I'm reading Aarts': Plane and Solid Geometry. Here:
How did he go from $b_{12}$ to $\tan 2\varphi=\cfrac{2a_{12}}{a_{11}-a_{22}}$? I have tried a few things on paper but I throw it all away because there was no progress, but I basically used the formulas for double angles and $\sin^2(x)+\cos^2(x)=1$. I've even speculated that one must be high to understand it. The tag of the formula is $(4.20)$, perhaps this is no coincidence but I smoked some joints and nothing happened.
You have: $$b_{12} = -\frac{1}{2}a_{11}\sin 2\varphi + a_{12}\cos 2\varphi + \frac{1}{2}a_{22}\sin 2\varphi.$$
Division by $\cos 2\varphi$ gives: $$b_{12}\frac{1}{\cos 2\varphi} = -\frac{1}{2}a_{11}\tan 2\varphi + a_{12} + \frac{1}{2}a_{22}\tan 2\varphi.$$ From here: $$\tan 2\varphi = \frac{2\left(a_{12} - \frac{b_{12}}{\cos 2\varphi} \right)}{a_{11} - a_{22}}.$$ Surely you're not too high to conclude now?