Answer
Here are the results for each case:
- **Case (i):** \( A = 60^\circ \), \( B = 30^\circ \)
- \( \cos(A - B) = 0.8660 \)
- \( \cos A - \cos B = -0.3660 \)
- \( \cos A \cdot \cos B + \sin A \cdot \sin B = 0.8660 \)
- **Case (ii):** \( A = 110^\circ \), \( B = 50^\circ \)
- \( \cos(A - B) = 0.5000 \)
- \( \cos A - \cos B = -0.9848 \)
- \( \cos A \cdot \cos B + \sin A \cdot \sin B = 0.5000 \)
- **Case (iii):** \( A = 225^\circ \), \( B = 135^\circ \)
- \( \cos(A - B) = 0.0000 \)
- \( \cos A - \cos B = 0.0000 \)
- \( \cos A \cdot \cos B + \sin A \cdot \sin B = 0.0000 \)
These calculations confirm the trigonometric identity \( \cos(A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B \) for the given angles.
Solution
Certainly! Let's evaluate each part for the given angles \( A \) and \( B \) using a calculator. We'll provide results up to four decimal places for precision.
---
### **Case (i): \( A = 60^\circ \), \( B = 30^\circ \)**
**(a) \( \cos(A - B) \)**
\[
\cos(60^\circ - 30^\circ) = \cos(30^\circ) \approx 0.8660
\]
**(b) \( \cos A - \cos B \)**
\[
\cos(60^\circ) - \cos(30^\circ) \approx 0.5000 - 0.8660 = -0.3660
\]
**(c) \( \cos A \cdot \cos B + \sin A \cdot \sin B \)**
\[
\cos(60^\circ) \cdot \cos(30^\circ) + \sin(60^\circ) \cdot \sin(30^\circ) \approx (0.5000)(0.8660) + (0.8660)(0.5000) = 0.8660
\]
---
### **Case (ii): \( A = 110^\circ \), \( B = 50^\circ \)**
**(a) \( \cos(A - B) \)**
\[
\cos(110^\circ - 50^\circ) = \cos(60^\circ) \approx 0.5000
\]
**(b) \( \cos A - \cos B \)**
\[
\cos(110^\circ) - \cos(50^\circ) \approx (-0.3420) - 0.6428 = -0.9848
\]
**(c) \( \cos A \cdot \cos B + \sin A \cdot \sin B \)**
\[
\cos(110^\circ) \cdot \cos(50^\circ) + \sin(110^\circ) \cdot \sin(50^\circ) \approx (-0.3420)(0.6428) + (0.9397)(0.7660) \approx -0.2198 + 0.7198 = 0.5000
\]
---
### **Case (iii): \( A = 225^\circ \), \( B = 135^\circ \)**
**(a) \( \cos(A - B) \)**
\[
\cos(225^\circ - 135^\circ) = \cos(90^\circ) = 0.0000
\]
**(b) \( \cos A - \cos B \)**
\[
\cos(225^\circ) - \cos(135^\circ) \approx (-0.7071) - (-0.7071) = 0.0000
\]
**(c) \( \cos A \cdot \cos B + \sin A \cdot \sin B \)**
\[
\cos(225^\circ) \cdot \cos(135^\circ) + \sin(225^\circ) \cdot \sin(135^\circ) \approx (−0.7071)(−0.7071) + (−0.7071)(0.7071) = 0.5000 - 0.5000 = 0.0000
\]
---
### **Summary of Results**
| **Case** | **\( \cos(A - B) \)** | **\( \cos A - \cos B \)** | **\( \cos A \cdot \cos B + \sin A \cdot \sin B \)** |
|----------|-----------------------|----------------------------|-------------------------------------------------------|
| (i) \( 60^\circ, 30^\circ \) | 0.8660 | -0.3660 | 0.8660 |
| (ii) \( 110^\circ, 50^\circ \) | 0.5000 | -0.9848 | 0.5000 |
| (iii) \( 225^\circ, 135^\circ \) | 0.0000 | 0.0000 | 0.0000 |
---
**Observation:**
- For each case, \( \cos(A - B) \) and \( \cos A \cdot \cos B + \sin A \cdot \sin B \) yield identical results, which aligns with the trigonometric identity:
\[
\cos(A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B
\]
Feel free to ask if you need further clarification or assistance!
Reviewed and approved by the UpStudy tutoring team
Explain
Simplify this solution