- Thread starter
- #1

- Thread starter shen07
- Start date

- Thread starter
- #1

- Thread starter
- #2

- Thread starter
- #3

- Admin
- #4

- Jan 26, 2012

- 4,043

Hi shen07 ,Got the answer..

Welcome to MHB! Sorry we couldn't help you quickly enough this time. I'm sure that in the future you'll find guidance with something you are stuck on. Care to share your answer so others may see it?

Jameson

- Jan 17, 2013

- 1,667

Hence we have

\(\displaystyle \Large (1+e^{ix})^n= e^{\frac{ixn}{2}}\left(e^{\frac{-ix}{2}}+e^{\frac{ix}{2}}\right)^n= 2^n e^{\frac{ixn}{2}}\cos^n \left(\frac{x}{2}\right)\)

Clearly the answer is the real part of the above expression .