Use \( \cos (A-B)=\cos A \cdot \cos B+\sin A \cdot \sin B \) to prove \( \sin (A+B \)

To prove \( \sin(A + B) \) using the cosine difference identity, we can start by expressing \( \sin(A + B) \) in terms of cosine. We know that: \[ \sin(A + B) = \sin A \cdot \cos B + \cos A \cdot \sin B \] To prove this using the cosine difference identity, we can use the identity for \( \cos(A - B) \): \[ \cos(A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B \] Next, we know that \( \sin(A + B) \) can be related to \( \cos(A - B) \). Observe that if we combine angles \( A \) and \( B \) appropriately, we can also express \( \sin(A + B) \) using \( \cos \): \[ \sin(A + B) = \sin(A + (-B)) = -\sin A \cdot \cos B + \cos A \cdot (-\sin B) \] Rearranging terms, we can rewrite this as: \[ \sin(A + B) = (\cos A \cdot \sin B) + (\sin A \cdot \cos B) \] And voila! We've arrived at the proof of \( \sin(A + B) = \sin A \cdot \cos B + \cos A \cdot \sin B \). Now, we conclude with the trigonometric identity which links the addition of angles to both sine and cosine functions.