Show that $\sin ^{2} 7 \theta-\sin ^{2} 4 \theta=\sin 11 \theta \sin 3 \theta$.

Step 1 ：Rewrite the given equation using the identity \(\sin^2(x) - \sin^2(y) = (\sin(x) + \sin(y))(\sin(x) - \sin(y))\):

Step 2 ：\((\sin(7\theta) + \sin(4\theta))(\sin(7\theta) - \sin(4\theta))\)

Step 3 ：Use the sum-to-product identities to rewrite the equation:

Step 4 ：\(4\sin(11\theta)\cos(3\theta)\cos(11\theta)\sin(3\theta)\)

Step 5 ：Use the product-to-sum identity to rewrite the equation:

Step 6 ：\(\sin(14\theta)^2 - \sin(8\theta)^2\)

Step 7 ：Rewrite the equation using the identity \(\sin^2(x) - \sin^2(y) = (\sin(x) + \sin(y))(\sin(x) - \sin(y))\):

Step 8 ：\((\sin(14\theta) + \sin(8\theta))(\sin(14\theta) - \sin(8\theta))\)

Step 9 ：Use the sum-to-product identities to rewrite the equation:

Step 10 ：\(4\sin(11\theta)\cos(3\theta)\cos(11\theta)\sin(3\theta)\)

Step 11 ：\(\boxed{\sin^2(7\theta) - \sin^2(4\theta) = \sin(11\theta)\sin(3\theta)}\)