On nonnegative-order first-kind Bessel functions with large argument

[IridescentRain]

Hello.

I'm not terribly proficient with Bessel functions, but I know that those of the first kind are given by
$$\begin{eqnarray} J_n(x) & = & \left(\frac{x}{2}\right)^n\,\sum_{\ell=0}^\infty\frac{(-1)^\ell}{\ell!\,\Gamma(n+\ell+1)}\,\left(\frac{x}{2}\right)^{2\ell}, \end{eqnarray}$$where $\Gamma$ is the gamma function.

I found in a book that for large $x$ the following is true:
$$\begin{eqnarray} J_n(x) & = & \sqrt{\frac{2}{\pi x}}\,\cos\left(x-\frac{n\pi}{2}-\frac{\pi}{4}\right)\,\left[1+\mathcal{O}(x^{-1})\right]\\ & \approx & \sqrt{\frac{2}{\pi x}}\,\cos\left[x-(2n+1)\,\frac{\pi}{4}\right]. \end{eqnarray}$$The second line of that last equation is obvious (they just removed every term except the first one from the first line because $1/x^\ell$ is negligible for large $x$ and $\ell\ge1$), but where does the cosine come from? How would I begin to prove this?

I'm working with $n\in\mathbb{N}\cup\{0\}$ only, so perhaps this expression is only true for Bessel functions of natural (or zero) order; I wouldn't know. All I know is that for such values of $n$ the gamma function becomes $\Gamma(n+\ell+1)=(n+\ell)!$.

Thanks in advance for any help!

 

[Millennial]

$$\cos(x) = \sum^{\infty}_{k=0} (-1)^{k} \frac{x^{2k}}{(2k)!}$$

Try using that.

 

[IridescentRain]

Thanks, Millennial. I will.