Derive formuli for $$\sin(\alpha+\beta)$$ and $$\cos(\alpha+\beta)$$ by using matrices of linear maps.

• #### Hint

Determine the matrix of the composed mapping.

• #### Resolution

When $$f$$ and $$g$$ are the rotations by angle $$\alpha$$ and $$\beta$$, then their matrices are:

$$[f]_{KK}= \begin{pmatrix} \cos \alpha & -\sin\alpha \\ \sin \alpha & \cos\alpha \\ \end{pmatrix}$$ a $$[g]_{KK}= \begin{pmatrix} \cos \beta & -\sin\beta \\ \sin \beta & \cos\beta \\ \end{pmatrix}$$

The composition of $$f$$ and $$g$$ is the rotation by angle $$\alpha+\beta$$. Its matrix is:

$$[g\circ f]_{KK}= \begin{pmatrix} \cos (\alpha+\beta) & -\sin(\alpha+\beta) \\ \sin (\alpha+\beta) & \cos(\alpha+\beta) \\ \end{pmatrix}$$

Simultaneously, the following must hold:

$$[g\circ f]_{KK}=[g]_{KK}[f]_{KK}= \begin{pmatrix} \cos \beta & -\sin\beta \\ \sin \beta & \cos\beta \\ \end{pmatrix} \begin{pmatrix} \cos \alpha & -\sin\alpha \\ \sin \alpha & \cos\alpha \\ \end{pmatrix} =\\= \begin{pmatrix} \cos\beta\cos\alpha -\sin\beta\sin\alpha & -\cos\beta\sin\alpha -\sin\beta\cos\alpha \\ \sin\beta\cos\alpha +\cos\beta\sin\alpha & -\sin\beta\sin\alpha +\cos\beta\cos\alpha \\ \end{pmatrix}$$

As both matrices should be qual, we get the desired formuli:
$$\sin (\alpha+\beta) = \sin\beta\cos\alpha +\cos\beta\sin\alpha$$,
$$\cos(\alpha+\beta) = -\sin\beta\sin\alpha +\cos\beta\cos\alpha$$.