Understanding the law of cosines, directly from the formula I understand the derivation of the law of cosines:
$$c^2 = a^2 + b^2 - 2ab \cos C,$$ where $\angle C$ is opposite side $c$.
By dropping a perpendicular $h$ to side $b$ from the angle opposite side $b$, then finding the length $h$ with $a\sin C$, and part of the length $b$ with $a \cos C$, we can use Pythagoras to find $c$:
$$c^2 = (a \sin C)^2 + (b - a\cos C)^2.$$
This use of Pythagoras sense to me. But then, expanding, we can use identity $s^2+c^2=1$ to simplify to the standard formula:
$$c^2 = a^2 + b^2 - 2ab\cos C.$$
This looks a bit like Pythagoras with an adjustment, but this doesn't really make sense to me - is there a way to make sense of it in this way directly (or is going via the derivation the only way?) Or perhaps umderstand it in some other direct way?
[ A secondary sub-question is whether understanding it "directly" is helpful? Or is going through the proof good enough? ]
 A: Think of it in terms of vectors. In $\triangle ABC$, we have $\vec{AB}=\vec{AC}-\vec{BC}$. Squaring this, we get
$$
\vec{AB}\cdot\vec{AB}=\vec{AC}\cdot\vec{AC}-2\vec{AC}\cdot\vec{BC}+\vec{BC}\cdot\vec{BC},
$$
in other words,
$$
\|AB\|^2=\|AC\|^2-2\vec{AC}\cdot\vec{BC}+\|BC\|^2.
$$
It is known (and not very hard to prove) that
$$
|\vec{AC}\cdot\vec{BC}|\le\|AC\|\|BC\|.
$$
Then
$$
\frac{\vec{AC}\cdot\vec{BC}}{\|AC\|\|BC\|}=\frac{\vec{AC}}{\|AC\|}\cdot\frac{\vec{BC}}{\|BC\|}\in[-1,1].
$$
This is the inner product of unit vectors in the directions of $\vec{AC}$ and $\vec{BC}$. Thus, it does not depend on the lengths $\vec{AC}$ and $\vec{BC}$, but only on their directions, so it is invariant under translation. In fact, you can also show that it is invariant under any rotation (and also under reflection, but that's a bonus), so it is indeed the function of just the angle $\angle ACB$. Let us call this function the cosine of $\angle ACB$ and denote it by $\cos\angle ACB$, i.e.
$$
\cos\angle ACB=\frac{\vec{AC}\cdot\vec{BC}}{\|AC\|\|BC\|}.
$$
(From this, we define all other trigonometric functions.) Therefore,
$$
\|AB\|^2=\|AC\|^2-2\|AC\|\|BC\|\cos\angle ACB+\|BC\|^2.
$$
A: Notice that when $C=\pi/2=90^\circ$, the law of cosines reduces to the Pythagorean theorem. Try looking at special cases when $(a,b,c)$ is equilateral or isosceles.
I'm not sure what you mean by a "direct" understanding. Try looking at the various proofs and see how they all relate to each other. I consider the above proof to be the most succinct and descriptive.
