Use angle sum ident to verify cos(2x)=2cos^2(x)-1

sum identity for cosine is ...
cos(x+y) = cosx*cosy - sinx*siny

now, cos(2a) = cos(a+a) ... ball is in your court, play it.
 
Hi adpcane!

I'm learning Trig myself but I think I've solved this one! :D

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 \(\displaystyle Cos\,(2A)\,=\,Cos^2A\,-\,Sin^2A\)

\(\displaystyle \L 2cos^2A\,-\,1=Cos\,(2A)\)
 
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