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

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

It follows that:

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

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

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

Since the cosine function is even:

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

and the sine function is odd:

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

we have:

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

We conclude:

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