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Home > Mathematics > Trigonometry > Trigonometry addition formula cos(a-b)=cos a cos b + sin a sin b
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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:
$$\cos (-b)=\cos b$$
$$\sin (-b)=-\sin b$$
we have:
$$\cos (a-b)=\cos a \cos b - \sin a \times - \sin b$$
We conclude:
$$\forall a,b \in \mathbb{R},\cos (a-b)=\cos a \cos b + \sin a \sin b$$