infinity sum Bessel-Type about $\sum _{k=-\infty }^{\infty } (-i)^k \cos (k) J_k(x)$ I would like to propose a problem following the sequence of series could find the general term when Co(x)^n it seem easy but i can not find 
$$\sum _{k=-\infty }^{\infty } (-i)^k \cos (k) J_k(x)=\cos (x \cos (1))-i \sin (x \cos (1))$$
$$\sum _{k=-\infty }^{\infty } (-i)^k \cos ^2(k) J_k(x)=\frac{1}{2} (\cos (x)+\cos (x \cos (2))-i (\sin (x)+\sin (x \cos (2))))$$
$$\sum _{k=-\infty }^{\infty } (-i)^k \cos ^8(k) J_k(x)=\frac{1}{128} (35 \cos (x)+56 \cos (x \cos (2))+28 \cos (x \cos (4))+8 \cos (x \cos (6))+\cos (x \cos (8))-i (35 \sin (x)+56 \sin (x \cos (2))+28 \sin (x \cos (4))+8 \sin (x \cos (6))+\sin (x \cos (8))))$$
 A: We wish to find an expression for
$$\sum^\infty_{k = -\infty} (-i)^k \cos^n (k) J_k (x),$$
where $J_k (x)$ is the Bessel function of the first kind of order $k$, $i$ is the imaginary unit, and $n \in \mathbb{N}$. 
To do this we will make use of the Jacobi-Anger expansion
$$e^{i x \cos \phi} = \sum^\infty_{k = -\infty} i^k J_k (x) e^{ik \phi}.$$
As
$$\cos (k) = \frac{1}{2} (e^{ik} + e^{-ik}),$$
on applying the binomial theorem we have
$$\cos^n (k) = \left (\frac{e^{ik} + e^{-ik}}{2} \right )^n = \frac{1}{2^n} \sum^n_{m = 0} \binom{n}{m} e^{i(2m - n)k}.$$
Thus
\begin{align*}
\sum^\infty_{k = -\infty} (-i)^k \cos^n (k) J_k (x) &= \frac{1}{2^n} \sum^n_{m = 0} \binom{n}{m} \sum^\infty_{k = -\infty} (-i)^k J_k (x) e^{i(2m - n)k}\\
&= \frac{1}{2^n} \sum^n_{m = 0} \binom{n}{m} \sum^\infty_{k = -\infty} i^k (-1)^k J_k (x) e^{i(2m - n)k}\\
&= \frac{1}{2^n} \sum^n_{m = 0} \binom{n}{m} \sum^\infty_{k = -\infty} i^k J_k (-x) e^{i(2m - n)k}, \quad {\rm since} \,\, J_k(-x) = (-1)^k J_k (x)\\
&= \frac{1}{2^n} \sum^n_{m = 0} \binom{n}{m} e^{-ix \cos (2m - n)},
\end{align*}
where in the last line we have made use of the Jacobi-Anger expansion. Hence
$$\sum^\infty_{k = -\infty} (-i)^k \cos^n (k) J_k (x) = \frac{1}{2^n} \sum^n_{m = 0} \binom{n}{m} \Big{(}\cos [x \cos (2m - n)] - i \sin [x\cos (2m - n)] \Big{)},$$
and is the desired expression you seek.
