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# Trigonometry addition formula cos(a-b)=cos a cos b + sin a sin b

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 $y=-b$

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

$$\cos (-b)=\cos b$$

$$\sin (-b)=-\sin b$$

we have:

$$\cos (a-b)=\cos a \cos b - \sin a \times - \sin b$$

We conclude:

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