14
$\begingroup$

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,

$\endgroup$
53
$\begingroup$

My trigonograph for the Law of Cosines may help:

enter image description here

$\endgroup$
  • 6
    $\begingroup$ 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. $\endgroup$ – Henning Makholm Oct 22 '17 at 10:48
  • 2
    $\begingroup$ @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!) $\endgroup$ – Blue Oct 22 '17 at 11:08
  • $\begingroup$ Nice. Do you also have a picture for the case $\angle C>\pi /2$? $\endgroup$ – DanielWainfleet Oct 23 '17 at 18:15
  • 2
    $\begingroup$ @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". $\endgroup$ – Blue Oct 23 '17 at 22:54
10
$\begingroup$

Based on Pythagorean theorem and Pythagorean trigonometric identity in this triangle

enter image description here

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

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • 6
    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Oct 22 '17 at 7:25
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.