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

### 3.1 Express as a single trigonometry ratio #### 3.1.1 We have \[ \cos 2x \cdot \cos 3x - \sin 2x \cdot \sin 3x. \] Using the cosine addition formula \[ \cos(A+B) = \cos A \cos B - \sin A \sin B, \] set \( A = 2x \) and \( B = 3x \). Thus, the expression becomes \[ \cos(2x+3x) = \cos 5x. \] #### 3.1.2 We are given \[ \sin 2x \cdot \cos x + \cos 2x \cdot \sin x. \] Using the sine addition formula \[ \sin(A+B) = \sin A \cos B + \cos A \sin B, \] with \( A = 2x \) and \( B = x \), we have \[ \sin(2x+x) = \sin 3x. \] --- ### 3.2 Determine the values without using a calculator #### 3.2.1 The expression is \[ \sin 85^\circ \cdot \cos 25^\circ - \cos 85^\circ \cdot \sin 25^\circ. \] Recognizing the sine difference formula \[ \sin(A-B) = \sin A \cos B - \cos A \sin B, \] set \( A = 85^\circ \) and \( B = 25^\circ \) so that it becomes \[ \sin(85^\circ - 25^\circ) = \sin 60^\circ. \] Since \[ \sin 60^\circ = \frac{\sqrt{3}}{2}, \] the value is \[ \frac{\sqrt{3}}{2}. \] #### 3.2.2 We are given \[ \cos 160^\circ \cdot \cos 10^\circ + \sin 160^\circ \cdot \sin 10^\circ. \] Using the cosine difference formula \[ \cos(A-B) = \cos A \cos B + \sin A \sin B, \] with \( A = 160^\circ \) and \( B = 10^\circ \), we obtain \[ \cos(160^\circ - 10^\circ) = \cos 150^\circ. \] Since \[ \cos 150^\circ = -\cos 30^\circ = -\frac{\sqrt{3}}{2}, \] the value is \[ -\frac{\sqrt{3}}{2}. \]