Note that Fourier inversion (recovering a function from its Fourier transform) is only possible for certain classes of functions, see, for instance,
http://en.wikipedia.org/wiki/Fourier_inversion_theorem. The functions in your first example and Sin[5*tt]*UnitStep are not even integrable on [-Inf; +Inf], so the Fourier transforms don't exist in the "conventional" sense. It is thus not too surprising that transforming these functions forth and back doesn't end up with the original function.
The results Mathematica comes up with are interesting, though. Directly calculating the forward transform via the integral does not work, of course:
In[19]:= Integrate[Sin[\[Alpha] t] UnitStep[t] Exp[I \[Omega] t], {t, -\[Infinity], +\[Infinity]}]
Out[19]= ConditionalExpression[\[Alpha]/(\[Alpha]^2 - \[Omega]^2),
Abs[Im[\[Alpha]]] < Im[\[Omega]]]
In[21]:= Integrate[Sin[\[Alpha] t] UnitStep[t] Exp[ I \[Omega] t], {t, -\[Infinity], +\[Infinity]},
Assumptions -> {\[Alpha] \[Element] Reals, \[Omega] \[Element] Reals}]
During evaluation of In[21]:= Integrate::idiv: Integral of E^(I t \[Omega]) Sin[t \[Alpha]] UnitStep[t] does not converge on {-\[Infinity],\[Infinity]}.
Out[21]= Integrate[ E^(I t \[Omega]) Sin[t \[Alpha]] UnitStep[t], {t, -\[Infinity], \[Infinity]}, Assumptions -> {\[Alpha] \[Element] Reals, \[Omega] \[Element] Reals}]
Of course. What does FourierTransform do?
In[30]:= ff[\[Omega]_] :=FourierTransform[Sin[5.0 t] UnitStep[t], \[Omega], t]
In[29]:= ff[\[Pi]]
Out[29]= FourierTransform[Sin[5 t] UnitStep[t], \[Pi], t]
Doesn't work either.
Now if one performs a forward *and* a backward transform, it seems Mathematica somehow feels obliged to present the user with some result
Albeit, the result seems to depend on whatever clever tricks Mathematica tries to play:
In[59]:= Module[{g, G},
g = Sin[5*tt]*UnitStep[tt];(*Continuous function*)
G = FourierTransform[g, tt, omega](*FT of function*);
FullSimplify[InverseFourierTransform[G, omega, tt]]]
Out[59]= 1/2 (1 + Sign[tt]) Sin[5 tt]
In[60]:= Module[{g, G},
g = Sin[5.*tt]*UnitStep[tt];(*Continuous function*)
G = FourierTransform[g, tt, omega](*FT of function*);
FullSimplify[InverseFourierTransform[G, omega, tt]]]
Out[60]= 0.5 Sign[tt] Sin[5. tt]
Hmmm. We are missing a "1" in the second case.
