Geometric proof of the subtraction theorem for the cosine. $$ \cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta $$
Have a look at the following diagram. There are four points on the unit circle:
Note that the angle between \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\) is the same as that between \(\overrightarrow{OC}\) and \(\overrightarrow{OS}\), namely \(\alpha - \beta\). So the red chords must have the same length. This is true for any angles \(\alpha\) and \(\beta\). For example, if we'd swap A and B in the diagram above, \(\alpha - \beta\) would be negative and C in the fourth quadrant, but the length of the chord wouldn't be affected.
There are two ways of calculating the chord length via the Euclidean distance.
The distance between A and B is
$$
\overline{AB} = \sqrt{(\cos \alpha - \cos \beta)^2 + (\sin \alpha - \sin \beta)^2 }
$$
and the distance between C and \((1, 0)\) is
$$
\overline{CS} =\sqrt{(\cos (\alpha - \beta) - 1)^2 + (\sin (\alpha - \beta) - 0)^2}
$$
Square both sides to get rid of the square roots.
$$
(\cos (\alpha - \beta) - 1)^2 + \sin^2 (\alpha - \beta) = (\cos \alpha - \cos \beta)^2 + (\sin \alpha - \sin \beta)^2
$$
Expand the squared sums into individual terms.
$$
\begin{align*}
&\cos^2 (\alpha - \beta) - 2 \cos (\alpha - \beta) + 1 + \sin^2 (\alpha - \beta)\\
&= \cos^2 \alpha - 2 \cos \alpha \cos \beta + \cos^2 \beta + \sin^2 \alpha - 2 \sin \alpha \sin \beta + \sin^2 \beta
\end{align*}
$$
Note that, by the unit circle definition of sine and cosine, \(\cos^2 \theta + \sin^2 \theta = 1\).
So we'll group terms by angle to get the following equation.
$$
\cancel{2} - 2 \cos (\alpha - \beta) = \cancel{2} - 2 \cos \alpha \cos \beta - 2 \sin \alpha \sin \beta
$$
Divide both sides by \(-2\).
Now set \(\overline{CS} = \overline{AB}\) and with a bit of algebraic manipulation, we arrive at the equation from the beginning....
$$ \cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta $$