Challenge Level

time-dependent current $i = V_0\sin(\omega t)/R$

time-dependent power $P = iv = V_0^2\sin^2(\omega t)/R$

$\sin^2(\omega t) = 1/2 - 1/2\times \cos(2\omega t)$

$$

\begin{align}

P_{average} &= 1/T\times \int^T_0{P}dt\quad\left(\mbox{where T is the period, }2\pi/\omega\right)\\

&= \frac{V_0^2}{RT}\left[\frac{t}{2} - \frac{1}{4\omega}\sin(2\omega t)\right]^T_0 = \frac{V_0^2}{2R}

\end{align}

$$

The DC power formula for voltage is $P = V^2/R$, so the good representative (RMS) voltage is $V_0/\sqrt{2}\;.$

If you draw out the power graph for the bulb, the frequency of the power is twice the frequency of the voltage, and the graph sits just above the $P=0$ line. The graph of the light output will follow this frequency, but will be raised up on the $y$-axis by some fixed amount, as of course the filament stays hot between cycles! There will also be a shift to the right because the filament will keep heating beyond the peak power for a short time.