Solution
= Solution
Reverse the second half of the DFT sum by putting $m=2N-n-1$. Since $\omega_{2N}^{2Nk}=1$,
$$
\begin{aligned}
Y_k
&=\sum_{n=0}^{N-1}x_n
\left(\omega_{2N}^{nk}+\omega_{2N}^{-(n+1)k}\right)\\
&=2\omega_{2N}^{-k/2}\sum_{n=0}^{N-1}x_n
\cos\left[\frac\pi N\left(n+\frac12\right)k\right].
\end{aligned}
$$
This is the required identity.