INVESTIGATING COMPOUND ANGLES AND THEREFORF, DOUBLE ANGLES, QUESTION I 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) $$ \cos A \cos B+\sin A \sin B $$ Compare the values of $$ \cos (A-B) $$ with the values of $$ \cos A-\cos B $$ and 1.2 $$ \cos A \cdot \cos B+\sin A \cdot \sin B $$ Using your $$ \operatorname{compurison} $$ in Question 1.2 ., what general $$ \operatorname{conclusion~can~be~made~regarding~} $$ (2) $$ \cos (A-B) $$, $$ \cos A-\cos B $$ and $$ \cos A $$. $$ \cos B+\sin A . \sin B $$ for any values of $$ A $$ and $$ B $$ ?

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UpStudy AI · Accepted Answer

Certainly! Let's tackle **Question I** systematically by evaluating each part step-by-step for the given angle pairs.

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### **1.1. Evaluations for Given Angle Pairs**

We are provided with three pairs of angles:

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

For each pair, we'll evaluate the following expressions:

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

Let's proceed with the calculations using a calculator (ensure it's set to degree mode).

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#### **Pair (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) $$ \cos A \cdot \cos B + \sin A \cdot \sin B $$:**
   $$
   (0.5)(0.8660) + (\frac{\sqrt{3}}{2})(0.5) \approx 0.4330 + 0.4330 = 0.8660
   $$

---

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

1. **(a) $$ \cos(A - B) $$:**
   $$
   \cos(110^{\circ} - 50^{\circ}) = \cos(60^{\circ}) \approx 0.5
   $$

2. **(b) $$ \cos A - \cos B $$:**
   $$
   \cos(110^{\circ}) - \cos(50^{\circ}) \approx (-0.3420) - 0.6428 = -0.9848
   $$

3. **(c) $$ \cos A \cdot \cos B + \sin A \cdot \sin B $$:**
   $$
   (-0.3420)(0.6428) + (0.9397)(0.7660) \approx -0.2196 + 0.7201 = 0.5005 \approx 0.5
   $$

---

#### **Pair (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) $$ \cos A \cdot \cos B + \sin A \cdot \sin B $$:**
   $$
   (-0.7071)(-0.7071) + (-0.7071)(0.7071) \approx 0.4999 - 0.4999 = 0
   $$

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### **Comparison of Values**

Let's tabulate the results for clarity:

| Pair | $$ \cos(A - B) $$ | $$ \cos A - \cos B $$ | $$ \cos A \cdot \cos B + \sin A \cdot \sin B $$ |
|------|--------------------|------------------------|-----------------------------------|
| (i)  | 0.8660             | -0.3660                | 0.8660                            |
| (ii) | 0.5000             | -0.9848                | 0.5005 ≈ 0.5000                   |
| (iii)| 0                  | 0                      | 0                                 |

**Observations:**

1. **$$ \cos(A - B) $$** and **$$ \cos A \cdot \cos B + \sin A \cdot \sin B $$** yield **identical or nearly identical** results for all pairs.
2. **$$ \cos A - \cos B $$** does **not** match $$ \cos(A - B) $$ except in a specific case (Pair (iii)).

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### **1.2. General Conclusion**

From the comparisons above, we can infer a **general trigonometric identity**:

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

This identity holds true **for any values of $$ A $$ and $$ B $$**. The term $$ \cos A - \cos B $$, however, does not generally relate to $$ \cos(A - B) $$ in a consistent manner and does not represent a standard trigonometric identity.

**Key Takeaways:**

- The expression $$ \cos(A - B) $$ **always equals** $$ \cos A \cdot \cos B + \sin A \cdot \sin B $$ regardless of the angle measures.
- The difference $$ \cos A - \cos B $$ does **not** equate to $$ \cos(A - B) $$ except in specific instances where the angles cause the terms to cancel out or align coincidentally.

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### **Further Insights**

Understanding these identities is crucial in simplifying complex trigonometric expressions and solving equations involving compound angles. Recognizing that **$$ \cos(A - B) $$** aligns perfectly with the sum of the products of cosines and sines emphasizes the interconnectedness of trigonometric functions and their fundamental identities.
For any angles $$ A $$ and $$ B $$, the following trigonometric identity holds: $$ \cos(A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B $$ This identity is consistent across different angle pairs and does not generally relate to $$ \cos A - \cos B $$.

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