<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] 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:
$$\cos4x=2\cos^22x-1=2(2\cos^2x-1)^2-1=8\cos^4x-8\cos^2x+1.$$
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