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: \( 3 \cos \theta \cdot \cos 22^{\circ}-3 \sin \theta \sin 22^{\circ}=-1 \)
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To deduce that \( \sin (A+B)=\sin A \cos B+\cos A \sin B \), we can utilize the given cosine identity. If we take \( B = -C \), we apply: \[ \cos(A - (-B)) = \cos A \cos(-B) + \sin A \sin(-B) \] This leads to: \[ \cos(A + B) = \cos A \cos B - \sin A \sin B \] Using the fundamental identity \( \sin^2 x + \cos^2 x = 1 \) gives the relationship between sine and cosine: \[ \sin(A+B) = \sin A \cos B + \cos A \sin B \] Now, for the general solution to the equation: \[ 3 \cos \theta \cdot \cos 22^{\circ} - 3 \sin \theta \sin 22^{\circ} = -1 \] First, factor out the 3: \[ 3(\cos \theta \cdot \cos 22^{\circ} - \sin \theta \cdot \sin 22^{\circ}) = -1 \] Then divide both sides by 3: \[ \cos \theta \cdot \cos 22^{\circ} - \sin \theta \cdot \sin 22^{\circ} = -\frac{1}{3} \] Recognizing that the left side can be rewritten using the cosine addition formula: \[ \cos(\theta + 22^{\circ}) = -\frac{1}{3} \] The general solution for \( \cos x = k \) is given by: \[ x = \pm \cos^{-1}(k) + 2n\pi \quad (n \in \mathbb{Z}) \] Thus: \[ \theta + 22^{\circ} = \pm \cos^{-1}\left(-\frac{1}{3}\right) + 2n\pi \] Finally, isolating \( \theta \): \[ \theta = -22^{\circ} \pm \cos^{-1}\left(-\frac{1}{3}\right) + 2n\pi \] This results in the complete general solution for \( \theta \).