VESTIGATING COMPOUND ANGLES AND THEREFORE, DOUBLE, ANGLES, UESTION 1 1. Given: (i) $$ A=60^{\circ} $$ and $$ B=30^{\circ} $$ (ii) $$ A=110^{\circ} $$ and $$ B=50^{\circ} $$ (iii) $$ A=225^{\circ} $$ and $$ B=135^{\circ} $$ Use a calculator to evaluate each of the following: (a) $$ \cos (A-B) $$ (b) $$ \cos A-\cos B $$ (c) $$ \cos A \cdot \cos B+\sin A \cdot \sin B $$ 1.2 $$ \operatorname{compare} $$ the values of $$ \cos (A-B) $$ with the values of $$ \cos A-\cos B $$ and $$ \cos A \cdot \cos B+\sin A \cdot \sin B $$ 1.3 Using your $$ \operatorname{comparison} $$ in Question 1.2 ., what general $$ \operatorname{conclusion} \operatorname{can} $$ be made regarding $$ \cos (A-B), \cos A-\cos B $$ and $$ \cos A \cdot \cos B+\sin A \cdot \sin B $$ for any values of $$ A $$ and $$ B $$ ?

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Let's tackle the problem step by step to understand the relationships between the given trigonometric expressions.

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### **Question 1.1: Evaluation of Expressions**

Given three sets of angles:

1. **Case (i):** $$ A = 60^\circ $$, $$ B = 30^\circ $$
2. **Case (ii):** $$ A = 110^\circ $$, $$ B = 50^\circ $$
3. **Case (iii):** $$ A = 225^\circ $$, $$ B = 135^\circ $$

We need to evaluate the following for each case:

- **(a)** $$ \cos(A - B) $$
- **(b)** $$ \cos A - \cos B $$
- **(c)** $$ \cos A \cdot \cos B + \sin A \cdot \sin B $$

Let's compute these step by step.

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#### **Case (i): $$ A = 60^\circ $$, $$ B = 30^\circ $$**

1. **(a) $$ \cos(A - B) = \cos(60^\circ - 30^\circ) = \cos(30^\circ) \approx 0.8660 $$**
   
2. **(b) $$ \cos A - \cos B = \cos(60^\circ) - \cos(30^\circ) = 0.5 - 0.8660 = -0.3660 $$**
   
3. **(c) $$ \cos A \cdot \cos B + \sin A \cdot \sin B = (0.5 \times 0.8660) + (\sin 60^\circ \times \sin 30^\circ) \approx (0.4330) + (0.8660 \times 0.5) = 0.4330 + 0.4330 = 0.8660 $$**

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#### **Case (ii): $$ A = 110^\circ $$, $$ B = 50^\circ $$**

1. **(a) $$ \cos(A - B) = \cos(110^\circ - 50^\circ) = \cos(60^\circ) = 0.5 $$**
   
2. **(b) $$ \cos A - \cos B = \cos(110^\circ) - \cos(50^\circ) \approx (-0.3420) - 0.6428 = -0.9848 $$**
   
3. **(c)**
   $$
   \begin{align*}
   \cos A \cdot \cos B + \sin A \cdot \sin B &= \cos(110^\circ) \cdot \cos(50^\circ) + \sin(110^\circ) \cdot \sin(50^\circ) \
   &\approx (-0.3420) \times 0.6428 + 0.9397 \times 0.7660 \
   &\approx -0.2196 + 0.7194 \
   &= 0.4998 \approx 0.5
   \end{align*}
   $$

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#### **Case (iii): $$ A = 225^\circ $$, $$ B = 135^\circ $$**

1. **(a) $$ \cos(A - B) = \cos(225^\circ - 135^\circ) = \cos(90^\circ) = 0 $$**
   
2. **(b) $$ \cos A - \cos B = \cos(225^\circ) - \cos(135^\circ) \approx (-0.7071) - (-0.7071) = 0 $$**
   
3. **(c)**
   $$
   \begin{align*}
   \cos A \cdot \cos B + \sin A \cdot \sin B &= \cos(225^\circ) \cdot \cos(135^\circ) + \sin(225^\circ) \cdot \sin(135^\circ) \
   &\approx (-0.7071) \times (-0.7071) + (-0.7071) \times 0.7071 \
   &\approx 0.5 - 0.5 \
   &= 0
   \end{align*}
   $$

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### **Question 1.2: Comparison of Values**

Let's summarize the computed values:

| **Case** | **$$ \cos(A - B) $$** | **$$ \cos A - \cos B $$** | **$$ \cos A \cdot \cos B + \sin A \cdot \sin B $$** |
|----------|-----------------------|----------------------------|-------------------------------------------------------|
| **(i)**  | 0.8660                | -0.3660                    | 0.8660                                                |
| **(ii)** | 0.5                   | -0.9848                    | 0.5                                                    |
| **(iii)**| 0                     | 0                          | 0                                                      |

**Observations:**

1. **$$ \cos(A - B) $$ and $$ \cos A \cdot \cos B + \sin A \cdot \sin B $$ are equal in all cases.**
2. **$$ \cos A - \cos B $$ differs from the other two expressions, except in Case (iii), where all expressions are zero.**

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### **Question 1.3: General Conclusion**

From the calculations and comparisons above, we can draw a general conclusion based on trigonometric identities.

**Identity:** 
$$
\cos(A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B
$$

**Conclusion:**

- **For any angles $$ A $$ and $$ B $$, the expression $$ \cos(A - B) $$ is always equal to $$ \cos A \cdot \cos B + \sin A \cdot \sin B $$.**
  
- **The expression $$ \cos A - \cos B $$ does not generally equal $$ \cos(A - B) $$ or $$ \cos A \cdot \cos B + \sin A \cdot \sin B $$, except in specific cases (e.g., when $$ A - B = 0^\circ $$ or $$ 180^\circ $$, leading to both expressions equating to zero).**

This demonstrates the fundamental trigonometric identity and highlights the distinct nature of each expression involving trigonometric functions.
For any angles $$ A $$ and $$ B $$, $$ \cos(A - B) $$ is always equal to $$ \cos A \cdot \cos B + \sin A \cdot \sin B $$.

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