Art Of Electronics - How does the phasor diagram explain away 3db verses 6db confusion
This is because db does account for phase, db is a magnitude.

for this circuit in phasor notation:

V=(-jwC)/(R-jwC)

at the corner frequency:

V=(-jR)/(R-jR)

||V||=(R)/(R*sqrt(2))=-3db

To relate it to the resistor divider circuit, if you were to separate the equation to its real and imaginary components you would get this:

V=(1/2)*(1-j)

Separate out the Real (in phase power):

Re{V}=(1/2)

||Re{V}||=1/2=-6db

So for this circuit at the -3db corner frequency you would see 1/sqrt(2) the voltage (-3db)  with a 45deg phase shift on an oscilloscope. The 45deg phase shift reduces the in phase voltage by another 1/sqrt(2). So the REAL in phase voltage magnitude at the output is half (-6db).

Ive noticed Phasor noticed is often explained poorly and used incorrectly.