Inverse Kinematics Computation -- why are alpha angle values not included?

Check out Equation 3.6:

$$ ^{i-1}_iT = left[ begin{array}{c} ctheta_i & -stheta_i & 0 & r_{i-1} \ stheta_icalpha_{i-1} & ctheta_icalpha_{i-1} & -salpha_{i-1} & -salpha_{i-1}d_i \ stheta_isalpha_{i-1} & ctheta_isalpha_{i-1} & calpha_{i-1} & calpha_{i-1}d_i \ 0 & 0 & 0 & 1 end{array} right] $$

and Figure 3.21, "Link parameters of the PUMA 560":

$$ begin{array}{c} i & alpha_{i-1} & r_{i-1} & d_i & theta_i \ \ 1 & 0 & 0 & 0 & theta_1 \ 2 & -90^{circ} & 0 & 0 & theta_2 \ 3 & 0 & a_2 & d_3 & theta_3 \ 4 & -90^{circ} & a_3 & d_4 & theta_4 \ 5 & 90^{circ} & 0 & 0 & theta_5 \ 6 & -90^{circ} & 0 & 0 & theta_6 \ end{array} $$

Here I've used $r$ instead of $a$ to prevent any confusion between $alpha$ and $a$. In Equation 3.6, $s$ means $sin$ and $c$ means $cos$.

To your question, $alpha$ is more related to the orientation of two joints relative to each other as opposed to the motion of the joints relative to each other. That is, $alpha$ is generally fixed by physical construction and thus, as a parameter, isn't time-dependent.

So hopefully you can see that, from Figure 3.21, all of the $alpha$ values are either +/- 90 degrees or zero. Equation 3.6 then uses it only in sine and cosine terms, meaning that they can all be reduced to +/-1 or zero.

I really, really hate when authors jump through simplifications like this without a heads up, so I totally understand the confusion.