Why is $\cos(\alpha+\beta) = \cos(\alpha)\cos(\beta)−\sin(\alpha)\sin(\beta)$

Question

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.

Answer 1

1. The rotation of $\mathbb R^2$ through angle $\alpha$ is a linear transformation with matrix $\left( \begin{matrix} \cos\alpha & -\sin\alpha\\ \sin\alpha &\cos\alpha \end{matrix}\right)$

2. The rotation of $\mathbb R^2$ through angle $\beta$ is a linear transformation with matrix $\left( \begin{matrix} \cos\beta & -\sin\beta\\ \sin\beta &\cos\beta \end{matrix}\right)$

3. The rotation of $\mathbb R^2$ through angle $\alpha+\beta$ is a linear transformation with matrix $\left( \begin{matrix} \cos(\alpha+\beta) & -\sin(\alpha+\beta)\\ \sin(\alpha+\beta) &\cos(\alpha+\beta) \end{matrix}\right)$

4. The rotation of $\mathbb R^2$ through angle $\alpha+\beta$ is the composition of the rotation of $\mathbb R^2$ through angle $\alpha$ and the rotation of $\mathbb R^2$ through angle $\beta$.

5. The matrix of the composition of two linear transformations is the product of matrices of these transformations.

6. So $$ \left( \begin{matrix} \cos\alpha & -\sin\alpha\\ \sin\alpha &\cos\alpha \end{matrix}\right)\cdot \left( \begin{matrix} \cos\beta & -\sin\beta\\ \sin\beta &\cos\beta \end{matrix}\right) = \left( \begin{matrix} \cos(\alpha+\beta) & -\sin(\alpha+\beta)\\ \sin(\alpha+\beta) &\cos(\alpha+\beta) \end{matrix}\right)$$

7. So, in particular, $$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.$$