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!
 A: For example:  $$8\cos^4x-8\cos^2x+1=1-8\cos^2x\sin^2x=1-2\sin^22x=\cos4x.$$
Also, we can use your idea:
$$\cos4x=2\cos^22x-1=2(2\cos^2x-1)^2-1=8\cos^4x-8\cos^2x+1.$$
A: 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) $0=0$
$\therefore 1 = 2$.
A: Your idea can work but it's not very well-explained. You want to show
$$\cos(4x)=8\cos^4(x)-8\cos^2(x)+1.$$
You first apply the double angle formula on the RHS, so it suffices to show
$$\cos^2(2x)-\sin^2(2x)=8\cos^4(x)-8\cos^2(x)+1.$$
Now you don't like the $\sin^2(2x)$ so you replace it with $1-\cos^2(2x)$ so that it suffices to show
$$2\cos^2(2x)=8\cos^4(x)-8\cos^2(x)+2.$$
Now you can divide both sides by $2$, rewrite the RHS as a square, apply another double angle identity, and arrive at the tautology $\cos^2(2x)=\cos^2(2x)$.
The important thing from the logical point of view is that all your steps need to be two-way, in order to prove the desired statement from the tautology and trig identities, rather than assuming what you want to show and deriving a tautology.  Thus you should not write $\Rightarrow$ in a derivation like this, since it makes it look like your steps are one-way when they're actually not.
If you're going to connect your statements with arrows at all, they should probably be $\Leftrightarrow$ (in principle $\Leftarrow$ would work too, but that would be a very weird way to explain things). Alternatively, you can write some steps in the reverse direction and then reverse all the steps to write your actual proof (which is also helpful for making sure that each of your steps really is reversible).
A: Well your proof is not "wrong" because each of the step you did was two way ie. Each of the two consecutive steps were equivalent. You should start at last step of you "argument". This "method" is often used in schools to prove some trivial logarithmic inequalities. But remember that this "method" doesn't work in general. Each of the steps must be reversible. So start the proof from the last step.
