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:

- \(S = (1, 0)\), the start point at an angle of 0
- \(A = (\cos \alpha, \sin \alpha)\) at an angle \(\alpha\),
- \(B = (\cos \beta, \sin \beta)\) at an angle \(\beta\), and
- \(C = (\cos (\alpha - \beta), \sin (\alpha - \beta))\) at an angle \(\alpha - \beta\).

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 \(-2\).

$$ \cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta $$