in a linear phased array it is possible to move the peak angle ("beam steering") by applying proper phase shift between the excitation currents:
The second picture represents the array factor for different beams angle. Last picture taken from here. What I see in the second picture is that beam steering increases all the lobes width but does not change the lobe peak. So, the peak value of the array factor is (or at least seems to be) the same for each scan angle.
Always there it is written that if beam steering is used, there will be some "scan losses":
Another characteristic of all active antennas is the loss of aperture gain as the beam is steered away from the boresight direction — defined as Ɵ=0. This characteristic, called scan loss, follows 10*log(cosN(Ɵ)) power, where Ɵ is the scan angle off boresight and N is a numeric value, typically in the 1.3 range, which accounts for the non-ideal isotropic behavior of the embedded element gain. Fig. 6 plots scan loss in dB vs. scan angle, measured in degrees. Note, at the origin, where the boresight angle is zero, there is no scan loss. As the scan angle is increased to 45 degrees, there is 2 dB scan loss. If you increase scan angle to a practical limit of 60 degrees, there is 4 dB scan loss.
So, which is the scan loss? From my initial picture it doesn't seem to be caused by the array factor. But, if it were caused by the single element pattern, how could we say that it is linked to the scan angle with a fixed equation not dependent on the single element pattern?