Talk:Pythagorean trigonometric identity

Trigonometric identity?

Do this subject have any content that shouldn't just be part of Trigonometric identity? If this stays as a separate article, it needs a 'lot' more explanatory text. ike9898 21:06, Nov 23, 2004 (UTC)

use of square roots in the proof is DODGY

basically the way things are shown here only holds for angles in the base region despite the fact the identity is valid for all angles Plugwash 23:36, 21 Dec 2004 (UTC)

Explanation needed

"They can be derived from (1) using simple algebra" How do the 2nd and 3rd identities come from (1)?, show me the algebra required to convert them.Bdodo1992 13:43, 27 July 2007 (UTC)

If you look at the bottom of the article now, you can see that they divided them up. Garrett247 (talk) 04:36, 27 October 2010 (UTC)

Last two sections not pertinent

The sections "Using power series" and "Using differential equation" are germane to articles on the trig functions, but not to the Pythagorean trigonometric identity. Brews ohare (talk) 17:25, 5 June 2010 (UTC)

These sections are OK. I added some steps and conclusions drawing the connections. Brews ohare (talk) 18:51, 8 June 2010 (UTC)

Formulas vs. Formulae

According to Wikipedia, it is appropriate to use either of the words. However, since the english wikipedia is based in the united states, it is more appropriate to use "formulas"...right? Either way, I changed it. Correct it if I'm wrong. Garrett247 (talk) 04:37, 27 October 2010 (UTC)

I think you are wrong I'm afraid: I don't think it's a US/UK issue, though that in itself is not reason to change an article (WP is an international encyclopaedia, not a US-centric one - see WP:ENGVAR). The mathematics manual of style says

The plural of formula is either formulae or formulas. Both are acceptable, but articles should be consistent with themselves. If an article is consistent, then editors should not change the article from one style to another.

So I've changed it back.--JohnBlackburne^words_deeds 09:20, 27 October 2010 (UTC)

Unexplained properties of Pythagoreans theorem for all triangles

according to Pythagoreans theorem these are an unexplained properties of all simplified triangles. simplified versions of the triangles when base equals 1 ,the lenghts of altitudes or sides are measures of angles.

$(c\times \sin B)^{2}+(b\times \cos C)^{2}=b^{2}$

$(c\times \sin A)^{2}+(a\times \cos C)^{2}=a^{2}$

$(c\times \sin B)^{2}+(c\times \cos B)^{2}=c^{2}$

$(c\times \sin A)^{2}+(c\times \cos A)^{2}=c^{2}$

$(\sin B)^{2}+(\cos B)^{2}=c^{2}$

$(\sin A)^{2}+(\cos A)^{2}=c^{2}$

$(b\times \cos C)+(c\times \cos B)=a$

$(a\times \cos C)+(c\times \cos A)=b$

$(a\times \cos B)+(b\times \cos A)=c$

$(c\times \cos B)+(b\times \cos C)=$

$(c\times \cos A)+(a\times \cos C)=b$

and to find one of the altitude:

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

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

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

And

$(\sin A)^{2}+(\sin B)^{2}+(\sin C)^{2}+(\cos A)^{2}+(\cos B)^{2}+(\cos C)^{2}=3$ and they apply for all triangles.208.98.222.24 (talk) 13:35, 26 January 2019 (UTC)

Calcea Johnson and Ne'Kiya Jackson

Apparently, Trigonometry is independent of Geometer, as proven by these two young mathematicians. See the article published as today (May 19, 2023): High School Seniors in New Orleans Solved a 2,000-Year-Old Mathematical Problem. [1]

And here is the AMS about the published article by these two young women: [2]