Hi adpcane!
I'm learning Trig myself but I think I've solved this one!
Since
Cos2A=Cos(A+A) we can use: \(\displaystyle \L Cos\,(A\,+\,B)\,=\,CosA\,CosB\,-\,SinA\,SinB\)
Except we'll manipulate it to
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
Cos(2A)=Cos2A−Sin2A
\(\displaystyle \L 2cos^2A\,-\,1=Cos\,(2A)\)