Unlike people that responded to you, I encourage you to do that. If your device is enclosed, disposition of heat generation is pretty important. Heat will generate on resistors, indeed, but if you put it a bit further from your regulator, you will split the heat more evenly and you will save a bit your active component. Of course at least passive cooling is nice to have on your regulator (dependable on voltage/current you are having). Keep on mind (thanks dim), that part of your current will be spent in voltage divider. If you are not ready to pay that price (dependable on your circuit or usage, sometimes it can be sufficient solution), you can try with distributing dissipation on serial resistor, as someone suggested (note that in that case you will tune voltage drop with value of that resistor). Another (better and more expensive) solution would be cascade regulators.

First of all, it is good to spread out heat dissipation across multiple components. That's why for sure you should use some passive components for lowering voltage before your regulator. Lets take the further example:

First of all, you should know how to treat voltage regulator from aspect of power consumption. Note that your regulator is taking some of your input current, so just for our current perspective, we can take next model for voltage regulator:

So, current for R1 and R5, wont be the same. One part of R1 current will go through the R6 (consumption of regulator). You can find this current in datasheet of the regulator you are using.

Now, if you know that you want to have 3.3V on output (Vout) of particular regulator, that means that next equation will do:

R2*Iout = 3.3V

On the other side of regulator, you want to lower your Vin voltage which will be:

Vin = V1 - R1*I1

Note also relations of current, where I1 = Iout + Iu, where Iu is load current of regulator (according to simplified model, current through R6 resistor).

Now, when you know V1, you need to decide what should your Iout be (consult your datasheet for available range). For example, you want it to be 500mA:

R2*500mA = 3.3
V1 - R1*(500mA+Iu) = Vin

You would like Vin (in this case) to be something between 4 and 5V, so from two upper equations you can choose your R1 resistor, presumably you know what your load (R2) is.

@dim: Is this more suitable then my previous answer?

Unlike people that responded to you, I encourage you to do that. If your device is enclosed, disposition of heat generation is pretty important. Heat will generate on resistors, indeed, but if you put it a bit further from your regulator, you will split the heat more evenly and you will save a bit your active component. Of course at least passive cooling is nice to have on your regulator (dependable on voltage/current you are having). Another (better and more expensive) solution would be cascade regulators.

Unlike people that responded to you, I encourage you to do that. If your device is enclosed, disposition of heat generation is pretty important. Heat will generate on resistors, indeed, but if you put it a bit further from your regulator, you will split the heat more evenly and you will save a bit your active component. Of course at least passive cooling is nice to have on your regulator (dependable on voltage/current you are having). Keep on mind (thanks dim), that part of your current will be spent in voltage divider. If you are not ready to pay that price (dependable on your circuit or usage, sometimes it can be sufficient solution), you can try with distributing dissipation on serial resistor, as someone suggested (note that in that case you will tune voltage drop with value of that resistor). Another (better and more expensive) solution would be cascade regulators.