Proof of the cosine subtraction formula

All the addition formulae descend from a single fundamental identity, the one for $\cos(\alpha-\beta)$. It is proved by computing in two different ways the same distance on the unit circle.

Proof — Formula for $\cos(\alpha-\beta)$

Consider two points on the unit circle: $P_\alpha(\cos\alpha;\sin\alpha)$ and $P_\beta(\cos\beta;\sin\beta)$. The distance $d$ between them can be computed in two ways.

First way (Cartesian coordinates):

d^2 &= (\cos\alpha-\cos\beta)^2 + (\sin\alpha-\sin\beta)^2 \\ &= \cos^2\alpha - 2\cos\alpha\cos\beta + \cos^2\beta + \sin^2\alpha - 2\sin\alpha\sin\beta + \sin^2\beta \\ &= 2 - 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta). \end{aligned}$$ **Second way** (cosine rule in the triangle $OP_\alpha P_\beta$, with $O$ the origin): the sides $OP_\alpha$ and $OP_\beta$ both have length $1$ (the radius of the circle) and the angle between them is $\alpha-\beta$, hence $$d^2 = 1 + 1 - 2\cdot 1\cdot 1\cdot \cos(\alpha-\beta) = 2 - 2\cos(\alpha-\beta).$$ Equating the two expressions for $d^2$ and simplifying: $$\cos(\alpha-\beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta. \qquad \blacksquare$$

Construction for the proof of $\cos(\alpha-\beta)$: the chord $d=P_\alpha P_\beta$ of the triangle $OP_\alpha P_\beta$ has a length computable in two ways, from which the formula follows.

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Topics: Trigonometric formulae
Concepts: Unit circle · Addition formulae · Cosine rule
Functions: Cosine · Sine
Skills: Proving