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

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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:

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

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