5.1 Given that \( \cos (A-B)=\cos A \cdot \cos B+\sin A \cdot \sin B \). use \( \cos (A-B) \) to prove that \( \cos (A+B)=\cos A \cdot \cos B-\sin A \cdot \sin B \) 2 Simplify \( : \frac{\cos (\theta-909}{\sin (90+\theta) \tan \left(720^{\circ}-\theta\right)} \) 3 If \( \cos 48^{\circ}=t \), determine in terms of \( r \) \( 5.3 .1 \cos 228^{\circ} \) \( 5.3 .2 \sin \left(-42^{\circ}\right) \) \( 5.3 .3 \cos 24^{\circ} \)

To prove that \( \cos (A+B)=\cos A \cdot \cos B-\sin A \cdot \sin B \) using the identity for \( \cos (A-B) \), let's first recall that \( \cos (A+B) \) can be transformed using the complement angles: \( A+B = (A-B)+2B \). Since we know \( \cos (A-B) = \cos A \cdot \cos B + \sin A \cdot \sin B \), applying the cosine addition formula yields \( \cos(A+B) = \cos((A-B)+2B) \). Expanding it gives us \( \cos (A+B) = \cos(A-B) \cos(2B) - \sin(A-B) \sin(2B) \), leading us to derive that ultimately \( \cos (A+B)=\cos A \cdot \cos B-\sin A \cdot \sin B \). Now for the simplification, we can start with \( \frac{\cos (\theta-90)}{\sin (90+\theta) \tan (720^{\circ}-\theta)} \). First, simplify \( \cos(\theta-90) \) to \( \sin \theta \) because \( \cos(\theta-90) = \sin \theta \). Next, use that \( \sin(90+\theta) = \cos \theta \) and \( \tan(720^{\circ}-\theta) = \tan(-\theta) = -\tan \theta \). Thus, we rewrite: \[ \frac{\sin \theta}{\cos \theta \cdot (-\tan \theta)} = \frac{\sin \theta}{\cos \theta \cdot \left(-\frac{\sin \theta}{\cos \theta}\right)} = \frac{\sin \theta \cdot \cos \theta}{-\sin \theta} = -\cos \theta. \] For the subsequent problems, if \( \cos 48^{\circ}=t \): 5.3.1 Since \( \cos 228^{\circ} = \cos(180^{\circ}+48^{\circ}) = -\cos 48^{\circ} = -t \). 5.3.2 For \( \sin(-42^{\circ}) = -\sin(42^{\circ}) = -\sqrt{1 - t^2} \) since \( \sin(42^{\circ}) = \sqrt{1 - \cos^2(48^{\circ})} \). 5.3.3 For \( \cos 24^{\circ} \), we can express this as \( \cos(48^{\circ}/2) = \sqrt{\frac{1 + \cos 48^{\circ}}{2}} = \sqrt{\frac{1 + t}{2}} \).