Cfrgtkky

I'm trying to get the inverse Fourier transform of a raised cosine filter.
The frequency domain equation of the filter is:

$$
begin{equation}
H(f)=
begin{cases}
1 &|f| < 0.8a \
cos left(frac{2pi|f|}{0.8a}right) & 0.8a leq |f| leq 1.2a \
0 & otherwise
end{cases}
end{equation}
$$

Doing the calculations manually first I end up with :

$$
h_1(t) = 0.4a cdot sinc(0.4api t)cos(2pi a t) \ + 0.2acdot sinc left[0.4api left(t-frac{1}{0.8a}right) right]cosleft[2pi a left( t - frac{1}{0.8a} right) right] \ + 0.2a cdot sinc left[0.4api left(t+frac{1}{0.8a}right) right]cosleft[2pi a left( t + frac{1}{0.8a} right) right]
$$

Apparently I should be able to simplify it to an equation which was given for cash in a textbook:
$$
h_2(t) = 2acdot sinc(2pi at) left[ frac{0.4pi a t}{1 - 0.64 a^2 t^2} right ]
$$

with
$$
sinc(x) = frac{sin(x)}{x}
$$

I checked with Maple and it seem that indeed the solution I found is equal to the solution answered with the textbook equation. Now, I wonder how in the world I could simplify $h_1(t)$ to make it equal to $h_2(t)$. Seems impossible to me... but there must be a way since those two equations are litterally equivalent. My question is mostly about how to carry on this simplification.

Thanks for any help in advance