UPDATED:How to do Band Pass Filter Analysis for the current circuit
To be able to find the pole frequency we need to find the equivalent resistance seen from the \$C_2\$ capacitor point of view. In short, we need to find the Thevenin's resistance seen from \$C_2\$ capacitor terminals. So, to do this we "short" all the voltage sources \$V_1\$ and \$C_1\$. 







simulate this circuit &ndash; Schematic created using CircuitLab

And from the inspection we see that \$R_{th} = R_3+R_1||R_2\$  

And the pole frequency is: 

$$F = \frac{1}{2 \pi * C_2 * R_{th}} \approx \frac{0.16}{C_2 R_{th}}$$

And you can do the same thing if you want to find \$C_1\$ pole. 

But this time we have an open circuit at the end of the voltage divider. 

Hence, if we ignore the voltage divider load we can see that \$R_{th}\$ for hign pass filter is \$R1+R2\$    

We can use this approximation method because \$C_1\$ >>\$C_2\$