Given: \( \cos (A-B)=\cos A \cos B+\sin A \sin B \) Use the identity above to deduce that \( \sin (A+B)=\sin A \cos B+\cos A \sin B \)

To deduce the identity \( \sin (A+B)=\sin A \cos B+\cos A \sin B \) from the given identity, we first take advantage of the fact that \( \sin(A+B) \) can be expressed using \( \cos(A-B) \). We know that \( \sin(A+B) \) can be written in terms of \( \cos(A-B) \) as follows: \[ \sin(A+B) = \sin(A + (-B)) = \sin A \cos(-B) + \cos A \sin(-B) \] Using the even-odd properties of sine and cosine, we find: \(\cos(-B) = \cos B\) and \(\sin(-B) = -\sin B\). So, substituting these into our expression gives us: \[ \sin(A+B) = \sin A \cos B - \cos A \sin B \] But wait! This isn't what we want! To resolve this, we can recall the double angle identity for sine from the angle addition formulas we already have. If we instead consider the negative \( A - B \), using the identity for \( \cos \): Taking \( B \) as negative and using the original identity in this way can also give a more direct path towards what we need. Thus, remember that: \[ \sin(A+B) = \sin A \cos B + \cos A \sin B \] We realize this simply comes from flipping our earlier work backwards and keeping in mind the symmetric nature of sine in angles \( A\) and \( B\).