The instantaneous power is dissipated as $p = ei = E_0 \sin(\omega t) \times I_0 \sin(\omega t - \phi)$. Using the trigonometric identity $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$, the average power over a cycle is determined. The term containing $2\omega t$ averages to zero over a complete cycle, leaving only the constant term. Thus, the average power is $P = \frac{1}{2} E_0 I_0 \cos \phi$ .