Given: \( \cos (A-B)=\cos A \cos B+\sin A \sin B \) Use the identity above to deduce that \( \sin (A+B)=\sin A \cos B+\cos A \sin B \) Determine the general solution of the equation: \( 3 \cos \theta \cdot \cos 22^{\circ}-3 \sin \theta \sin 22^{\circ}=-1 \)

We start with the given identity \[ \cos (A-B)=\cos A\,\cos B+\sin A\,\sin B. \] If we wish to deduce an identity for \(\sin(A+B)\), we write \[ \sin(A+B)=\cos\Bigl(90^\circ-(A+B)\Bigr) \] and note that \[ 90^\circ-(A+B)=(90^\circ-A)-B. \] Then using the cosine difference formula we have \[ \cos\Bigl[(90^\circ-A)-B\Bigr]=\cos(90^\circ-A)\cos B+\sin(90^\circ-A)\sin B. \] But using the co-function identities \[ \cos(90^\circ-A)=\sin A\quad\text{and}\quad \sin(90^\circ-A)=\cos A, \] it follows that \[ \sin(A+B)=\sin A\,\cos B+\cos A\,\sin B. \] Now we turn to the equation we want to solve: \[ 3\cos\theta\,\cos22^\circ-3\sin\theta\,\sin22^\circ=-1. \] Notice that the left‐hand side can be written as \[ 3\bigl(\cos\theta\,\cos22^\circ-\sin\theta\,\sin22^\circ\bigr). \] Recall the cosine addition formula: \[ \cos\bigl(\theta+22^\circ\bigr)=\cos\theta\,\cos22^\circ-\sin\theta\,\sin22^\circ. \] Thus the equation becomes \[ 3\cos\bigl(\theta+22^\circ\bigr)=-1. \] Dividing both sides by 3 gives \[ \cos\bigl(\theta+22^\circ\bigr)=-\frac{1}{3}. \] The general solution for \(\cos x = -\frac{1}{3}\) is: \[ x = \pm\arccos\Bigl(-\frac{1}{3}\Bigr)+360^\circ k,\quad k\in\mathbb{Z}. \] Here, we set \[ x=\theta+22^\circ, \] so that \[ \theta+22^\circ=\pm\arccos\Bigl(-\frac{1}{3}\Bigr)+360^\circ k,\quad k\in\mathbb{Z}. \] Solving for \(\theta\) we obtain \[ \theta = -22^\circ +\arccos\Bigl(-\frac{1}{3}\Bigr)+360^\circ k\quad\text{or}\quad \theta = -22^\circ -\arccos\Bigl(-\frac{1}{3}\Bigr)+360^\circ k,\quad k\in\mathbb{Z}. \]