Define the triangle waves $S(x) = \arcsin(\sin x)$ and $C(x) = \arcsin(\cos x)$, which are analogues of $\sin x, \cos x$. Indeed, $S(x), \sin x$ are tangent to each other for $x = 2k \pi$, and the same holds for $C(x), \cos x$ for $x = (2k+1)\pi$, $k \in \mathbb Z$. Given any function, I want to decompose it as a linear combination of these triangle waves: $\sum_{n = 0}^\infty a_n \ S(n \pi x/L) + \sum_{n = 0}^\infty b_n \ C(n \pi x/L)$, analogous to Fourier series.
Now it turns out that:
$$\int_{-L}^L C \left({m \pi x} \over L \right) S \left({n \pi x} \over L \right) dx = 0.$$
for arbitrary integer $m, n$. If we define $f(x) = C \left({m \pi x} \over L \right) S \left({n \pi x} \over L \right)$, this is because $f(x)$ is odd: $S(-x) = -S(x)$ as both $\arcsin x, \sin x$ are odd, and similarly, $C(-x) = C(x)$, so $-f(-x) = -C(-x) S(-x) = C(x) S(x) = f(x)$.
Next we consider the integrals $\int_{-L}^L C \left({m \pi x} \over L \right) C \left({n \pi x} \over L \right) dx$ and $\int_{-L}^L S \left({m \pi x} \over L \right) S \left({n \pi x} \over L \right) dx$. Defining the integrands to be $g(x), h(x)$ respectively, from before, we know that $g(x), h(x)$ are both even. Now, when $m$ is an odd integer, $\cos(m \pi (L - x)/L) = \cos(m \pi - m \pi x / L) = -\cos(m \pi x/L)$ (recall $\cos(\pi - x)$ $=-\cos x$ and periodicity), and for the same reason, $\cos(m \pi (L - x)/L) = \cos(m \pi x/L)$ when $m$ is even. Similarly, $\sin(m \pi (L - x)/L) = \sin(m \pi x/L)$ when $m$ is odd, and $\sin(m \pi (L - x)/L) = -\sin(m \pi x/L)$ when $m$ is even. As $\arcsin$ is odd, we can take the minus sign out of the brackets, and all these properties also hold for $C(x)$ and $S(x)$.
Putting this all together, $\int_{-L}^L C \left({m \pi x} \over L \right) C \left({n \pi x} \over L \right) dx = 2 \int_0^L C \left({m \pi x} \over L \right) C \left({n \pi x} \over L \right) dx$ and $\int_{-L}^L S \left({m \pi x} \over L \right) S \left({n \pi x} \over L \right) dx$ $ = 2 \int_0^L S \left({m \pi x} \over L \right) S \left({n \pi x} \over L \right) dx $. Then we have two cases:
$m, n$ have different parity: the functions will multiply to give a negative sign when we apply the transformation $x \to L-x$ on each function. Hence $g(x)$ is odd around $x = L/2$ which means the integral from $0$ to $L$ is $0$, and hence the entire integral from $-L$ to $L$ is also $0$.
$m, n$ have the same parity: either way, two positives or two negatives multiply out to a positive. Hence $h(x)$ is even around $x = L/2$ which means the integral will not necessarily be $0$ (it seems that it is never $0$ according to Desmos testing).
So unlike with Fourier series where besides $m = n = 0$ and $m = n \ne 0$, all the other integrals are $0$, if we want to represent a function as a sum of triangle waves, say, $f(x) = \sum_{n = 1}^{\infty} a_n \ C(n \pi x / L)$, and if we switch the sum and integral here which may not be justified, we get $\int_{-L}^L f(x) \ C(m \pi x / L)$ $ = \sum_{n = 1}^{\infty} a_n \int_{-L}^L C(n \pi x / L) C(m \pi x / L)$, and we have to find the sum of $\int_{-L}^L C(n \pi x / L) C(m \pi x / L)$ over all odd or even $n$ to be able to recover $a_n$. It is not at all obvious how to do this, but I have a conjecture:
$$\sum_{n=-\infty, n=\text{odd}}^\infty \int_{-L}^L C \left({\pi x} \over L \right) C \left({n \pi x} \over L \right) dx = 4L$$
which looks plausible using Desmos. It would be great if someone could verify this conjecture (note that for $\sum_{n=-\infty}^\infty \int_{-L}^L S \left({\pi x} \over L \right) S \left({n \pi x} \over L \right) dx$, $S(x)$ is an odd function so this evaluates to $0$, and we cannot recover $a_n$ this way). For a start, the summation above is twice the summation from $0$ to $L$ due to the even-ness of $C(x)$, and for the same reason, it is twice the answer for positive $n$ only, so a better version of the conjecture would be:
$$\sum_{n=0, \ n = \text{odd}}^\infty \int_{0}^L C \left({\pi x} \over L \right) C \left({n \pi x} \over L \right) dx = L$$
Here is some progress so far:
The conjecture is false. After the substitution $x/L \mapsto x$ and finding the closed forms of the integrals for $n = 1, n = 3$ and so on, it was pretty clear that all of the answers were going to have a $\pi^2$ from finding the equations of the lines, and then adding each of the sections up between the sharp points of $\cos(n \pi x)$. Notably, the integral from $0$ to $L$ for $m = n = 1$ is $\pi^2/12 \cdot L$.
After some more experimentation with Desmos, I found that for the revised conjecture, the given sum is equal to:
$$\pi^2 L \sum_{k = 0}^\infty \frac{1}{(2 + 4k)(6 + 12k)} = \frac{\pi^2 L}{12} \sum_{k = 0}^\infty \frac{1}{(1+2k)^2} = \frac{\pi^2 L}{12} \frac{\pi^2}{8} = \frac{\pi^4 L}{96}$$
and indeed, $\frac{\pi^4}{96}$ is fairly close to $1$ (note that we got $\pi^2/8$ by separating the Basel sum into even and odd terms, and the sum with only even terms is a copy of the original sum, which results in $\frac{\pi^2}{12} (1 - (1/2)^2)$.
Also, here are two more conjectures. It seems that: $$\int_0^{k/n} C(\pi x) \ C(n \pi x) = 0$$
for all integer $k$ such that $0 < k < n$, $k$ even and this statement has a corresponding statement with $x/L$ instead of just $x$, I'm sure.
It also seems that:
$$\int_{0}^{L}C\left(\frac{m\pi x}{L}\right)C\left(\frac{n\pi x}{L}\right)\ dx = \int_{0}^{L} S\left(\frac{m\pi x}{L}\right) S\left(\frac{n\pi x}{L}\right)\ dx $$
when $m, n$ are both odd and $m + n \pmod 4 \ne 0$, and that:
$$\int_{0}^{L}C\left(\frac{m\pi x}{L}\right)C\left(\frac{n\pi x}{L}\right)\ dx = - \int_{0}^{L} S\left(\frac{m\pi x}{L}\right) S\left(\frac{n\pi x}{L}\right)\ dx $$
when $m, n$ are odd and $m + n \pmod 4\equiv 0$. Note that the transformation $m \mapsto km, n \mapsto kn$ for some integer $k$ does not change either integral, as the integral from $0$ to $L/k$ is reduced by a factor of $1/k$, but there are $k$ identical copies due to the evenness of the function around $x = nL/k$, $n \in \mathbb Z$, so the even case is covered in the odd case for when $k = 2$. More generally, we can require that $m, n$ be coprime.