QUESTION 3 Applications 3.1 Express the following as single trigonometry ratio: 3.1.1 \( \cos 2 x \cdot \cos 3 x-\sin 2 x \cdot \sin 3 x \) 3.1.2 \( \sin 2 x \cdot \cos x+\cos 2 x \cdot \sin x \) \( \qquad \) 3.2 Determine the values of the following without using a calculator. (2) \( 3.2 .1 \sin 85^{\circ} \cdot \cos 25^{\circ}-\cos 85^{\circ} \cdot \sin 25^{\circ} \) \( \qquad \) \( \qquad \) \( 3.2 .2 \cos 160^{\circ} \cdot \cos 10^{\circ}+\sin 160^{\circ} \cdot \sin 10^{\circ} \) (4) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) [11] Grade 12 Mathematics SBA 2025 Page 15 of 29

1. **For 3.1:** 1.1. The expression \[ \cos 2x \cdot \cos 3x - \sin 2x \cdot \sin 3x \] can be written in the form \(\cos A \cdot \cos B - \sin A \cdot \sin B\), which equals \(\cos (A+B)\). Here, \(A = 2x\) and \(B = 3x\). Therefore, \[ \cos 2x \cdot \cos 3x - \sin 2x \cdot \sin 3x = \cos(2x+3x)=\cos5x. \] 1.2. The expression \[ \sin 2x \cdot \cos x + \cos 2x \cdot \sin x \] matches the form \(\sin A \cos B + \cos A \sin B\), which equals \(\sin (A+B)\). Here, \(A = 2x\) and \(B = x\). Thus, \[ \sin 2x \cdot \cos x + \cos 2x \cdot \sin x = \sin(2x+x)=\sin3x. \] 2. **For 3.2:** 2.1. Evaluate \[ \sin 85^{\circ} \cdot \cos 25^{\circ} - \cos 85^{\circ} \cdot \sin 25^{\circ}. \] Recognize the identity \(\sin A \cos B - \cos A \sin B = \sin (A-B)\) with \(A = 85^{\circ}\) and \(B = 25^{\circ}\). Therefore, \[ \sin 85^{\circ} \cdot \cos 25^{\circ} - \cos 85^{\circ} \cdot \sin 25^{\circ} = \sin(85^{\circ} - 25^{\circ})=\sin 60^{\circ}. \] We know that \(\sin 60^{\circ}=\frac{\sqrt{3}}{2}\). 2.2. Evaluate \[ \cos 160^{\circ} \cdot \cos 10^{\circ} + \sin 160^{\circ} \cdot \sin 10^{\circ}. \] Recognize the identity \(\cos A \cos B + \sin A \sin B = \cos (A-B)\) with \(A = 160^{\circ}\) and \(B = 10^{\circ}\). Thus, \[ \cos 160^{\circ} \cdot \cos 10^{\circ} + \sin 160^{\circ} \cdot \sin 10^{\circ} = \cos(160^{\circ} - 10^{\circ})=\cos150^{\circ}. \] Since \(\cos150^{\circ}=\cos(180^{\circ}-30^{\circ})=-\cos30^{\circ}\) and \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\), we have \[ \cos150^{\circ}=-\frac{\sqrt{3}}{2}. \]