Talk:Law of cosines

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WikiProject Echo has identified Law of cosines as a foreign language featured article. You may be able to improve this article with information from the French language Wikipedia.

1 Phythagorean Theorem also Proved by Law of Cosines?
2 Circular proof?
3 Section moving
4 Another use for Law of Cosines
5 We?
6 Rewrite
7 Faster demonstrations
8 Alternative proof using Power of a Point Theorem

[edit] Phythagorean Theorem also Proved by Law of Cosines?

Okay, in the article, it states the since cos 90 is 0 the Law of Cosines is reduced to the Phythagorean Theorem. I don't think that's the case cause if $a 2 = b 2 + c 2 - 2 b c c o s A$ , then wouldn't $a 2 = b 2 + c 2$ ? —Preceding unsigned comment added by 70.107.165.59 (talk • contribs)

Yes, and that's the Pythagorean theorem (in the case where the right angles is the angle between the sides of lengths b and c). Michael Hardy 23:20, 1 March 2007 (UTC)

[edit] Circular proof?

In the dot product article, a dot b = a b cos theta is proved using the Law of Cosines. And the Law of Cosines is proved using vector dot products. Shouldn't somebody fix this? --Orborde 07:48, 9 September 2005 (UTC)

The dot product proof should be removed. It's circular. Law of cosines comes first historically and all of vector calculus is assumed to depend on it, not vice versa. Might as well use vectors to prove the pythagorean theorem. Pfalstad 10:22, 9 September 2005 (UTC)

I have hopefully solved this problem, by stating the law of cosines is equivalent to the dot product formula from theory of vectors. --345Kai 10:45, 30 March 2006 (UTC)

Wow that's pretty darn smart. --M1ss1ontomars2k4 | T | C | @ 04:04, 21 May 2006 (UTC)

If you define the dot product (a,b)*(c,d) as ac+bd (so it is linear) and then prove (a,b)*(c,d)=|(a,b)|*|(c,d)|*cos(α) then there is nothing circular. The definition used is a matter of taste, I guess some others are possible. I like the argument with dot product because it gives a different light on the cosine law and I think it deserves a place on this article. Ricardo sandoval 23:20, 12 April 2007 (UTC)

I never saw the story of the dot product but I would guess it came from the scalar product of physics (as in the definition of work done in a particle) is that right? Ricardo sandoval 23:20, 12 April 2007 (UTC)

I guess we could also define (a,b)*(c,d) as |(a,b)|*|(c,d)|*cos(α) prove the linearity by geometrical arguments and use it to prove the cosine law. right? Ricardo sandoval 23:20, 12 April 2007 (UTC)

[edit] Section moving

I removed the following section because this is said in the first paragraph and follows from elementary algebra:

[edit] Another use for Law of Cosines

The Law of Cosines can also be used to find the measure of the three angles in a triangle if you know the length of the three sides. This is how you do it: &nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp

a 2 + b 2 - 2 a b cos C = c 2

- 2 a b cos C = - a 2 - b 2 + c 2

cos C = ( - a 2 - b 2 + c 2) / ( - 2 a b)

cos C = (a 2 + b 2 - c 2) / (2 a b)

C = arccos[(a 2 + b 2 - c 2) / (2 a b)]

Now you can find the measure of angle C!

— Sverdrup (talk) 21:00, 25 Feb 2004 (UTC)

If noone objects, I'm going to put the vector-based proof first and move the other one down since the former is more simple and universal as opposed to the latter.. or someone else could do it.. or whatever... - Evil saltine 00:33, 22 May 2005 (UTC)

As noted above, the vector-based proof is circular: the proof that the dot product of two vectors has a geometric interpretation in Euclidian space is itself based on the law of cosines. So using the geometric interpretation of a dot product to prove the law of cosines is a bit problematic... --Delirium 03:17, 13 November 2005 (UTC)

see above --345Kai 10:45, 30 March 2006 (UTC)

[edit] We?

