Prove the sum formula for cosines with binomial coefficients

[shen07]

$$\sum_{k=0}^{k=n}(nCk * cos(kx)) = cos(nx/2)*(2cos(x/2))^n$$

 

[shen07]

I also know that i have to use

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

 

[shen07]

Got the answer..

 

[Jameson]

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

 

[alyafey22]

The answer could be obtained by choosing the real part after substituting $z=e^{ix}$

Hence we have

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

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