# Multiply signal $x[k]$ with $\cos(2\pi\nu_0k)$, then given $X(\nu)$ draw resulting function in frequency domain?

Let $$y[k]=x[k]\cdot \cos(2\pi\nu_0k) .\tag{1}$$

Then, given a signal $$x[k]$$ with the DTFT $$X(\nu)$$ according to the following figure

what will the frequency domain for $$Y(\nu)$$ look like for a given $$\nu_o$$ in the range $$0<\nu_0?

According to my calculation, the resulting function $$Y(\nu)$$ is

$$Y(\nu)= \frac{1}{2}\left ( X(\nu-\nu_0) +X(\nu+\nu_0) \right ). \tag{2}$$

This since the DTFT of cosine is two unit impulses in $$\pm\nu_0$$.

Is this correct? And how can this resulting $$Y(\nu)$$ be drawn in the frequency domain? Thank you! Some explanation with mathematical elements would be appreciated.

Would the following figure for $$Y(\nu)$$ be correct?