Prove the identity. \( \begin{array}{l}1-\cos 5 \theta \cos 3 \theta-\sin 5 \theta \sin 3 \theta=2 \sin ^{2} \theta \\ \text { To verify the identity, start with the left side and transform it to obtain the right side. Choose the correct step and transform the expression according to the step chosen. } \\ 1-\cos 5 \theta \cos 3 \theta-\sin 5 \theta \sin 3 \theta \\ =\end{array} \)

To transform the left side, we can utilize the cosine addition formula, which states that \(\cos(a + b) = \cos a \cos b - \sin a \sin b\). Using this identity, we rewrite \(\cos 5\theta \cos 3\theta + \sin 5\theta \sin 3\theta\) as \(\cos(5\theta - 3\theta) = \cos 2\theta\). Therefore: \[ 1 - \cos 5\theta \cos 3\theta - \sin 5\theta \sin 3\theta = 1 - \cos(2\theta) \] Next, we remember that \(1 - \cos(2\theta) = 2 \sin^2(\theta)\). So we have: \[ 1 - \cos(2\theta) = 2\sin^2(\theta) \] Thus, the left side simplifies to the right side, confirming the identity: \[ 1 - \cos 5\theta \cos 3\theta - \sin 5\theta \sin 3\theta = 2\sin^2(\theta) \] Q.E.D.!