A complicated sum of cosines involving sums inside the cosine arguments

Question

Let $n$ be an integer greater than $1$. How do I show that $$\sum_{L_{1}=0}^{n-2}\sum_{L_{2}=0}^{L_{1}}\left(-1\right)^{L_{1}+L_{2}}\left(1+\left(-1\right)^{L_{1}+L_{2}}\cos\left(\pi\sum_{k=L_{2}+1}^{L_{1}+1}\frac{k}{2n+1}+k\right)\right)\left(1+\left(-1\right)^{n+L_{1}}\cos\left(\pi\sum_{k=L_{1}+2}^{n}\frac{\left(-1\right)^{k}k}{2n+1}+k\right)\right)=\frac{n-1}{4}?\qquad\qquad(\star)$$ I've verified $(\star)$ numerically up to about $n=50$ and proved it directly for smaller $n$, say $n=2,3,4$. However, I still don't have any idea how to approach this sum generally. I came across a similar-looking identity that simplifies very nicely as well: $$\sum_{L_{1}=0}^{n-1}\left(-1\right)^{L_{1}}\left(1-\left(-1\right)^{n+L_{1}}\cos\left(\pi\sum_{k=L_{1}+1}^{n}\frac{\left(-1\right)^{k}k}{2n+1}+k\right)\right)=\frac{1}{2}.$$ However, the proof for this is just rearranging the terms in the sum $\sum_{k=0}^{n}\left(-1\right)^{k}\cos\left(\frac{k\pi}{2n+1}\right)=\frac{1}{2}$. I tried using a number of different trig identities to simplify the summand in $(\star)$, but none of them seemed to help. Those sums inside the cosines make this very confusing for me. Any advice on how to approach this would be greatly appreciated.