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

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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
       
       $$
       \cos150^{\circ}=-\frac{\sqrt{3}}{2}.
       $$
1. **For 3.1:** 1.1. $$ \cos 2x \cdot \cos 3x - \sin 2x \cdot \sin 3x = \cos 5x $$ 1.2. $$ \sin 2x \cdot \cos x + \cos 2x \cdot \sin x = \sin 3x $$ 2. **For 3.2:** 2.1. $$ \sin 85^{\circ} \cdot \cos 25^{\circ} - \cos 85^{\circ} \cdot \sin 25^{\circ} = \sin 60^{\circ} = \frac{\sqrt{3}}{2} $$ 2.2. $$ \cos 160^{\circ} \cdot \cos 10^{\circ} + \sin 160^{\circ} \cdot \sin 10^{\circ} = \cos 150^{\circ} = -\frac{\sqrt{3}}{2} $$

QUESTION 3
Applications
3.1 Express the following as single trigonometry ratio:
3.1.1
3.1.2

3.2 Determine the values of the following without using a calculator.
(2)

(4)

[11]
Grade 12 Mathematics SBA 2025
Page 15 of 29

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QUESTION 3 Applications 3.1 Express the following as single trigonometry ratio: 3.1.1 3.1.2 3.2 Determine the values of the following without using a calculator. (2) (4) [11] Grade 12 Mathematics SBA 2025 Page 15 of 29