 Solution

Apply the product-to-sum identity directly to the stated modulation:The carrier has angular frequency $\omega$ and the two modulation sidebands have angular frequencies $\omega \pm {\omega }_m$, each with amplitude $b/2$.

Propagation qualification. In the literal formula, all three terms have the same spatial wavenumber. That identity is correct, but a time-only amplitude applied everywhere is not generally a freely propagating wave at speed $c$. For $k=\omega /c$, applying ${\partial }_t^2-c^2{\partial }_x^2$ to the stated field gives $A_0^{\prime \prime }\cos (\omega t-kx)-2\omega A_0^{\prime }\sin (\omega t-kx)$, which at $x=t=0$ equals $-b{\omega }_m^2\ne 0$ when the modulation is nontrivial.

A physical modulation launched at a boundary and travelling without dispersion has envelope $a+b\cos [{\omega }_m(t-x/c)]$. The same expansion then has sideband wavenumbers $k_{\pm }=(\omega \pm {\omega }_m)/c$, while its value at $x=0$ agrees with the given time-modulated signal. This distinguishes the requested frequency decomposition from an unsupported fixed-wavenumber propagation claim.