A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Explain this identity.
anonymous
 4 years ago
Explain this identity.

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm... is that e^(ix) + e^(ix) so = 2e^(ix) on the right side?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\cos(x)=\frac{e^{ix}+e^{ix}}{2} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that means I can split cos(x) into two different functions right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what do you mean split up cosx ? that looks a bit like the hperbolic coshx...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well 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
 4 years ago
Best ResponseYou've already chosen the best response.0Can you at least show me how \[\cos(x)=\frac{e^{ix}+e^{ix}}{2} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is guess, but try doing the series expansion for e^(ix) + e^(ix). i have a feeling the imaginary terms will cancel.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Write e^ix =cos(x)+i sin(x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0e^(ix) = cos x + i sin x (Euler's Formula) e^(ix) = cos(x) + i sin (x) = cos x i sin x Add them.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.