Editing John Napier (section)

===Trigonometry===
[[File:Spherical_trigonometry_Napier_right-angled.svg|center|thumb|300x300px]]

When one of the angles, say ''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 ''a'', ''b'', ''c'', ''A'', ''B''.

Napier<ref>{{cite book |last=Napier |first=J |url=https://books.google.com/books?id=VukHAQAAIAAJ |title=Mirifici Logarithmorum Canonis Constructio |year=1614 |page=50 |author-link=John Napier |access-date=14 May 2016 |archive-url=https://web.archive.org/web/20130430105056/http://books.google.com/books?id=VukHAQAAIAAJ |archive-date=30 April 2013 |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=3 March 2020}}</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  a  gives ''aCbAcB''. Next replace the parts that are not adjacent to C (that is ''A, c, B'') by their complements and then delete the angle 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 <math>a</math> we have:

:<math> \sin a = \tan(\pi/2-B)\,\tan b = \cos(\pi/2-c)\, \cos(\pi/2-A)
=\cot B\,\tan b = \sin c\,\sin A. </math>
The full set of rules for the right spherical triangle is (Todhunter,<ref name=todhunter>{{cite book
|last = Todhunter
|first = I.
|author-link = Isaac Todhunter
|title = Spherical Trigonometry
|year = 1886
|publisher = MacMillan
|edition = 5th
|url = http://www.gutenberg.org/ebooks/19770
|access-date = 28 July 2013
|archive-date = 14 April 2020
|archive-url = https://web.archive.org/web/20200414233849/http://www.gutenberg.org/ebooks/19770
|url-status = live
}}</ref> Art.62)

<math>
\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> 

[[File:Spherical_trigonometry_Napier_quadrantal_01.svg|center|thumb|300x300px|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  ''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 ''A'B'C'<nowiki/>''  with sides ''a',b',c'<nowiki/>''  such that ''A'<nowiki/>'' = ''{{pi}}''&nbsp;−&nbsp;''a'',&nbsp; ''a'''  =&nbsp;''{{pi}}''&nbsp;−&nbsp;''A'' etc. The results are:

<math>
\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>

'''Logarithm''' 

Given a positive [[real number]] ''b'' such that ''b'' ≠ 1, the ''[[logarithm]]'' of a positive real number ''x'' with respect to base ''b'' is the exponent by which ''b'' must be raised to yield ''x''. In other words, the logarithm of ''x'' to base ''b'' is the unique real number ''y'' such that ''b''<sup>''y''</sup> = ''x''.

The logarithm is denoted "log<sub>''b''</sub> ''x''" (pronounced as "the logarithm of ''x'' to base ''b''", "the base-''b'' logarithm of ''x''", or most commonly "the log, base ''b'', of ''x''").

An equivalent and more succinct definition is that the function <math>f \colon x\to log_{b} x</math> is the [[inverse function]] to the function <math>f \colon x\to b^{x}.</math>