We all know Jacobi–Anger expansion as follows:
$$\cos(z\cdot \cos(\theta))=J_0(z)+2\sum_{n=1}^{\infty}(-1)^n J_{2n}(z) \cos(2n\theta)$$
$$\sin(z\cdot \cos(\theta))=-2\sum_{n=1}^{\infty}(-1)^n J_{2n-1}(z) \cos((2n-1)\theta)$$
$$\cos(z\cdot \sin(\theta))=J_0(z)+2\sum_{n=1}^{\infty}J_{2n}(z) \cos(2n\theta)$$
$$\sin(z\cdot \sin(\theta))=2\sum_{n=1}^{\infty}J_{2n-1}(z) \sin((2n-1)\theta)$$
If we want to analyze the spectral of $\sin(A_1\cos(\omega_1 t+\phi_1))$, we use Bessel Function instead of the Fourier transform.
$$\sin(A_1 \cos(\omega_1 t +\phi_1))=2J_1(A_1)\cos(\omega_1 t+\phi_1)-2J_3(A_1)\cos(3\omega_1 t+3\phi_1)$$
Therefore we know the amplitude of $f_1$ is $2J_1 (A_1)$, the phase of $f_1$ is $\phi_1$.
It also works for $\cos(A_1\cos(\omega_1 t+\phi_1)+A_2\cos(\omega_2 t+\phi_2))$, but $cos(\sum_{n} A_n\cos(n*\Delta\omega t+n\phi))$ does not work. The expansion is too complex to analyse.
So how can I get the spectrum of $$\cos \left[\sum_{n} A_n\cos(n\Delta\omega t+n\phi)\right]$$?
Thanks very much for any response