Q:

$\sum\limits_{k=0}^{\lfloor{n/2}\rfloor}(-1)^k\binom{n-k}{k}=\frac{2}{3}\left(\cos\frac{\pi(n-1)}{3}+\cos\frac{\pi n}{3}\right)$

Be $n\in\mathbb{N}_0$ .
This is a simple formula but not obviously :
$$\sum\limits_{k=0}^{\lfloor{n/2}\rfloor}(-1)^k\binom{n-k}{k}=\frac{2}{3}\left(\cos\frac{\pi(n-1)}{3}+\cos\frac{\pi n}{3}\right)$$  
Can be found a short proof for that ?

A:

Chebyshev polynomials of the second kind have the following representation:
$$ U_n(x)=\sum_{r\geq 0}\binom{n-r}{r}(-1)^r (2x)^{n-2r} \tag{1}$$
hence the wanted sum is just $U_n\left(\frac{1}{2}\right)$, and since $\frac{1}{2}=\cos\frac{\pi}{3}$,
$$ U_n\left(\frac{1}{2}\right) = \frac{\sin((n+1)\pi/3)}{\sin(\pi/3)}\tag{2} $$
and the claim is a straightforward consequence of $(2)$.

