Gain

[Cujo]

How to find the gain as the ratio of the amplitudes of the output and the input voltage from a sinusoid?

[bob91343]

Divide the output amplitude by the input amplitude.  It can be rms or peak to peak or just peak.

[CaptDon]

And how do you want that gain expressed? In decibels or as a simple fractal ratio??
There are so many on-line Decibel calculators I am surprised you posted the question here.
I often just 'approximate' the gain remembering power ratios of double =3db  quadruple
=6Db and tenfold =10Db. But remember voltage ratios are a bit trickier, A 10 to 1 increase
in voltage is 20Db in power all else being equal, 10 times the voltage causes 10 times
higher current flow and the power increases 100 times or 20Db (10X10=100 or a 20Db
power ratio). In radar we use DbK (decibels over 1 kilowatt) for transmit powers, but still
use receiver merit ratings of Db below 1 milliwatt with typical echo signals running lower
than -120Dbm or 120 decibels below 1 milliwatt.

[TimFox]

Strictly speaking, dB signifies the ratio of two power values:  dB = 10 x log(P₂/P₁).
However, it is common to assume (absent a specification to the contrary) that the impedance levels at the two measurement points are equal, giving the other common equation  dB = 20 x log(V₂/V₁).
The voltage ratio version is correct for equal impedance (e.g. 50 ohms).  In a typical voltage amplifier, where the input impedance is high and the load impedance unknown, it is still a useful definition.  So long as you use the same voltage measurement (rms, peak, peak-to-peak, etc.) for both voltages, the ratio is independent of the measurement type.
Where things get interesting is with opto-electronics, where the output current is proportional to the input optical power, so a 3 dB reduction in light power gives a 6 dB reduction in output current.