Phase-shift between $\sum_{k=0}^n \cos\pi(x)$ and $\sum_{k=0}^n \cos\pi(x-k)$

Question

Is there an algebraic way of representing the phase-shift between the following two summations?

$$\sum_{k=0}^n \cos\pi( x)$$

$$\sum_{k=0}^n \cos\pi(x-k)$$

Thanks in advance :-)

Answer 1

Note that

$$\sum_{k=0}^n \cos(\pi x)=(n+1)\cos(\pi x)$$

while

$$\sum_{k=0}^n \cos(\pi (x-k))=\begin{cases} \cos(\pi x)&,n\,\text{even}\\\\0&,n\,\text{odd}\end{cases}$$

since $\cos(\pi(x-k))=(-1)^k\cos(\pi x)$.