If cos θ = 1/3, then what is the value of sin(θ/2) sin(3θ/2)?

Using product-to-sum: sin(θ/2) sin(3θ/2) = (1/2)[cos(θ/2 - 3θ/2) - cos(θ/2 + 3θ/2)] = (1/2)[cos(-θ) - cos(2θ)] = (1/2)[cos θ - cos 2θ]. With cos θ = 1/3, cos 2θ = 2(1/9) - 1 = -7/9. So (1/2)[1/3 - (-7/9)] = (1/2)[3/9 + 7/9] = (1/2)(10/9) = 5/9.