<s>[INST] Eliminate $\alpha$ by combining two integrals I have a function $f(t,x)$:
$$
f(t, x) = | \sin(2 \cdot \pi \cdot (t - x)) - \sin(2 \cdot \pi \cdot t) |
$$
and two values for $x$:
$$
l = \cos(\alpha)\cdot b \\ m = \sin(\alpha)\cdot b
$$
Can I "combine" $f(t, l)$ and $f(t, m)$ via some function $g(u, v)$ such that
$$
r = g(\int_0^{\frac12} f(t,l) \, dt, \int_0^{\frac12} f(t,m) \, dt)
$$
is independent of the value of $\alpha$? (I still need it to depend on both $u$ and $v$ in order to be able to say something about $b$ in the end; so "trivial" solutions such as $g(u,v) = 0$ or $g(u,v) = b$ are not what I'm looking for.)
I know that $cos^2(\alpha) + sin^2(\alpha) = 1$, but don't know if I can make use of this somehow. I've plotted both integrals for different values of $\alpha$ and somehow feel that there should be a way ...
For reference, here's how the two integrals look plotted with varying $\alpha$ values (on horizontal axis, in degree): 
Background are calculations about the difference in amplitude between multiple (spacial, 2D) points with respect to a wave. The spacial difference is transferred into a "time difference" (because speed of wave is known) $x$. Depending on the direction of impact of the wave ($\alpha$) that $x$ is different; of interest are differences between a point $p_l = (-b, 0)$ and the origin as well as between a point $p_m = (0, b)$ and the origin. I assume that I can get rid of $\alpha$ somehow (I don't care really about the direction atm).
A bit more background: I basically get 2 values ($u$ and $v$, the results of the two integrals) and (atm) don't know $\alpha$. But I need a (constant, independent of $\alpha$) value $r = g(u,v)$.
 [/INST] The first step would be to solve the integral before trying to find such a function $g$.
Since $f(t,x+1) = f(t,x)$, we can assume that $x \in (-\frac{1}{2},\frac{1}{2}]$. Then for $t \in (0, \frac{1}{2}]$ we have $f(t,x) = 0$ if either $ x \in \mathbb{Z}$ or $t = t_*  = \frac{1}{4} + \frac{x}{2}$. In particular, you cannot differentiate between $b$ and $b+1$ with any function $g$.
If $x = 0$, then the desired integral is $0$, so let assume for now that $x \neq 0$. Then the $f(\cdot,x)$ has precisely one zero in $(0,\frac{1}{2})]$, hence we can split the integral in two parts:
\begin{align*}
\int_0^{\frac{1}{2}} f(t,x) \text{d}t &= \int_0^{t_*} f(t,x) \text{d}t + \int_{t_*}^{\frac{1}{2}} f(t,x) \text{d}t \\
&= \pm \int_0^{t_*} \sin( 2 \pi t - 2 \pi x) \text{d}t \\
&\quad \mp \int_0^{t_*} \sin( 2 \pi t ) \text{d}t \\
& \quad \mp \int_{t_*}^{\frac{1}{2}} \sin( 2 \pi t - 2 \pi x) \text{d}t \\
& \quad \pm \int_{t_*}^{\frac{1}{2}} \sin( 2 \pi t) \text{d}t.
\end{align*}
Where the sign is determined by $x$. Solving the $4$ integrals we get, up to a constant $\frac{- 1}{2 \pi}$ for simplicity,
\begin{align*}
\int_0^{\frac{1}{2}} f(t,x) \text{d}t &\propto \pm \cos( 2 \pi t_* - 2 \pi x) \mp \cos( - 2 \pi x) \\
&\quad \mp  \cos( 2 \pi t_* ) \pm \cos(0) \\
& \quad \mp \cos( \pi - 2 \pi x) \pm \cos( 2 \pi t_* - 2 \pi x) \\
& \quad \pm \cos( \pi ) \mp \cos(2 \pi t_*) \\
&= \pm \cos(\frac{\pi}{2} - \pi x) \mp \cos( - 2 \pi x) \\
&\quad \mp  \cos( \frac{\pi}{2} + \pi x) \pm 1 \\
& \quad \mp \cos( \pi - 2 \pi x) \pm \cos( \frac{\pi}{2} - \pi x) \\
& \quad \pm (-1) \mp \cos( \frac{\pi}{2} + \pi x).
\end{align*}
If we group corresponding terms, we obtain
\begin{align*}
\int_0^{\frac{1}{2}} f(t,x) \text{d}t &\propto \pm 2 \cos(\frac{\pi}{2} - \pi x) \mp 2 \cos( \frac{\pi}{2} + \pi x) \\
&\quad \mp \cos( - 2 \pi x) \mp \cos( \pi - 2 \pi x).
\end{align*}
Since $\cos(\alpha) = - \cos(\pi + \alpha)$ and $\cos(\frac{\pi}{2} - \alpha) = -\cos(\frac{\pi}{2} + \alpha) =  \sin(\alpha)$ we find
\begin{align*}
\int_0^{\frac{1}{2}} f(t,x) \text{d}t &= \pm \frac{2}{\pi} \sin(\pi x).
\end{align*}
In particular, if $x = 0$, then $\frac{2}{\pi} \sin(\pi x) = 0$ which is precisely the value of $\int_0^{\frac{1}{2}} f(t,0) \text{d}t$, so for all $x \in (\frac{1}{2},\frac{1}{2}]$ we find
\begin{align}
\int_0^{\frac{1}{2}} f(t,x) \text{d}t &= \pm \frac{2}{\pi} \sin(\pi x).
\end{align}
You can now try to find a function $g$ which satisfies your desired properties.
</s>