Simple examples of $3 \times 3$ rotation matrices I'd like to have some numerically simple examples of $3 \times 3$ rotation matrices that are easy to handle in hand calculations (using only your brain and a pencil). Matrices that contain too many zeros and ones are boring, and ones with square roots are undesirable. A good example is something like
$$
M = \frac19 \begin{bmatrix}
1 & -4 & 8 \\
8 &  4 & 1  \\
-4 & 7 & 4
\end{bmatrix}
$$
Does anyone have any other examples, or a process for generating them?
One general formula for a rotation matrix is given here. So one possible approach would be to choose $u_x$, $u_y$, $u_z$ and $\theta$ so that you get something simple. Simple enough for hand calculations, but not trivial. Like the example given above.
 A: Here are some:
$$\left[ \begin {array}{ccc} 1/3&2/3&2/3\\ 2/3&-2/3&1/3\\ 2/3&1/3&-2/3\end {array} \right] 
$$
$$\left[ \begin {array}{ccc} 2/7&3/7&6/7\\ 3/7&-6/7&2/7\\ 6/7&2/7&-3/7\end {array} \right] 
$$
$$
\left[ \begin {array}{ccc} \frac{2}{11}&{\frac {6}{11}}&{\frac {9}{11}}\\ -{\frac {6}{11}}&-{\frac {7}{11}}&{\frac {6}{11}}
\\ {\frac {9}{11}}&-{\frac {6}{11}}&\frac{2}{11}\end {array}
 \right] 
$$
EDIT: here are the small positive integer solutions of $a^2 + b^2 + c^2 = d^2$ with
$a \le b \le c$ and $\gcd(a,b,c,d)=1$, in order of increasing $b^2 + c^2$:
$$\eqalign{1^2 + 2^2 + 2^2 &= 3^2\cr
2^2 + 3^2 + 6^2 &= 7^2\cr
4^2 + 4^2 + 7^2 &= 9^2\cr
1^2 + 4^2 + 8^2 &= 9^2\cr
6^2 + 6^2 + 7^2 &= 11^2\cr
2^2 + 6^2 + 9^2 &= 11^2\cr
3^2 + 4^2 + 12^2 &= 13^2\cr
2^2 + 10^2 + 11^2 &= 15^2\cr
2^2 + 5^2 + 14^2 &= 15^2\cr
8^2 + 9^2 + 12^2 &= 17^2\cr
1^2 + 12^2 + 12^2 &= 17^2\cr
}$$
A: Keep in mind that the product of two rational matrices is going to be a rational matrix too.  This allows you to build plenty of rational rotaton matrices by composing them.  For instance:
$$
\frac15 \begin{pmatrix}
3 & 4 & 0 \\
-4 & 3 & 0  \\
0 & 0 & 5
\end{pmatrix}
\cdot
\frac1{13} \begin{pmatrix}
5 & 0 & -12 \\
0 & 13 & 0 \\
12 & 0 & 5
\end{pmatrix} = 
\frac1{65} \begin{pmatrix}
15 & 52 & -36 \\
-20 & 39 & 48 \\
60 & 0 & 25
\end{pmatrix}
$$
and so the latter is also a rational rotation matrix.
A: Some entries of the rotation matrix, whether $2 \times 2$ or $3 \times 3$, are the trigonometric functions; to ensure that entries of the matrix are simple numbers that are less computationally expensive, pick integer multiples of $\pi$ on which the trigonometric functions are either $1, -1$ or $0$.
Is that what you meant?
A: Whenever I get a chance to teach Linear algebra I do things like the following to produce "nice" rotation matrices. The basic idea is that a composition of two reflections is always a rotation.  Restricting myself to 3D in what follows.
The reason why I think this fits the bill here is that reflections usually have nice
matrices. If we reflect $\Bbb{R}^3$ w.r.t. to the plane with normal $\vec{n}=(n_1,n_2,n_3)$, then that reflection $s$ is given by the recipe
$$
s(\vec{x})=\vec{x}-2\,\frac{\vec{x}\cdot\vec{n}}{\Vert\vec{n}\Vert^2}\vec{n}.
$$
If $\vec{n}$ has rational components, then the matrix of $s$ w.r.t. the standard basis will have rational entries. As we need two reflection to get a rotation, we can either multiply two such matrices, or may be use a very easy choice of $\vec{n}$ for the other.
For example, the reflection w.r.t. the plane $3x+2y+z=0$ with $\vec{n}=(3,2,1)$ sends 
$$
\begin{aligned}
(1,0,0)&\mapsto(1,0,0)-\frac37(3,2,1)=\frac17(-2,-6,-3),\\
(0,1,0)&\mapsto(0,1,0)-\frac27(3,2,1)=\frac17(-6,3,-2),\\
(0,0,1)&\mapsto(0,0,1)-\frac17(3,2,1)=\frac17(-3,-2,6).
\end{aligned}
$$
If we (post)compose this with the reflection $(x,y,z)\mapsto(-x,y,z)$, we get the rotation represented by the matrix
$$
R=\frac17\left(
\begin{array}{rrr}
2&6&3\\
-6&3&-2\\
-3&-2&6
\end{array}\right).
$$
The axis of rotation here has to be a vector that is perpendicular to both 
$\vec{n}$ and $(1,0,0)$ (the normal of the second plane of reflection).
The cross product $\vec{w}=(0,-1,2)$ is one such vector, and you can easily
verify that $R(0,-1,2)^T=(0,-1,2)^T$.
