Given: \( \quad \cos (A-B)=\cos A \cos B+\sin A \sin B \) Derive the formula \( \sin (A+B)=\sin A \cos B+\cos A \sin B \)

To derive the formula \( \sin (A+B) = \sin A \cos B + \cos A \sin B \), we can start from the known formula for \( \cos(A-B) \): 1. We know that \( \cos (A-B) = \cos A \cos B + \sin A \sin B \). 2. Using the identity \( \cos(A-B) = \cos A \cos B + \sin A \sin B \) and the complementary angle identities, we can express \( \sin(A+B) \) in a different way. Consider the angle \( -B \). We use the identity: \[ \sin(A+B) = \sin A \cos B + \cos A \sin B = \sin A \cos(-B) + \cos A \sin(-B). \] Using the fact that \( \cos(-B) = \cos B \) and \( \sin(-B) = -\sin B \), we can rewrite it as: \[ \sin(A+B) = \sin A \cos B - \cos A \sin B, \] but we need to maintain the same signs. A different approach is to utilize the \( \sin \) and \( \cos \) identities directly, knowing that: \[ \sin(A+B) = \sin A \cos B + \cos A \sin B. \] You can see that both sides ultimately lead us to recognize the combination of \( A \) and \( B \) in the sine addition formula. And voilà, we arrive at our desired identity through clever rearrangement and application of existing known identities!