Finite sine-transform orthogonality at level ell

For an integer $r$ with $0<\mathrm{\mid }r\mathrm{\mid }<2\mathrm{\ell }$, put $C_r={\sum }_{m=1}^{\mathrm{\ell }-1}\cos (rm\pi \mathrm{/}\mathrm{\ell })$. If $r$ is even, then $z=e^{2\pi i(r\mathrm{/}2)\mathrm{/}\mathrm{\ell }}$ is a nontrivial $\mathrm{\ell }$-th root of unity; the geometric sum ${\sum }_{m=0}^{\mathrm{\ell }-1}{z}^m=0$ has real part $1+C_r$, so $C_r=-1$. If $r$ is odd, the geometric sum of $e^{i\pi rm\mathrm{/}\mathrm{\ell }}$ for $0\le m<2\mathrm{\ell }$ is zero. Pairing terms at $m$ and $2\mathrm{\ell }-m$ for $1\le m\le \mathrm{\ell }-1$ leaves twice $C_r$; the two unpaired real terms are $1+\cos (r\pi )=0$, so $C_r=0$. Also $C_0=\mathrm{\ell }-1$.

The product-to-sum identity gives the left side as $\frac{1}{2}(C_{a-b}-C_{a+b})$, with $C_0$ used when $a=b$. If $a=b$, then $a+b$ is nonzero even and less than $2\mathrm{\ell }$, so the value is $\frac{1}{2}((\mathrm{\ell }-1)-(-1))=\mathrm{\ell }\mathrm{/}2$. If $a\mathrm{\ne }b$ and $a,b$ have the same parity, both nonzero indices $a-b$ and $a+b$ are even and their $C$ values are $-1$; if they have opposite parity, both indices are odd and their values are $0$. In both off-diagonal cases the difference is zero, proving the formula.