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) 3.2.1 \( \sin 85^{\circ} \cdot \cos 25^{\circ}-\cos 85^{\circ} \cdot \sin 25^{\circ} \) \( \qquad \) \( \qquad \) \( \qquad \) 3.2.2 \( \cos 160^{\circ} \cdot \cos 10^{\circ}+\sin 160^{\circ} \cdot \sin 10^{\circ} \) (4) \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \)

We can use the well‐known angle addition and subtraction formulas to simplify the expressions. 1. For 3.1.1, we have:   cos 2x · cos 3x – sin 2x · sin 3x Recall that cos(A + B) = cos A · cos B – sin A · sin B. Thus, with A = 2x and B = 3x,   cos 2x · cos 3x – sin 2x · sin 3x = cos(2x + 3x) = cos 5x. 2. For 3.1.2, we have:   sin 2x · cos x + cos 2x · sin x Recall that sin(A + B) = sin A · cos B + cos A · sin B. Thus, with A = 2x and B = x,   sin 2x · cos x + cos 2x · sin x = sin(2x + x) = sin 3x. 3. For 3.2.1, we have:   sin 85° · cos 25° – cos 85° · sin 25° Again, using the sine subtraction formula: sin A · cos B – cos A · sin B = sin(A – B). Thus, with A = 85° and B = 25°,   sin 85° · cos 25° – cos 85° · sin 25° = sin(85° – 25°) = sin 60°. 4. For 3.2.2, we have:   cos 160° · cos 10° + sin 160° · sin 10° Recall the cosine subtraction formula: cos A · cos B + sin A · sin B = cos(A – B). Thus, with A = 160° and B = 10°,   cos 160° · cos 10° + sin 160° · sin 10° = cos(160° – 10°) = cos 150°. So the simplified results are: • 3.1.1: cos 5x • 3.1.2: sin 3x • 3.2.1: sin 60° • 3.2.2: cos 150°