I'd treat it as two separate circuits, an input circuit to drive an opto-isolator's LED, output circuit using the opto-isolator's photo-transistor to switch the trigger voltage.
The input side probably wants a voltage divider (certainly if you're triggering from mains voltage! Consider the maximum current that could flow on e.g. 480V and size the divider resistors accordingly!), followed by a bridge rectifier (so you don't have too much worry over polarity) and a capacitor to give a transient with the change in voltage on its input if you want to look at edges (i.e. mains on and off), straight to the LED if you want to trigger from mains waveforms, hence a pulse for triggering that drives the LED (i.e. with power switch on/off Vs at a consistent point in the continuous waveform) - a Zener diode clamping the LED voltage to something it can handle is probably a good idea!
The output side will need a power source - the Fluke uses a PP3 9 volt battery, can't see anything wrong with that! The opto-transistor can be in the +ve line feeding a resistive voltage divider to get down to the 1.2v the 'scope requires and limit the current through the transistor.
Remember that attaching it to mains requires some serious attention to insulation and PCB design, several mm of air-gap between the two sides of the circuit, ideally an earthed strip between, and it's a good idea to thoroughly clean off any flux residue (which becomes conductive once it absorbs moisture from the air) and apply a conformal coating to reduce the likelihood of flash-over
An useful data sheet:https://www.fairchildsemi.com/application-notes/AN/AN-3001.pdf
- replace C1 with a bleed resistor and add a capacitor (with a switch to short it for waveform measurements - with sufficient insulation for the incoming voltage!) between the bridge rectifier and the diode to give a pulse/step with a change in input (AC or DC) voltage, perhaps add a Zener across the LED for reliability, could be worth a try?
If you want it to trigger from millivolt signals, you're going to need some kind of amplification up front on the input side, which will then need its own, separate, isolated power supply - that's when it starts getting complicated...
Hope this helps!
Dave H. (the other one)