# Cosine Difference Identity

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 $$\overline{CS} =\sqrt{(\cos (\alpha - \beta) - 1)^2 + (\sin (\alpha - \beta) - 0)^2}$$

Now set $$\overline{CS} = \overline{AB}$$ and with a bit of algebraic manipulation, we arrive at the equation from the beginning....

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 $$-2$$.

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