INVESTIGATING COMPOUND ANGLES AND THEREFORE, DOUBLE ANGLES. QUESTION 1 1.1. Given: (i) $$ \quad A=60^{\circ} $$ 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) $$ \cos A \cdot \cos B+\sin A \cdot \sin B $$ $$ 1.2 \quad $$ Compare the values of $$ \cos (A-B) $$ with the values of $$ \cos A-\cos B $$ and $$ \cos A \cdot \cos B+\sin A \cdot \sin B $$ $$ 1.3 \quad $$ Using your $$ \operatorname{comparison} $$ in Question 1.2 ., what general conclusion 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 ?

Question

UpStudy AI · Accepted Answer

Certainly! Let's work through each part of the problem step by step.

## **Question 1**

### **1.1 Evaluations**

We'll evaluate the expressions for each given pair of angles $$ A $$ and $$ B $$.

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

1. **(a) $$ \cos(A - B) $$**
   $$
   \cos(60^{\circ} - 30^{\circ}) = \cos(30^{\circ}) = \frac{\sqrt{3}}{2} \approx 0.8660
   $$

2. **(b) $$ \cos A - \cos B $$**
   $$
   \cos(60^{\circ}) - \cos(30^{\circ}) = 0.5 - \frac{\sqrt{3}}{2} \approx 0.5 - 0.8660 = -0.3660
   $$

3. **(c) $$ \cos A \cdot \cos B + \sin A \cdot \sin B $$**
   $$
   \cos(60^{\circ}) \cdot \cos(30^{\circ}) + \sin(60^{\circ}) \cdot \sin(30^{\circ}) = (0.5)(0.8660) + (0.8660)(0.5) = 0.4330 + 0.4330 = 0.8660
   $$

---

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

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

2. **(b) $$ \cos A - \cos B $$**
   $$
   \cos(110^{\circ}) - \cos(50^{\circ})
   $$
   - $$ \cos(110^{\circ}) = \cos(180^{\circ} - 70^{\circ}) = -\cos(70^{\circ}) \approx -0.3420 $$
   - $$ \cos(50^{\circ}) \approx 0.6428 $$
   $$
   -0.3420 - 0.6428 \approx -0.9848
   $$

3. **(c) $$ \cos A \cdot \cos B + \sin A \cdot \sin B $$**
   $$
   \cos(110^{\circ}) \cdot \cos(50^{\circ}) + \sin(110^{\circ}) \cdot \sin(50^{\circ}) = (-0.3420)(0.6428) + (0.9397)(0.7660) \approx -0.2197 + 0.7205 = 0.5008 \approx 0.5
   $$

---

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

1. **(a) $$ \cos(A - B) $$**
   $$
   \cos(225^{\circ} - 135^{\circ}) = \cos(90^{\circ}) = 0
   $$

2. **(b) $$ \cos A - \cos B $$**
   $$
   \cos(225^{\circ}) - \cos(135^{\circ})
   $$
   - $$ \cos(225^{\circ}) = \cos(180^{\circ} + 45^{\circ}) = -\cos(45^{\circ}) \approx -0.7071 $$
   - $$ \cos(135^{\circ}) = \cos(180^{\circ} - 45^{\circ}) = -\cos(45^{\circ}) \approx -0.7071 $$
   $$
   -0.7071 - (-0.7071) = 0
   $$

3. **(c) $$ \cos A \cdot \cos B + \sin A \cdot \sin B $$**
   $$
   \cos(225^{\circ}) \cdot \cos(135^{\circ}) + \sin(225^{\circ}) \cdot \sin(135^{\circ}) = (-0.7071)(-0.7071) + (-0.7071)(0.7071) = 0.5 - 0.5 = 0
   $$

---

### **Summary of Results**

| 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                                                  |

### **1.2 Comparison of Values**

From the table above, we observe the following:

1. **$$ \cos(A - B) $$** and **$$ \cos A \cdot \cos B + \sin A \cdot \sin B $$** yield the **same** values across all cases.
   
2. **$$ \cos A - \cos B $$** generally produces **different** values compared to the other two expressions, except in the third case where they all equal zero.

### **1.3 General Conclusion**

Based on the comparisons:

- **Trigonometric Identity Confirmation:**
  
  The consistent equality between $$ \cos(A - B) $$ and $$ \cos A \cdot \cos B + \sin A \cdot \sin B $$ across all cases confirms the trigonometric identity:
  $$
  \cos(A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B
  $$
  
- **Difference with $$ \cos A - \cos B $$:**
  
  The expression $$ \cos A - \cos B $$ does **not** generally equal $$ \cos(A - B) $$ or $$ \cos A \cdot \cos B + \sin A \cdot \sin B $$. It coincides only in specific instances (as seen in Case (iii)) where the values of $$ \cos A $$ and $$ \cos B $$ are equal.

**Final Conclusion:**

For any angles $$ A $$ and $$ B $$:

- **$$ \cos(A - B) $$ is always equal to $$ \cos A \cdot \cos B + \sin A \cdot \sin B $$.**
  
- **$$ \cos(A - B) $$ is generally not equal to $$ \cos A - \cos B $$, except in particular cases where specific relationships between $$ A $$ and $$ B $$ make the expressions equivalent.**

This illustrates the fundamental trigonometric identity for the cosine of a difference of two angles.
For any angles $$ A $$ and $$ B $$, the cosine of the difference $$ A - B $$ is always equal to the product of the cosines of $$ A $$ and $$ B $$ plus the product of the sines of $$ A $$ and $$ B $$. However, $$ \cos(A - B) $$ is generally not equal to $$ \cos A - \cos B $$, except in specific cases where both expressions yield the same value.

Real Tutor Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Trigonometry Questions