Proving $\cos(4x)=8\cos^4(x)-8\cos^2(x)+1$ without using LHS and RHS 
Prove that
$$\cos(4x)=8\cos^4(x)-8\cos^2(x)+1$$

My Attempt
$$\Rightarrow \cos^2(2x)-\sin^2(2x) =8\cos^4(x)-8\cos^2(x)+1$$
Add 1 to both sides
$$\cos^2(2x)-\sin^2(2x) +1=8\cos^4(x)-8\cos^2(x)+2$$$$\Rightarrow \cos^2(2x)=4\cos^4(x)-4\cos^2(x)+1$$
$$\Rightarrow \cos^2(2x)=(2\cos^2(x)-1)^2$$ $$\Rightarrow \cos^2(2x)=(\cos^2(x)-\sin^2(x))^2=(\cos(2x))^2$$
$$\therefore \cos(4x)=8\cos^4(x)-8\cos^2(x)+1$$
My professor says that this is an invalid proof as it was not proven using LHS and RHS and assumed that they were equivalent to add 1 to each side. My question: are there any proofs that would not work if you assumed that the two side were equivalent? Any examples will be much appreciated!

Your proof is not wrong, it's just backwards. You need to start with $\cos^2(2x) = \cos^2(2x)$, then do the same manipulations you did in reverse order to end up with $ \cos(4x)=8\cos^4(x)-8\cos^2(x)+1$.
But your question is can a backwards proof ever be wrong? Yes it can. Consider this very simple "proof" that $1=2$:
(1): $1=2$
(2): $1*0=2*0$ (multiplying both sides by 0)
(3) $0=0$
$\therefore 1 = 2$.
