What is the most elegant and simple proof for the law of cosines?

Given 2 sides, and an angle between those two sides, what is the simplest proof you can come up with to find the measure of the 3rd side?

Solution 1:

$\hspace{5cm}$ Using Properties of the Dot Product

$\hspace{4.5cm}$ $$ \begin{align} |a-b|^2 &=(a-b)\cdot(a-b)\\ &=a\cdot a+b\cdot b-2\,a\cdot b\\ &=|a|^2+|b|^2-2|a||b|\cos(\theta) \end{align} $$ Justification of $\mathbf{a\cdot b=|a||b|cos(\theta)}$:

Using the formula for the cosine of a difference, we have $$ \begin{align} a\cdot b &=|a|\left(\cos(\alpha),\sin(\alpha)\right) \cdot |b|\left(\cos(\beta),\sin(\beta)\right)\\ &=|a||b|\left(\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)\right)\\ &=|a||b|\cos(\alpha-\beta)\\ \end{align} $$

$\hspace{2cm}$

$$ \begin{align} c^2 &=(b\cos(\theta)-a)^2+(b\sin(\theta))^2\\ &=a^2+b^2(\sin^2(\theta)+\cos^2(\theta))-2ab\cos(\theta)\\ &=a^2+b^2-2ab\cos(\theta) \end{align} $$

Solution 2:

Here's a proof similar to the dot product proof in previous answers, but using complex numbers and Euler's formula:

$$ |a-b|^2 = (a-b) \overline{(a-b)} = a \overline{a} + b \overline{b} - (\overline{a} b + a \overline{b}) = \\ |a|^2 + |b|^2 - |a||b|(e^{i(\arg{b} - \arg{a})} + e^{i(\arg{a} - \arg{b})}) = \\ |a|^2 + |b|^2 - 2|a||b|\cos{\theta} $$

Solution 3:

Here is my take. Consider the formula and the following picture:

$$ {\color{darkgreen}{c^2}} = {\color{red}{a^2} + \color{darkorange}{b^2}} - {\color{darkgreen}{2ab\cos\theta}} $$

$\hspace{90pt}$

The yellow vectors are precisely $a$ and $b$, where $a > b$. The only tricky point is that the area of two small red squares sums up to the orange square (so that the big red square $\color{red}{a^2}$ does not overlap with the smaller orange $\color{darkorange}{b^2}$ one), but this is due the blue triangle and Pythagorean theorem. Of course one should not forget to apply Pythagorean theorem also to (big green square) = (orange square) + (middle red square), after this it's just a rearrangement of the rest.

Cheers!