Different Laplace transforms of $\cos^2(at)$ from different methods

Question

Forgive me, for I am a newbie to Laplace transforms, but I came across a problem recently that asked me to find the Laplace transform of $\cos^2(at)$. I did this by taking the derivative, $-2a\cos(at)\sin(at) = -a\sin(2at)$. Using the identity $\mathcal{L}(f'(t)) = sF(s)-f(0)$, I got $(-2a^2)/(s^2+4a^2)=sF(s)-1$, thus $F(s) = (1/s)+(-2a^2)/(s^3+4a^2s)$. However, the answer given was computed by using the trig identity $\cos^2(at) = (1/2)(1+\cos(2at))$, which has a Laplace transform of $(1/2)(\frac{1}{s}+\frac{s}{s^2+4a^2})$. I've been looking over this for a while and I can't seem to find an algebra mistake nor do the two answers seem equivalent, am I missing something here? Have I done something illegal?