Express the given product as a sum containing only sines or cosines.
\[
\sin (6 \theta) \sin (2 \theta)
\]

Step 1 ：We are given the product of two sine functions: \(\sin (6 \theta) \sin (2 \theta)\).

Step 2 ：We can express this product as a sum or difference of cosine functions using the product-to-sum identities in trigonometry.

Step 3 ：The identity we will use is: \(\sin(A) \sin(B) = \frac{1}{2} [\cos(A - B) - \cos(A + B)]\).

Step 4 ：In this case, A = 6θ and B = 2θ. So we can substitute these into the identity to get the expression as a sum of cosines.

Step 5 ：Substituting A and B into the identity, we get: \(\sin(6\theta) \sin(2\theta) = \frac{1}{2} [\cos(6\theta - 2\theta) - \cos(6\theta + 2\theta)]\).

Step 6 ：Simplifying the above expression, we get: \(\sin(6\theta) \sin(2\theta) = \frac{1}{2} [\cos(4\theta) - \cos(8\theta)]\).

Step 7 ：\(\boxed{\sin (6 \theta) \sin (2 \theta) = \frac{1}{2} [\cos(4 \theta) - \cos(8 \theta)]}\) is the final answer.