<s>[INST] 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!
 [/INST] 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) $0=0$
$\therefore 1 = 2$.
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