# Cosine of Difference

## Theorem

$\map \cos {a - b} = \cos a \cos b + \sin a \sin b$

where $\sin$ denotes the sine and $\cos$ denotes the cosine.

## Proof 1

 $\ds \map \cos {a - b}$ $=$ $\ds \cos a \map \cos {-b} - \sin a \map \sin {-b}$ Cosine of Sum $\ds$ $=$ $\ds \cos a \cos b - \sin a \map \sin {-b}$ Cosine Function is Even $\ds$ $=$ $\ds \cos a \cos b + \sin a \sin b$ Sine Function is Odd

$\blacksquare$

## Proof 2

Consider two radii $OP$ and $OQ$ of a unit circle whose center is at the origin of a Cartesian plane.

Let:

 $\ds \angle xOP$ $=$ $\ds B$ $\ds \angle xOQ$ $=$ $\ds A$

Then the coordinates of $P$ and $Q$ are given by:

 $\ds P$ $=$ $\ds \tuple {\cos B, \sin B}$ $\ds Q$ $=$ $\ds \tuple {\cos A, \sin A}$

Hence:

 $\ds PQ^2$ $=$ $\ds \paren {\cos B - \cos A}^2 + \paren {\sin B - \sin A}^2$ $\ds$ $=$ $\ds \cos^2 B - 2 \cos A \cos B + \cos^2 A + \sin^2 B - 2 \sin A \sin B + \sin^2 A$ multiplying out $\ds$ $=$ $\ds 2 - 2 \cos A \cos B - 2 \sin A \sin B$ Sum of Squares of Sine and Cosine $\ds$ $=$ $\ds 1 + 1 - 2 \map \cos {A - B}$ Law of Cosines, as $\angle POQ = A - B$ $\ds \leadsto \ \$ $\ds \map \cos {A - B}$ $=$ $\ds \cos A \cos B + \sin A \sin B$ simplifying

$\blacksquare$

## Historical Note

The Cosine of Sum formula and its corollary were proved by François Viète in about $1579$.