obtained by applying the rules for the right spherical triangle of the previous section to the polar triangle {{math|△''A'B'C' ''}} with sides {{mvar|a', b', c'}} such that {{math|1=''A' ''= {{pi}} − ''a''}}, {{math|1=''a' ''= {{pi}} − ''A''}} etc. The results are:
<math display="block">\begin{alignat}{4}
  &\text{(Q1)}&\qquad  \cos C &= -\cos A\,\cos B,
&\qquad\qquad
  &\text{(Q6)}&\qquad \tan B &= -\cos a\,\tan C,\\
  &\text{(Q2)}& \sin A &= \sin a\,\sin C,
&&\text{(Q7)}& \tan A &= -\cos b\,\tan C,\\
  &\text{(Q3)}& \sin B &= \sin b\,\sin C,
&&\text{(Q8)}& \cos a &= \sin b\,\cos A,\\
  &\text{(Q4)}& \tan A &= \tan a\,\sin B,
&&\text{(Q9)}& \cos b &= \sin a\,\cos B,\\
  &\text{(Q5)}& \tan B &= \tan b\,\sin A,
&&\text{(Q10)}& \cos C &= -\cot a\,\cot b.
\end{alignat}</math>

===Five-part rules===
Substituting the second cosine rule  into the first and simplifying gives:
<math display=block>\begin{align}
  \cos a &= (\cos a \,\cos c + \sin a \,\sin c \,\cos B) \cos c + \sin b \,\sin c \,\cos A \\[4pt]
  \cos a \,\sin^2 c &= \sin a \,\cos c  \,\sin c \,\cos B + \sin b \,\sin c \,\cos A
\end{align}</math>
Cancelling the factor of {{math|sin ''c''}} gives
<math display=block>\cos a \sin c = \sin a \,\cos c \,\cos B + \sin b \,\cos A</math>

Similar substitutions in the other cosine and supplementary cosine formulae give a large variety of 5-part rules. They are rarely used.

===Cagnoli's Equation===
Multiplying the first cosine rule  by {{math|cos ''A''}}  gives
<math display=block>\cos a \cos A = \cos b \,\cos c \,\cos A + \sin b \,\sin c - \sin b \,\sin c \,\sin^2 A.</math>
Similarly multiplying the first supplementary cosine rule  by {{math|cos ''a''}}  yields
<math display=block>\cos a \cos A = -\cos B \,\cos C \,\cos a + \sin B \,\sin C - \sin B \,\sin C \,\sin^2 a.</math>
Subtracting the two and noting that it follows from the sine rules that <math>  \sin b \,\sin c \,\sin^2 A = \sin B \,\sin C \,\sin^2 a </math>  produces Cagnoli's equation
<math display=block>\sin b \,\sin c + \cos b \,\cos c \,\cos A = \sin B \,\sin C - \cos B \,\cos C \,\cos a</math>
which is a relation between the six parts of the spherical triangle.<ref>{{cite book
|last = Chauvenet
|first = William
|title = A Treatise on Plane and Spherical Trigonometry
|publisher = J. B. Lippincott & Co.
|location = Philadelphia
|year = 1867
|page = 165
|url = https://books.google.com/books?id=d8E8AAAAYAAJ
|access-date = 2021-07-11
|archive-date = 2021-07-11
|archive-url = https://web.archive.org/web/20210711194635/https://books.google.com/books?id=d8E8AAAAYAAJ
|url-status = live
}}</ref>