## Romero 3 years ago Explain this identity.

1. dpaInc

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

2. Romero

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

3. Romero

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

4. dpaInc

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

5. Romero

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.

6. Romero

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

7. dpaInc

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

8. shivam_bhalla

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

9. estudier

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