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March 16

help with Latin translation

Hi friendly reference desk folks. I've been trying to read (very small bits of) Thomas Fincke's Geometria rotundi (1583, in Latin). I don't read Latin so this is a serious slog. Anyway.

Fincke has, in modern notation:
30. ${\textstyle \sec x-\tan x=\tan {\tfrac {1}{2}}{\bigl (}{\tfrac {1}{2}}\pi -x{\bigr )}.}$
31. ${\textstyle \sec x+\tan x=\tan {\tfrac {1}{2}}{\bigl (}{\tfrac {1}{2}}\pi +x{\bigr )}.}$

Or if we want to get very literal Fincke's version is closer to:
30. ${\textstyle \sec x=\tan x+\tan {\tfrac {1}{2}}{\bigl (}{\tfrac {1}{2}}\pi -x{\bigr )}.}$
31. ${\textstyle \sec x+\tan x=\tan {\bigl [}x+{\tfrac {1}{2}}{\bigl (}{\tfrac {1}{2}}\pi -x{\bigr )}{\big ]}\qquad x<{\tfrac {1}{4}}\pi .}$

Original Latin:
30. Secans peripheriae aequatur tangenti & datae & semissis complementi.
31. Secans peripheriae, quadrantis peripheriae semisse minoris, cum tangente eiusdem aequatur tangenti datae semisse sui complementi auctae

How's my English translation? Can anyone make this more accurate to the original or clearer?

30. The secant of an arc is equal to the sum of the tangents of the arc and its half-complement.
31. The secant of an arc, less than half a quadrant, and the tangent of the same arc are together equal to the tangent of the sum of the given arc and half its complement.

You can see the original with figures in context here:
https://archive.org/details/den-kbd-pil-130018099382-001/page/n101/mode/1up

See also Wu, Rex H. (2019). "Proof Without Words: Revisiting Two Trigonometric Figures and Two Identities from Bressieu and Fincke". Mathematics Magazine. 92 (4): 302–304. doi:10.1080/0025570X.2019.1603732.

–jacobolus (t) 06:41, 16 March 2023 (UTC)[reply]

Bressieu (1581, also in Latin) also included these, with essentially the same figures but labeling the angles a bit diferently
https://archive.org/details/bub_gb_PAsx_rxQpgUC/page/n54/

–jacobolus (t) 06:52, 16 March 2023 (UTC)[reply]

I made a minor correction to the transcription. I suspect that in (30) the first ampersand was inserted by the type setter by mistake and that the text should have tangenti datae just as in (31). I find it curious to see that the agent of the passive verb form aequatur takes the genitive case. Elsewhere I saw occasionally the expected ablative, as on p. 117 in angulus etiam i o u aequatur angulo u o e ex fabrica. --Lambiam 09:39, 16 March 2023 (UTC)[reply]

Oops, tangenti is the dative or ablative of tangens – although, according to Wiktionary, the latter only when used adjectivally, but here the sense is nominal. --Lambiam 18:34, 16 March 2023 (UTC)[reply]

I thought maybe the "tangenti &" there had to do with this being the sum of two distinct tangents. –jacobolus (t) 15:21, 16 March 2023 (UTC)[reply]

I missed that. Indeed, in Latin et X et Y can be used for "X as well as Y", so then we get, "the tangent of the given [quantity] as well as of half the complement". It is then up to the reader to understand that this means the respective tangents, and that these have to be added together to equal the secant. --Lambiam 16:54, 16 March 2023 (UTC)[reply]

I think for someone who reads Latin and tries to follow the proof with reference to the figure, it should hopefully become clear enough. But yeah, mathematical notation makes statements much easier to interpret precisely. –jacobolus (t) 18:27, 16 March 2023 (UTC)[reply]

Here is another one, from book X: https://archive.org/details/den-kbd-pil-130018099382-001/page/n303/

ut semissis summae terminorum rationis sinuum ad differentiam semissis ac termini alterius, sic tangens semissis summae datae ad tangentem peripheriae peripheriarum qua minor quaesitarum semisse summae minor, aut major, major est.

I think this is the identity:

{\frac {{\frac {1}{2}}(\sin x+\sin y)}{{\tfrac {1}{2}}(\sin x+\sin y)-\sin y}}={\frac {\tan {\tfrac {1}{2}}(x+y)}{\tan {\bigl (}{\tfrac {1}{2}}(x+y)-y{\bigr )}}}.

But I am not sure I am interpreting it correctly. The phrasing is very confusing to me as a non-reader of Latin.

I would instead write this identity (if that is what it is saying) as:

{\frac {\sin x+\sin y}{\sin x-\sin y}}={\frac {\tan {\tfrac {1}{2}}(x+y)}{\tan {\tfrac {1}{2}}(x-y)}}.

It's possible it's instead aiming for an identity equivalent to:

{\frac {\sin(x+y)}{\sin(x-y)}}={\frac {\tan x+\tan y}{\tan x-\tan y}}.

(or something else entirely) –jacobolus (t) 19:19, 16 March 2023 (UTC)[reply]

Oh wait, I am not quite getting the setup right, though I am still pretty sure the identity is the one expressed. The given data are the sum and difference of the angles and the ratio of the sines. So I think the "terminorum rationis sinuum" are the two parts of that ratio. Like we know

\sin x:\sin y::p:q,

and then we use the identity (I think?)

{\frac {{\tfrac {1}{2}}(p+q)}{{\tfrac {1}{2}}(p-q)}}={\frac {\tan {\tfrac {1}{2}}(x+y)}{\tan {\tfrac {1}{2}}(x-y)}}

If I am interpreting it right, this is getting close to the law of tangents. (It's frustrating Google Translate is so incredibly bad for Latin, considering it works pretty well now for modern languages.) –jacobolus (t) 22:00, 16 March 2023 (UTC)[reply]

Cicero would not have been able to make sense of these writings either. --Lambiam 13:19, 17 March 2023 (UTC)[reply]