# Where does the $-2ab$ term come from in the cosine law?

I understand that in the cosine rule i.e. $c^2 = a^2 + b^2 - 2ab \cos C$, the cosine function acts to bring down the value of $c^2$ for acute angles ($\cos C>0$, $-2ab\cos C<0$ ) and increase the value of $c^2$ for obtuse angles ($\cos C <0$, $-2ab\cos C > 0$). I still wonder where the $2ab$ term comes from? Any ideas about the intuition behind that?

Regards,

• Welcome to MSE. Please use MathJax. Oct 22 '17 at 7:06
• Reference Wolfram mathworld: Law of Cosines Oct 22 '17 at 7:07
• Oct 22 '17 at 7:07
• You can think of the $bc$ factor as providing the correct units, were you to measure $a,b,c$ in, say, meters. The factor of $2$ can be argued to be there by the requirement that $c=0$ if $\theta=\pi$ and $a=b$. Oct 22 '17 at 11:58

My trigonograph for the Law of Cosines may help:

• Nice proof-without-words, +1! But it would be easier to read if the squares in the calculation below were not filled in, since they correspond just to the square outlines in the diagram. Oct 22 '17 at 10:48
• @HenningMakholm: Thanks for catching that. The filled-in squares are a bit of a typo-graphical error. In previous versions I've created, the squares are outlines. (Oh, I just realized that I'd uploaded a previous version in this answer ... over five years ago!)
– Blue
Oct 22 '17 at 11:08
• Nice. Do you also have a picture for the case $\angle C>\pi /2$? Oct 23 '17 at 18:15
• @DanielWainfleet: The obtuse case is drawable, but it takes some artistic finesse to keep the overlapping elements from creating a visual mess. The "equation of boxes" also isn't quite as pretty, because it's tricky to represent the effect of a negative cosine. (I get away with that kind of thing in trigonographs like this one, but that's because I can readily reconcile absolute values in the real equations.) So, I prefer to leave the obtuse case as the proverbial "exercise to the reader".
– Blue
Oct 23 '17 at 22:54

Based on Pythagorean theorem and Pythagorean trigonometric identity in this triangle

we have $$c^2 = (a-b\cos C)^2 + (b \sin C)^2 \\ = a^2 - 2ab \cos C + b^2\cos^2 C + b^2 \sin^2 C \\ = a^2 - 2ab \cos C + b^2$$

• This equation can easly be seen by (scaled) unit circle and (scaled) sin, cos. One can also imagine one fixed line $a = AB$ and other line $b = BC$ moving around a point $B$ to form third side $c$ when connecting points $A, C$. So, $c^2_{min} = (a - b)^2, c^2_{mid} = a^2 + b^2, c^2_{max} = (a + b)^2$ and here is visible "cosine effect".
– 1b3b
Jul 3 '20 at 0:05

Let's accept that $c^2=a^2+b^2$ for a right Euclidean triangle. Then for a degenerate obtuse triangle, where we take angle $C\to\pi$, we have $c^2 \to (a+b)^2 = a^2+b^2+2ab$. On the other hand, as we take $C\to0$, we find $c^2\to(a-b)^2 = a^2+b^2-2ab$.

It is apparent that length $c^2$ is a function of angle $C$ between sides $a$ and $b$. We see that $c^2=a^2+b^2-2ab\cos C$ modulates between our boundary cases an provides every value in between.

• I would have started an "intuitive" explanation exactly the same way (+1). Another point may be that the cosine needs to be multiplied by something quadratic in $a,b,c$. Either for dimensional reasons (all the terms must be square meters, or square feet in the anglosaxon world). Or, if we scale the triangle up by a certain factor $k$, then $a^2,b^2,c^2$ will scale by a factor $k^2$. The cosine won't change because the angle stays the same, so $k^2$ needs to come from somewhere else. Oct 22 '17 at 7:25

On the real line $\mathbb R$ we define the absolute value of a number as

$\tag 1 |x| = \sqrt{x^2}$

The distance between any two numbers $a$ and $b$ on the line is defined as $|a - b|$.

The binomial theorem is useful:

$\tag 2 (a + b)^2 = a^2 + b^2 +2ab$

We also have

$\tag 3 |(a + b)^2| = |a + b|^2 =|a|^2 + |b|^2 \pm 2 |a||b|$

and since $|b - a| \text{ (distance) } = |b + (-a)| = |(-a) + b|$,

$\tag 4 |b - a|^2 =|a|^2 + |b|^2 \pm 2 |a||b|$

When you move from the real line to $\mathbb R \times R$, you want to bring along this idea of distance. Using graphs paper and a ruler, it won't be long before you conclude that for line segment lengths $a$, $b$ and $c$ (distance) forming a triangle in the plane that

$\tag 5 c^2 = a^2 + b^2 + \gamma a b \text{ with } -1 \le \gamma \le 1$

better work.