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,
On
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.
On
Based on Pythagorean theorem and Pythagorean trigonometric identity in this triangle
we have $$c^2 = (a-b\cos C)^2 + (b \sin C)^2 \\ = a^2 - 2ab \cos C + b^2\cos^2 C + b^2 \sin^2 C \\ = a^2 - 2ab \cos C + b^2$$
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.