Is the usage of "we" throughout this article proper? Shouldn't "we can easily prove" be "can be proved" (wlong with some sentence rearrangement). Broken Segue 04:00, 30 March 2006 (UTC)

[edit] Rewrite

I have rewritten the article and expanded it considerably. I have taken a lot of material from the French article, according to the above suggestion. I hope you like it!!! Please improve further... (and sorry that in the history of this page the big change was signed "Euklid". That was me and and I know we shouldn't use these kinds of names on Wikipedia, so I've changed my login. --345Kai 10:45, 30 March 2006 (UTC)

Great work!! The French article was way better by any standards, I fell bad I can't understand it. I only see the general flow of ideas. Now it is all clear I will read in finer detail, Thanks!!! Ricardo sandoval 23:42, 12 April 2007 (UTC)

[edit] Faster demonstrations

The last demonstrations using the power of a point can be simplified, all of them can be worked out with just one application of the power point theorem, and the the first two cases can be made one by considering b<c(always can be done by flipping the sides). Someone willing to make new pictures? Ricardo sandoval 08:02, 16 April 2007 (UTC)

If you need any diagrams drawn, altered, fixed etc. Please post a request at the Graphics Lab. Please read instructions at the top of the page before posting. Thank you! XcepticZP 18:20, 15 November 2007 (UTC)

[edit] Alternative proof using Power of a Point Theorem

This proof along with that using Pythagoras theorem are of considerable historical significance since both have their origins in Euclid's Elements and both are referred to in the ground breaking work of Nicolaus Copernicus: "De Revolutionibus Orbium Coelestium". On Page 20 and Page 21 of Book 1 Copernicus describes two techniques for determining angles given all three sides of a triangle. These techniques correspond respectively to the Pythagorean and Power of Points derivations of the Law of Cosines.

Replica of diagrams in Book 1, Page 21 of "De Revolutionibus Orbium Coelestium"

Fig 1 and Fig 2 are replicas of the diagrams from Page 21 of De Revolutionibus Orbium Coelestium.In Fig 1 triangle ABC is drawn with B an acute angle. With C as centre and BC (=a) as radius semi-circle DBF is constructed with DF a diameter and D a point on AC. The circle meets side AB at E and EC is joined forming isosceles triangle ECB with EB = 2acos(B).Power of Points Theorem (intersecting secants theorem) is applied to point A outside the circle:

$\begin{array}{lcl} \quad\quad AD \times AF = AB \times AE\\ \Rightarrow (b-a)(b+a)=c(c-2a\cos\hat{B})\\ \Rightarrow \quad b^2-a^2=c^2-2ac.\cos\hat{B}\\ \Rightarrow \quad \quad b^2=a^2+c^2-2ac\cos\hat{B} \end{array}$

It will be no different in the case of Fig 2 where angle B is obtuse:

$\begin{array}{lcl} \quad\quad AD \times AF = AB \times AE\\ \Rightarrow (b-a)(b+a)=c(c+2a\cos(180-\hat{B}))\\ \Rightarrow \quad b^2-a^2=c^2-2ac.\cos\hat{B}\\ \Rightarrow \quad \quad b^2=a^2+c^2-2ac\cos\hat{B} \end{array}$

Proof of Law of Cosines using Power of a Point Theorem

In addition to the two cases dealt with above, we also need to consider the situation shown in the diagram where Power of Point theorem is applied about point B inside the construction circle.

Triangle ABC is drawn with side AB=c,BC=a and CA=b. With A as centre construct circle DCE with radius b and diameter DE passing through B. CB produced meets the circle at F. Since triangle CAF is isosceles:

$CF=2b\cos\hat{C}\quad$ .

Now apply the Power of a Point Theorem (intersecting chords theorem) to point B inside the circle:

$\begin{array}{lcl} \quad\quad BD \times BE = BC \times BF\\ \Rightarrow (b-c)(b+c)=a(2b\cos\hat{C}-a)\\ \Rightarrow \quad b^2-c^2=2ab.\cos\hat{C}-a^2\\ \Rightarrow \quad \quad c^2=b^2+a^2-2ab\cos\hat{C} \end{array}$ —Preceding unsigned comment added by Neil Parker (talk • contribs) 10:28, 17 December 2007 (UTC)