15. Find $a_{0}, a_{l, k}(1 \leqslant l<k \leqslant n)$, such that for any $x \in \mathbf{R}(\sin x \neq 0)$, we have $\frac{\sin ^{2} n x}{\sin ^{2} x}=a_{0}+$
$$
\sum_{1 \leqslant l<k \leqslant n} a_{l, k} \cdot \cos 2(k-l) x \text {. }
$$
$$
\begin{array}{l}
\text { 15. Since } \sin 2 x+\sin 4 x+\sin 6 x+\cdots+\sin 2 n x=\frac{\sin n x \cdot \sin (n+1) x}{\sin x} \\
\cos 2 x+\cos 4 x+\cos 6 x+\cdots+\cos 2 n x=\frac{\sin n x \cdot \cos (n+1) x}{\sin x} \\
\text { therefore } \frac{\sin ^{2} n x}{\sin ^{2} x}=\frac{\sin ^{2} n x \cdot \sin ^{2}(n+1) x}{\sin ^{2} x}+\frac{\sin ^{2} n x \cdot \cos ^{2}(n+1) x}{\sin ^{2} x}=(\sin 2 x+\sin 4 x+\sin 6 x+\cdots+ \\
\sin 2 n x)^{2}+(\cos 2 x+\cos 4 x+\cos 6 x+\cdots+\cos 2 n x)^{2}=n+\sum_{1 \leqslant l<k \leqslant n}(2 \sin 2 l x \cdot \sin 2 k x+2 \cos 2 l x \cos 2 k x)= \\
n+2 \sum_{1 \leqslant l<k \leqslant n} \cos 2(k-l) x
\end{array}
$$

Therefore, \(a_{0}=n, a_{1, k}=2\) is the solution.
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