Cosine of Difference/Proof 2

Theorem

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

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 $

$=$

$\ds 2 - 2 \cos A \cos B - 2 \sin A \sin B$

$\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

$\blacksquare$