The actual transient (upon applying a sine wave) is a DC level that exponentially decays. It decays because the source is not a DC source but was forced to produce a DC current transient due to the inductor. Given that the source doesn't naturally produce DC, the circulating DC current exponentially decays to zero due to energy dissipation in the series resistor.
If there wasn't a series resistor and you applied a sine wave to a pure inductor, the DC transient would remain without decaying for all time. Then, if you applied a cosine wave, you'd find there is no DC transient. Try it.
Here's my simulation of the circuit that clearly indicates that the peak magnitude of the current is higher in the first half cycle of applied voltage than at any other time: -
And if I lowered the resistor value from 180 Ω to 100 Ω, it's even clearer: -
If I reduced the resistor to zero ohms, the inductor current will never go negative: -
The transient is pure DC and remains pure DC for all time because although the supply source cannot naturally sustain DC (it's an AC source), there is no resistor in the loop to dissipate the acquired DC energy hence, it remains.
If I applied a cosine wave to the inductor we see this: -
There is no transient because an inductor will naturally produce no transient when driven from a "sine" waveform that begins at its peak value. Note that to make this work in the simulator I needed to set the initial current condition of the inductor to zero (