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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: \( 3 \cos \theta \cdot \cos 22^{\circ}-3 \sin \theta \sin 22^{\circ}=-1 \)

Ask by Knight Chan. in South Africa
Mar 17,2025

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Answer

To deduce the identity for \(\sin(A+B)\), use the cosine difference formula with \(90^\circ - (A+B)\) and apply co-function identities to get: \[ \sin(A+B) = \sin A \cos B + \cos A \sin B \] For the equation \(3\cos\theta\cos22^\circ - 3\sin\theta\sin22^\circ = -1\), simplify to: \[ 3\cos(\theta + 22^\circ) = -1 \implies \cos(\theta + 22^\circ) = -\frac{1}{3} \] The general solution is: \[ \theta = -22^\circ \pm \arccos\left(-\frac{1}{3}\right) + 360^\circ k,\quad k \in \mathbb{Z} \]

Solution

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 \[ \theta = -22^\circ +\arccos\Bigl(-\frac{1}{3}\Bigr)+360^\circ k\quad\text{or}\quad \theta = -22^\circ -\arccos\Bigl(-\frac{1}{3}\Bigr)+360^\circ k,\quad k\in\mathbb{Z}. \]

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Bonus Knowledge

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 \( \theta \).

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