Select the correct option. Write \( \sin \left(\frac{7 \pi}{12}\right) \cos \left(\frac{5 \pi}{12}\right) \) as a sum or difference. \( \begin{array}{ll}\frac{1}{2} \cos (\pi)-\frac{1}{2} \cos \left(\frac{\pi}{6}\right) & \frac{1}{2} \sin (\pi)+\frac{1}{2} \sin \left(\frac{\pi}{6}\right) \\ \frac{1}{2} \cos (\pi)+\frac{1}{2} \cos \left(\frac{\pi}{6}\right) & \frac{1}{2} \sin (\pi)-\frac{1}{2} \sin \left(\frac{\pi}{6}\right)\end{array} \)

To rewrite \( \sin \left(\frac{7 \pi}{12}\right) \cos \left(\frac{5 \pi}{12}\right) \) as a sum or difference, we can use the product-to-sum identities. The specific identity we need is: \[ \sin a \cos b = \frac{1}{2} \left( \sin(a + b) + \sin(a - b) \right). \] In our case, let \( a = \frac{7\pi}{12} \) and \( b = \frac{5\pi}{12} \). Thus, applying the identity: \[ \sin \left(\frac{7 \pi}{12}\right) \cos \left(\frac{5 \pi}{12}\right) = \frac{1}{2} \left( \sin\left(\frac{7\pi}{12} + \frac{5\pi}{12}\right) + \sin\left(\frac{7\pi}{12} - \frac{5\pi}{12}\right) \right). \] Calculating the angles: \[ \frac{7\pi}{12} + \frac{5\pi}{12} = \frac{12\pi}{12} = \pi, \] \[ \frac{7\pi}{12} - \frac{5\pi}{12} = \frac{2\pi}{12} = \frac{\pi}{6}. \] Therefore, substituting back: \[ \sin \left(\frac{7 \pi}{12}\right) \cos \left(\frac{5 \pi}{12}\right) = \frac{1}{2} \left( \sin(\pi) + \sin\left(\frac{\pi}{6}\right) \right). \] Since \(\sin(\pi) = 0\) and \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\), the expression simplifies to: \[ \frac{1}{2} \left( 0 + \frac{1}{2} \right) = \frac{1}{4}. \] Now checking the options provided, we see that the correct answer is: \(\frac{1}{2} \sin(\pi) + \frac{1}{2} \sin\left(\frac{\pi}{6}\right)\). So, the correct option is: \(\frac{1}{2} \sin(\pi)+\frac{1}{2} \sin \left(\frac{\pi}{6}\right)\).