How do I prove that delta - sinc function is the same as an (-1)^n times the sinc

Question

$$\delta(n) - \frac{1}{2} \mbox{sinc} \left(\frac{n}{2}\right) = (-1)^n \frac{1}{2} \mbox{sinc} \left( \frac{n}{2} \right)$$

I tried to split it into two sequences: one for n even and one for n odd. I also tried to make use of the fact that $sin(-\frac{n\pi}{2}) = (-1)^nsin(\frac{n\pi}{2}), \forall n \in \mathbb Z$

But so far I haven't managed to tie everything together. Here is a photo showing what I've tried so far. enter image description here

Answer 1

$ \delta(n) - \frac{1}{2} sinc (\frac{n}{2}) = (-1)^n \frac{1}{2} sinc( \frac{n}{2}) $

Given that

$ \sin{\frac{-n\pi}{2}} = (-1)^n \sin{\frac{n\pi}{2}} $

$ \delta(n) - \frac{\frac{1}{2}\sin{\frac{\pi n}{2}} }{\pi \frac{n}{2}} $ =

$ \frac{\frac{1}{2}\sin{\frac{-\pi n}{2}}}{\pi \frac{n}{2}} $

$ \frac{1}{2}(-1)^n \frac{\sin{\frac{\pi n}{2}}}{\frac{\pi n}{2}} $

= $ \frac{1}{2}(-1)^n \sin{\frac{n}{2}} $

QED