Derivation of cutoff frequency and phase shift for RC low pass filter
Cut-off frequency or 3-dB frequency is defined as the frequency of the input signal at which, the magnitude of the output signal reduces to \$1/\sqrt2\$ of the input, or the power reduces to half ( i.e., by 3 dBs).
A simple RC circuit:

$$V_{out} = \frac {V_{in} * -jX_c}{R-jX_c}$$
By our above definition, at cut-off frequency \$f_o\$ , \$ \frac {-jX_c}{R-jX_c} \$ should be equal to \$1/\sqrt2\$
i.e.,
$$\frac{-jX_c}{R -jX_c} = \frac{1}{\sqrt2} $$
taking magnitude of the complex expression:
$$=&gt; \frac {X_c}{\sqrt {R^2+X_c^2}} = \frac{1}{\sqrt2} $$
$$ =&gt; \frac {1}{\sqrt{(R^2/X_c^2)+1}} = \frac{1}{\sqrt2} $$
$$ =&gt; (R^2/X_c^2) = 1 $$
$$ =&gt; R = X_c $$
$$ =&gt; R = \frac{1}{Cw_o}$$
$$ =&gt; w_o = \frac {1}{RC}$$
$$ =&gt; f_o = 1/2 \pi RC$$
for complex numbers, phase of \$a+ jb = tan^{-1}(b/a)\$ , \$V_{out}\$ is a complex expression, Hence its phase \$ \phi \$ would be:
$$ \angle \frac {tan^{-1}(-\infty)}{tan^{-1}(-X_c/R)} $$
$$ = \angle\frac {-tan^{-1}\infty}{-tan^{-1}(X_c/R)}   $$
$$ = -\frac{\pi}{2} + tan^{-1}(X_c/R) $$
using the expression, \$ tan^{-1}x + tan^{-1}(1/x) = \pi/2 \$
$$ =&gt; \phi =  -tan^{-1}(R/X_c) = -tan^{-1}(wRC)$$