Hi adpcane!
I'm learning Trig myself but I think I've solved this one!
Since \(\displaystyle Cos\,2A\,=\,Cos\,(A\,+\,A)\) we can use: \(\displaystyle \L Cos\,(A\,+\,B)\,=\,CosA\,CosB\,-\,SinA\,SinB\)
Except we'll manipulate it to \(\displaystyle Cos\,(A\,+\,A)\)
\(\displaystyle \L 2cos^2A\,-\,1=\,CosA\,CosA\,-\,SinA\,SinA\)
This left side can be rewritten as:\(\displaystyle \L \;Cos^2A\,-\,Sin^2A\)
Since \(\displaystyle Cos\,(2A)\,=\,Cos^2A\,-\,Sin^2A\)
\(\displaystyle \L 2cos^2A\,-\,1=Cos\,(2A)\)