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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? ]

hyperpallium
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  • You might be interested in this Mathologer video on the Pythagorean Theorem and variations. It includes a nice proof of the law of cosines. https://www.youtube.com/watch?v=p-0SOWbzUYI – awkward Mar 22 '18 at 13:21
  • Perhaps this answer, with my Law of Cosines trigonograph, is close to what you seek. – Blue Apr 08 '18 at 12:44

2 Answers2

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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, $$ \|AB\|^2=\|AC\|^2-2\|AC\|\|BC\|\cos\angle ACB+\|BC\|^2. $$

Alexander Burstein
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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.

terrygarcia
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  • By "direct", I mean from looking directly at the formula (not indirect, via the derivation steps). e.g. from a triangle with sides $a,b,c$, draw a right triangle with perpendicular side lengths $a,b$. And then see, somehow, that adjusting its hypotenuse squared by $$-2ab\cos C$$ gives you $c^2$. Maybe any kind of non-trivial proof will be "indirect", in having steps. I just want to understand - but I guess any authentic understanding will also be a proof, and so have these indirect steps. – hyperpallium Mar 22 '18 at 06:43