How to prove sinc function series?

Question

I am not sure how to prove this series for the sinc function: $$\text{sinc}(x)=2\cos\left(\frac{x}{2}\right)\sum_{n = 0}^\infty\frac{(-1)^n x^{2 n}}{2^{2 n + 1}(2 n + 1)!}.$$ Is there an elementary proof? Any ideas?

Answer 1

Start with $$\sin(x)=2\cos(x/2)\sin(x/2)$$ $$\operatorname{sinc}(x)=\frac2x\cos(x/2)\sin(x/2)$$ Then use $$\sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}$$ To get $$\operatorname{sinc}(x)=\frac2x\cos\frac{x}2\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\left(\frac{x}2\right)^{2n+1}$$ $$\operatorname{sinc}(x)=\frac2x\cos\frac{x}2\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2^{2n+1}(2n+1)!}$$ $$\operatorname{sinc}(x)=\cos\frac{x}2\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{2^{2n}(2n+1)!}$$ QED