We are going to show that for any angles a, b the trigonometry formula cos (a-b)=cos a cos b + sin a sin b

We consider the demonstration of cos (a+b)=cos a cos b - sin a sin b as established.

It follows that:

\[\forall x,y \in \mathbb{R}, \quad \cos (x+y)=\cos x \cos y-\sin x \sin y\]

In particular, by making the change of variable $x=a$, and $y=-b$

\[\cos(x+y)=\cos (a-b)=\cos a \cos (-b)-\sin a \sin (-b)\]

Since the cosine function is even:

\[\cos (-b)=\cos b\]

and the sine function is odd:

\[\sin (-b)=-\sin b\]

we have:

\[\cos (a-b)=\cos a \cos b - \sin a \times - \sin b\]

We conclude:

\[\forall a,b \in \mathbb{R},\cos (a-b)=\cos a \cos b + \sin a \sin b\]