Plotting phase of $\cos(\omega)$ and $\sin(\omega)$

Question

The actual problem I'm trying to figure out is plotting the phase of the DTFT of $x[n] = \delta[n+1] + \delta[n-1]$. I got that the DTFT is $ X(e^{jw}) = 2\cos(\omega)$ (I think), but what does it mean to plot the phase of $2\cos(\omega)$?

Answer 1

For magnitude, $$|X(e^{j\omega})|=2|\cos(\omega)|$$

For the phase, consider positive and negative cases. For positive cases, phase is constant zero. However, for the cases when $\cos(\omega)=-1$ since $-1=e^{j\pi}$ the phase is $\pi$.

$$\angle X(e^{j\omega})=\begin{cases}0,& 2k\pi\le \omega\le \pi/2+2k \pi \text{ or } 3\pi/2+2k\pi\le\omega\le2\pi+2k\pi\\\pi,&\pi/2+2k\pi<\omega<3\pi/2+2k\pi\end{cases}$$