Find the rotation matrix for +15° out of the rotation matrix +60° without using trigonometric functions Find the rotation matrix for +15° out of the rotation matrix +60° without using trigonometric functions. And as much as I would love to tell you what I did so far. I don't even know where to start with this exercise.. 
The matrix for +60° is this.
$$\left[
\begin{matrix} 
1/2 &  -\sqrt 3/2\\
\sqrt 3/2 & 1/2 \\
\end{matrix} \right]
$$
How do I even start?

The only approach I can see is to take the fourth root of the matrix, that is, find a $2 \times 2$ matrix $A$, such that $A^4=\left(\begin{matrix}
1/2 &  -\sqrt 3/2\\
\sqrt 3/2 & 1/2 \\
\end{matrix}\right)$.  If you expect that it is of the form $\left(\begin{matrix}
a &  -b\\
b & a \\
\end{matrix}\right)$ with $a^2+b^2=1$, it should go through.
Added:  If you first take a square root, you get $a^2-b^2=1/2, -2ab=\sqrt{3}/2$, which gives $b^2=\frac{3}{16a^2}, a^2-\frac{3}{16a^2}=\frac{1}{2}, a=\frac{\sqrt{3}}{2}, b=\frac{1}{2}$ (taking the positive root).  This give the $30^{\circ}$ matrix as $A^2=\left(\begin{matrix}
\sqrt 3/2 &  -1/2\\
1/2 & \sqrt 3/2 \\
\end{matrix}\right)$.  
Now the same approach gives $a^2-b^2=\sqrt 3/2, -2ab=1/2$ and you can get a quadratic in $a^2$ again.
