Op amp phase shifter with unity constant gain - How does this circuit shift the phase between -180 and 0?
For the DC current and at "low" signal frequency the capacitor reactance is equal to :
\$X_C = \infty  \$

Hence you circuit becomes a noninverting voltage follower with the gain of \$+ 1 V/V\$ with \$0^{\circ} \$ phase shift. 

The output voltage is superimposed on this two outputs (\$+2 + (-1) = +1V\$)
Noninverting amplifier with the gain of \$+2\$ and inverting amplifier with a gain of \$-1\$







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At high the capacitor reactance is \$X_C = 0\Omega\$  

And this time your circuit becomes a textbook example of an inverting amplifier with voltage gain equal to
\$-\frac{R_2}{R_1} = -1\$ 

So you have gain one but the output voltage is \$-180^{\circ}\$ out of phase shift.     







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And the transfer function for your circuit (All pass filter) is: 







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$$ A_V(s) = -\frac{R_2}{R_1} + (1 + \frac{R_2}{R_1}) \cdot\frac{1}{1+ sR_3 C} = \frac{1 - sRC}{1+ sRC}$$

And the magnitude becomes \$ 1\$ (pole is canceled by the zero) 

And the phase shift is 

\$\phi = -2arc tg (\omega RC) = -2arctan \left( \frac{F}{F_O}\right)\$ 

Where: 

\$F_O = \frac{1}{2 \pi R_3 C}\$

So for the frequency when \$F = F_O\$ the phase shift is 

\$ \phi = -2arctan \left( \frac{1}{1}\right) = -2arctan \left(1\right) = -2 \cdot 45^{\circ} = -90^{\circ}\$