Wave – Is the Amplitude of a Sine Wave Increasing or Decreasing in This Phase?

One way is to use a peak detector, but this has a diode voltage drop, which gives an error in the result.

You don't say anything about frequency, but I'll assume it is much lower than the ADC's sampling frequency.

I would use a resistive voltage adder to bring the full signal within the ADC's range. The resistive adder has the advantage over the capacitively coupled adder that you don't have to take frequency-dependent attenuation from the capacitor into account.

Suppose the ADC has a 0 V to 5 V range, and your signal is maximum 10 V peak-to-peak. Then you want Vout to be 0 V for -5 V input, and as close to +5 V as possible for +5 V input. The latter means that R1||R2 should be much smaller than R3. The former means that R1 should be equal to R2: if the input is zero then there's no current through R3, and R1 and R2 form a resistor divider between -5 V and +5 V. So if we choose R1 = R2 = 1 kΩ and R3 = 47 kΩ then the output signal will swing between 0 V and 4.95 V.

Now take enough samples to find the minimum and maximum values, and multiply the difference by 10 V/ 4.95 V, that will give you the peak-to-peak value.

The problem with analysing it formulaicly is that the diodes are very non-linear devices and this is implied in the diagram below (red curve is output): -

The blue curve shows a full-wave rectified signal with no capacitor on the output, just a load resistor. When you add a capacitor it charges very quickly due to the forward conduction of one of the diodes but as soon as it attains near-peak voltage (and the blue waveform turns back down towards zero volts), the capacitor remains charged thus reverse biasing the diode and no further diode conduction is seen until the next peak.

You have to think along these lines to adequately "model" what happens. The red curve shows a discharging element between peaks and this is the effect of drawing current into the load resistor. It's a reasonable approximation to regard this as a time-linear discharge but in fact it's an exponential discharge. However, for most decently designed (and used) bridge circuits, using a linear approximation is fine and the ripple voltage formula works pretty good.