Couldn't you hard-code in the angle that each wall is, then when you strike it just look it up in a list or something?

I haven't put a lot of work into this, but I think there's a way to calculate the ball's motion without calculating trig functions. The trick is realizing that trig functions are nothing but ratios, and that the numerical value of the angle is useless in this case. In two dimensions, motion is completely determined by two quantities: an angle (the direction) and the magnitude of the velocity.

Let the angle the ball's velocity vector makes with the positive x-axis to be $\phi_0$ and the angle the wall makes with the positive x-axis $\theta_0$. With these definitions, the ball's velocity can be written as $\vec{\gamma} = \cos(\phi_0) \hat{i} + \sin(\phi_0) \hat{j} = \alpha \hat{i} + \beta \hat{j}$and the velocity after the collision is $\bar{\vec{\gamma}} = \cos(\phi_1) \hat{i} + \sin(\phi_1) \hat{j} = \bar{\alpha} \hat{i} + \bar{\beta} \hat{j}$, so all that's needed is to calculate the new alpha and beta. Also define $\zeta_c = \cos(2\theta_0)$ and $\zeta_s = \sin(2\theta_0)$.

If the ball strikes the wall, we want to find the new angle it makes with the x-axis (let's call is $\phi_1$). It can be shown that this new angle is $\phi_1 = 2 \theta_0 - \phi_0$, and taking their sine/cosines yield

(1)
\begin{align} \bar{\alpha} = \cos(\phi_1) = \cos(2\theta_0) \cos(\phi_0) + \sin(2\theta_0) \sin(\phi_0) = \zeta_c \alpha + \zeta_s \beta \end{align}

(2)
\begin{align} \bar{\beta} = \sin(\phi_1) = \sin(2\theta_0)\cos(\phi_0) - \cos(2\theta_0)\sin(\phi_0) = \zeta_s \alpha - \zeta_c \beta \end{align}

The quantities alpha and beta are the current components of the velocity vector (excluding the magnitude) and so would be known, and if you were to hard-code in the angles (or better yet, the cosine/sine of those angles) you would have a simple calculation to determine the ball's velocity. All that would be left to do is calculate the magnitude(s).

(fixed part of your $\LaTeX$ - thornahawk)