The following transformation matrix describes a rotation $r_α :\mathbb{R}^2\to \mathbb{R}^2$ that rotates with angle $α$ to the left around the null vector with respect to the standard basis :
$M^B_B(r_α)$=$\left( \begin{array}{rrr}
\cos(α) & -\sin(α) \\
\sin(α) & \cos(α) \\
\end{array}\right)$.
If i first rotate with angle $α$ and then with angle $\beta$ it would be the same as $\alpha+\beta$.
How could i apply the given information to derive formulas to calculate $\cos(\alpha+\beta)$ and $\sin(\alpha+\beta)$ using $\cos(\alpha)$, $\sin(\alpha)$, $\cos(\beta)$, $\sin(\beta)$ ?
Thanks for any type of help 🙂
Best Answer
$$\begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}\begin{bmatrix} \cos \beta & -\sin \beta \\ \sin \beta & \cos \beta \end{bmatrix} =\begin{bmatrix} \cos( \alpha +\beta ) & -\sin( \alpha +\beta )\\ \sin( \alpha +\beta ) & \cos( \alpha +\beta ) \end{bmatrix}$$
Multiply.
Expanding on my answer.
We want rotation matrices to behave in a natural way. I.e if $$\mathbf{R}(\theta)=\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$$ Intuitively, a rotation of an angle $\alpha+\beta$ should be the same as a composition of a rotation by an angle $\alpha$ and then an angle $\beta$. Since composition of matrices corresponds to multiplication of them, this means we would expect $$\mathbf{R}(\alpha)\mathbf{R}(\beta)=\mathbf{R}(\alpha+\beta) \\ \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}\begin{bmatrix} \cos \beta & -\sin \beta \\ \sin \beta & \cos \beta \end{bmatrix}=\begin{bmatrix} \cos (\alpha+\beta) & -\sin (\alpha+\beta) \\ \sin (\alpha+\beta) & \cos (\alpha+\beta) \end{bmatrix}$$ If you multiply the two matrices on the left, this gives you $$\begin{bmatrix} \cos\alpha\cos\beta-\sin\alpha\sin\beta & -\sin\alpha\cos\beta-\cos\alpha\sin\beta \\ \sin\alpha\cos\beta+\cos\alpha\sin\beta & \cos\alpha\cos\beta-\sin\alpha\sin\beta \end{bmatrix}=\begin{bmatrix} \cos (\alpha+\beta) & -\sin (\alpha+\beta) \\ \sin (\alpha+\beta) & \cos (\alpha+\beta) \end{bmatrix}$$ I.e, matching the components of the two equivalent matrices, $$\cos (\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta \\ \sin (\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$$ Alternatively, reversing the sign of $\beta$ we get $$\cos (\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta \\ \sin (\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$$ Exercise: Try combining the expressions for $\cos(\alpha-\beta),\cos(\alpha+\beta)$ to get an expression for $\cos(\alpha)\cos(\beta)$ and similarly for $\sin$.