Explain this identity.

Question

anonymous · Answer

hmm... is that e^(ix) + e^(ix) 
so =  2e^(ix)
on the right side?

anonymous · Answer

$$\cos(x)=\frac{e^{ix}+e^{-ix}}{2} $$

anonymous · Answer

that means I can split cos(x) into two different functions right?

anonymous · Answer

what do you mean split up cosx ?
that looks a bit like the hperbolic coshx...

anonymous · Answer

Well there is a problem where I try to show if cos(x) is an eigenfunction of an operator. I did the operation and found out it wasn't but then my book says if it's not a the eigenfunction that it must be a superposition eigenfunction so it splits it into two functions. I just dont see how it splits or how this superposition will be the eigenfunction.

anonymous · Answer

Can you at least show me how
$$\cos(x)=\frac{e^{ix}+e^{-ix}}{2} $$

anonymous · Answer

this is guess, but try doing the series expansion for e^(ix) + e^(-ix).  i have a feeling the imaginary terms will cancel.

anonymous · Answer

Write e^ix =cos(x)+i sin(x)

anonymous · Answer

e^(ix) = cos x + i  sin x  (Euler's Formula)
e^(-ix) = cos(-x) + i sin (-x) = cos x- i sin x
Add them.