Difference between sinc and $\cos$ can be expressed using Bessel function $J_{3/2}$ I've tried to prove this property of Bessel function but I don't seem to be going anywhere
$$\sqrt{\frac 12 \pi x} J_\frac 32 (x) = \cfrac{\sin x}{x} - \cos x$$
I have tried substituting $\frac 32$ for $J_n (x)$ and then manipulating with the product but it doesn't seem to give me something similar with the series on my LHS. I don't know if there is another different approach which I must follow. 

Using the series expansion of $J_{3/2}(x)$ and the Legendre duplication formula for the gamma function, we find
$$
\sqrt {\frac{{\pi x}}{2}} J_{3/2} (x) = \sqrt {\frac{{\pi x}}{2}} \left( {\tfrac{1}{2}x} \right)^{3/2} \sum\limits_{n = 0}^\infty  {( - 1)^n \frac{{\left( {\frac{1}{4}x^2 } \right)^n }}{{n!\Gamma \left( {n + \frac{5}{2}} \right)}}} 
\\
 = \sqrt \pi  \sum\limits_{n = 0}^\infty  {( - 1)^n \frac{{x^{2n + 2} }}{{2^{2n + 2} n!\Gamma \left( {n + \frac{5}{2}} \right)}}}  = \sum\limits_{n = 0}^\infty  {( - 1)^n \frac{{(2n + 2)}}{{(2n + 3)!}}x^{2n + 2} } 
\\
 = \sum\limits_{n = 1}^\infty  {( - 1)^{n + 1} \frac{{2n}}{{(2n + 1)!}}x^{2n} }  = \sum\limits_{n = 1}^\infty  {\frac{{( - 1)^n }}{{(2n + 1)!}}x^{2n} }  - \sum\limits_{n = 1}^\infty  {( - 1)^n \frac{{x^{2n} }}{{(2n)!}}} 
\\
 = \sum\limits_{n = 0}^\infty  {\frac{{( - 1)^n }}{{(2n + 1)!}}x^{2n} }  - \sum\limits_{n = 0}^\infty  {( - 1)^n \frac{{x^{2n} }}{{(2n)!}}}  = \frac{{\sin x}}{x} - \cos x.
$$
