Derivation of basic rotation matrices in three dimensions

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I was looking at these matrices and it got me thinking:

Can these matrices be derived?

How do we know that in order to rotate around z for example, we have to $\cos\theta$, $\sin\theta$ in the first column and $0$)?

So far I have figured out only that whatever axis we rotate around, we place a 1 in the $R_{11}$, $R_{22}$ or $R_{33}$ element (corresponding to rotation around $x$, $y$ and $z$).

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Rotation matrix in $N$ dimensions are given by elements of the special orthogonal group $SO(N)$. A Lie group. Elements in this group satisfy the two relations:

  • $R^TR=\mathbf{1}_N$
  • $\mathrm{det}(R)=1$

The first one imply the rigidity of the rotation ($\mathbf{v}\cdot\mathbf{w}=\mathbf{v}R^T\cdot R\mathbf{w}$), while the second one select rotations without reflection about an axis. The case $N=2$ is a simple exercise in which you can find a parametrization in terms of cosine and sine for a general rotation matrix in two dimension. Try to use a generic matrix $$ R=\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$$ and impose the constraints.

See special orthogonal group on wikipedia.

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Note that the first, second, and third columns of each rotation matrix correspond the image of the $\hat{i}$, $\hat{j}$, and $\hat{k}$ unit vectors under the rotation.

When rotating $\theta$ radians around the $x$-axis, $\hat{i}$ remains at $$\left(\begin{array}{c}1\\ 0\\ 0\\ \end{array}\right)$$ while $\hat{j}$ is mapped to $$\left(\begin{array}{c} 0\\ \cos{\theta}\\ \sin{\theta}\\ \end{array}\right)$$ and $\hat{k}$ is mapped to $$\left(\begin{array}{c} 0\\ -\sin{\theta}\\ \cos{\theta}\\ \end{array}\right)$$

Ergo, the rotation matrix for rotation about the $x$-axis is $$\left(\begin{array}{c c c}1&0&0\\0&\cos{\theta}&-\sin{\theta}\\0&\sin{\theta}&\cos{\theta}\end{array}\right)$$

You can derive the other rotation matrices in a similar fashion.