TITLE: Why is $\cos(\alpha+\beta) = \cos(\alpha)\cos(\beta)−\sin(\alpha)\sin(\beta)$?
QUESTION [7 upvotes]: I already posted a question about transformation matrices and rotation. But I'm not satisfied with the answer.
They simply said

Composition of functions corresponds to multiplication of matrices.

I think I understand the concept but I'm still confused why exactly $$\cos(\alpha + \beta)=\cos(\alpha)\cos(\beta)−\sin(\alpha)\sin(\beta)$$
Is there a lemma or formula I have to use or does it simply derive from distributivity of matrix multiplication? I can't get my head around it.

REPLY [10 votes]: You can also simply prove it using complex  numbers :
$$ e^{i(\alpha + \beta)} = e^{i\alpha} \times e^{i\beta} 
 \Leftrightarrow \cos (a+b)+i \sin (a+b)=(\cos a+i \sin a) \times(\cos b+i \sin b) $$
Finally we obtain, after distributing :
$$ \cos (a+b)+i \sin (a+b) =\cos a \cos b-\sin a \sin b+i(\sin a \cos b+\cos a \sin b) $$
By identifying the real and imaginary parts we get
$$
\cos (a+b)=\cos a \cos b-\sin a \sin b
$$
$$ \sin (a+b)=\sin a \cos b+\cos a \sin b $$
Of course you need know basics about complex numbers but if you already know all of this it is usually really quick to proove most of this cumbersome trigonometric formulas with complex numbers like what I did here.