Verified answer

Your problem doesn't say how many times it oscillates between 20 and 100 over the 20 minute interval, so I'll assume it only oscillates one time.

Therefore the period for one oscillation is 20 minutes.

So in 20 minutes, it needs to cover 2π radians.

ω = 2π rad/20 minutes = π/10 rad/min

The range of oscillation is 100C - 20C = 80C, so you can think of this as an oscillation of 80C/2 = 40C above and below the mean value of (100 + 20)/2 = 120/2 = 60C. That is, at 0 minutes, the temperature is 60 - 40 = 20; at 5 minutes, the temperature is 60; at 10 minutes, it is 60 + 40 = 100; at 15 minutes, it's back to 60, and then at 20 minutes it's back down to 60 - 40 = 20.

Since the temperature starts at the low point, it's easier to use cosine (i..e. cos 0 = 1, but sin 0 = 0).

So one way to write it would be:

Temp(t) = 60 - cos(π/10 t)

where t is in minutes. The other way to make it work is to convert cosine to sin using the identity:

sin(π/2 - x) = cos(x)

So

Temp(t) = 60 - sin(π/2 - π/10 t)