Understanding the Cosine Rule: Deriving the Other Two Formulas

[Taylor_1989]

So here is my question, I understand how to derive the cosine rule from, both triangles acute and obtuse. My problem is the 3 formula you get from this equation.

When I derive from a triangle I get the formula: $c^2=a^2+b^2-2abcos∅$

so how do you derive the other two formula, I read that the letter are interchangeable so you can place them where every you like on the triangle. If that is the case, why do we show the other two formula? I presume people reading this post know of the other two formula I am referring to if not here they are: $a^2=c^2+b^2-2cbcos∅$

$b^2=a^2+c^2-2accos∅$.

This has been on my mind for a while, I would appreciate if someone could clarify my confusion. Big thanks in advance.

 

[HallsofIvy]

First, your formulas are not quite correct because you have the same angle, ∅, in each.
Instead use the usual convention that we label sides of the triangle by small letters and the opposite angle by the corresponding capital letter:

$a^2= b^2+ c^2- 2ab cos(A)$
$b^2= a^2+ c^2- 2ac cos(B)$
$c^2= a^2+ b^2- 2ab cos(C)$

Yes, you are correct that it does not matter how we label the sides so we don't really need all three. But all three are often given because some beginners have difficulty "changing" letters when they need to calculate several parts of the same triangle. That is, in a given problem, we are given that "c" is a specific side and many people just don't like the idea that we can then think of "c" representing a different side. Of course, the three equations are all gotten from the first by interchanging the letters.

 

[Taylor_1989]

Thanks for the reply back. I realized after I posted what I had done wrong but did not have any time to correct.

Once again thanks for clearing that up for me.

 

[THSMathWhiz]

Suppose we had $u,v\in\mathbb{R}^2$ such that $u=B-A$ and $v=C-A$, then $u-v=B-C$. Then $a=||u-v||$, $b=||v||$, and $c=||u||$. So $$||u-v||^2=||u||^2+||v||^2-2uv\cos A,$$ or $$(u-v)\cdot(u-v)=u\cdot u+v\cdot v-2u\cdot v.$$ This is true and can easily be checked..

 

[Taylor_1989]

Thanks for the reply, can sort of see how that is true, I have only just started studying vectors, so I bit slow on the uptake of that proof. I have see something similar, if memory serves; on wki. I am studying vectors today so will hopefully try to understand your method a little better by the end. I am under the assumption these are a vector notation? As I said just getting into mechanics and the use of vectors in pure math.

 

[HallsofIvy]

By the way, notice that if $\theta= 90$ degrees (or $\pi/2$ radians) the cosine is 0 giving the Pythagorean theorem: $c^2= a^2+ b^2$

If $\theta= 0$ degrees (or radians) cosine is 1: $c^2= a^2+ b^2- 2ab= a^2- 2ab+ b^2= (a- b)^2$. That is, rather than a triangle we have a single line segment. The angle has "closed" down so that the "third side" is just the distance from A to B.

If $\theta= 180$ degrees (or $\pi$ radians) cosine is -1: $c^2= a^2+ b^2+ 2ab= a^2+ 2ab+ b^2= (a+ b)^2$. Now, the angle has "opened" up so that we have a single line segment including both a and b.

 

[Taylor_1989]

Okay now I am with ya, see where you are coming from. It also makes the above proof easier to see. Thanks for that. Big help.

 

[Taylor_1989]

Okay so first I appolgse for reopening this thread, but I think I may have made an erro of judgment. See I assumed I actually knew what the cosine rule was all about, because I read the book and went yep that proof works. My problem cam up when I was looking at a question on a strange looking triangle, I knew that I would apply the cosine rule, which is easy enough; any Tom dick nd Harry can do that ( English expression I believe ). So what bugged me was when I tried to apply the proof, couldn't because, I only knew the proof for 2 specific triangles, which are normally the ones given, so I found and article on cut the knot which I believe explain it quite well. Now my gist of what I see is that, to prove for any trangle to find the length of the opposite side, you have to create a right hand triangle with the triangle given, then just follow normal protocol as you would to say prove a obtuse triangle.

I have posted the article for someone to have a quick look to see if I have got the general idea right?
http://www.cut-the-knot.org/pythagoras/cosine.shtml