# Cosine rule confusion

by Taylor_1989
Tags: confusion, cosine, rule
 P: 106 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.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,533 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.
 P: 11 Cosine rule confusion 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..
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,533 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.