The diffidulty in the given DE is the sin(psi1(t)) that makes it nonlinear. taylor will not expand in terms of psi1(t). Freezing psi1(t) (or just substituting it with a variable) causes taylor to replace essentially 0 because the differential terms now become 0.
I found I had to replace sin(psi1(t)) and cos(psi1(t)) with sin(spsi1) and cos(spsi1) and then do the taylor expansion:
which is not very elegant.
There is however a problem with this: Until you know something about psi1(t) you do not know whether this linearization makes any sense. dsolve will solve the homogeneous linearized DE easily:
but it will also solve the original DE:
and one now sees that the initial conditions are relevant here in assessing whether this linearization is applicable.
So it isn't quite clear (to me) what you are trying to achieve here...