As we've seen, Eq. (7.33) describes the beat pattern. Let's now derive a different version of that expression assuming that the two overlapping equal-amplitude cosine waves have angular spatial frequencies of $k_{c}+\Delta k$ and $k_{c}-\Delta k,$ and angular temporal frequencies of $\omega_{c}+\Delta \omega$ and $\omega_{c}-\Delta \omega,$ respectively. Here $k_{c}$ and $\omega_{c}$ correspond to the central frequencies. Show that the resultant wave is then
$$E=2 E_{01} \cos (\Delta k x-\Delta \omega t) \cos \left(k_{c} x-\omega_{c} t\right)$$
Explain how each term relates back to
$$E=2 E_{01} \cos \left(k_{m} x-\omega_{m} t\right) \cos (\bar{k} x-\bar{\omega} t)$$
Prove that the speed of the envelope, which is the wavelength of the envelope divided by the period of the envelope, equals the group velocity, namely, $\Delta \omega / \Delta k$.