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[SOLVED] Help with the Proof

shen07

Member
Aug 14, 2013
54
\(\displaystyle \sum_{k=0}^{k=n}(nCk * cos(kx)) = cos(nx/2)*(2cos(x/2))^n\)
 

shen07

Member
Aug 14, 2013
54
I also know that i have to use

\(\displaystyle
z=exp(ix)
\)
\(\displaystyle
(1+z)^n = 2^n cos^n (x/2) cos (nx/2) \)
 

shen07

Member
Aug 14, 2013
54
Got the answer..
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
Got the answer..
Hi shen07 (Wave),

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
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
The answer could be obtained by choosing the real part after substituting $z=e^{ix}$

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 .