Your question and your suggested answer of the last paragraph. Yes, it tracks the input voltage and controls the current it consumes. The model you have shown can select from 'constant resistance' and 'constant current' controls, and IIRC 'constant power' (down to a min voltage) as well. Once there's a microcontroller in there to do fancy stuff like that, then it can simulate current draw into rectifier loads, harmonicy or pulsey loads as well.
I have used a mains version of that pictured load at work. It's a fraction of the price of previously available alternatives, yet still much more than 1000 ($/£/Euro), and limited to mains-type frequencies, so obviously of no interest to you.
You are nearly there with your rectified DC load. However, the fact that you say it only draws spikes of current at the peaks suggests that your rectifier has the standard big electrolytic capacitors in it. If you remove those, then the current distortion all but disappears, you will be able to draw current at all phases of the waveform for where the voltage is larger than your diode drops. The missing volt or two in the middle may be acceptable. Removing the caps from the 'power' supply means that you will need an additional 'control' power supply to do the voltage sensing and bias the current-drawing components.
The frequency range of this method is not limited by transformers, or control laws buried in any PSU control chips. However, to go above a few hundred Hz, you must use appropriate rectifier diodes. The standard 1N540x series (and most 'mains' diode quads) are very slow, they manage mains frequencies and not much more. Buy 'fast' diodes to go much above mains frequencies.
I have used the variac+lamp load method on the bench, and it works very well, although due to the variac it is limited to a small band of frequencies from mains to a few times mains.
When building your load, use FETs as switches by all means, but don't dissipiate too much power in FETs in the linear mode, stay below 10% of rated power, while heatsinking well. The standard switching FET can only dissipate its rated power when switching between saturation and off. In the linear mode, the bias tempco of the FET's individual cells means they can 'unshare' current and burn out, even at relatively low powers. This concern is not the same as the sharing between multiple FETs, which in saturation share nicely. You can get 'linear rated' FETs (intended for audio amplifier output stages) but they are expensive and hard to find. Stick to using FETs to switch resistors, or <10% of rated dissipation, or BJTs.
As a general rule, it is much more scalable if you can switch your power out to lamp loads, or the heater element of a fan heater (with the fan rigged to keep going). Then your dissipation is not limited by how large your box is.
The old skool way of making AC loads was to lower some plate electrodes into an electrolyte bath. Under AC, the resistance of such a load will be more or less linear. Choose your electrolyte from 'clean' water to strong salt solution to control the range of conductivities available, and then the insertion depth of the plates for control. Power handling is of course excellent, into a large bath of water. A bit messy and not easy to miniaturise for bench use, and you need to vent off the small amount of hydrogen or chlorine produced to avoid hazards.
A modern way of approximating to a resistive load would be to use a simple DC load fed from a power factor corrected PSU. It is only an approximation because while a PFC corrected PSU works to draw a 'resistive load' waveform, it is only designed to meet power quality regulations, not 'instrumentation' quality specifications, so it gives up tracking at relatively high phase voltages, and most controllers would only work at around mains frequencies. It's unlikely that this would work for you.
Having given you all these alternatives, what exactly are you going to use the AC load for? If it's testing cores to see how hot they get while delivering power, then OK. If it's using a 'scope to look at their voltage and current waveforms, then you probably shouldn't use anything less perfect for the load than a real resistor, and if necessary, switched by real switches (or relays, or back-to-back saturated FETs). Otherwise, you'll see a wrinkle on the current trace, and then wonder whether that's a characteristic of the core, or your load misbehaving.
Best Answer
Convert them into phasor forms and THEN add them and then you will understand that that is where the 19 degrees comes from, you can determine it mathematically there is no need to estimate it.
Google "convert sinusoid to phasor form" and that should help. If you have questions please ask and I can try to help.
Look at the green circled part of the example but do it backwards. Imagine you are starting with the $$1.12cos(2\pi ft-2.68)$$ term and want to get to the top phasor term $$1.12\angle -153^o$$ (ignore the middle step, it is not relevant here). Do you see how the conversion is done, assuming the 2.68 in the cosine term is in radians? You want to do the same conversion for your two sin terms and then add them together and you will get your result!