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 $$ \cos A \cos B+\sin A \sin B $$

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Let's evaluate each part for the given angle pairs using a calculator. We'll then compare the results of $$ \cos(A - B) $$, $$ \cos A - \cos B $$, and $$ \cos A \cos B + \sin A \sin B $$.

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

### Calculations:

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

- **(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) $$ \cos A \cos B + \sin A \sin B $$:**
  $$
  \cos(60^\circ) \cos(30^\circ) + \sin(60^\circ) \sin(30^\circ) \approx (0.5000)(0.8660) + (0.8660)(0.5000) = 0.4330 + 0.4330 = 0.8660
  $$

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

- **(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) $$ \cos A \cos B + \sin A \sin B $$:**
  $$
  \cos(110^\circ) \cos(50^\circ) + \sin(110^\circ) \sin(50^\circ) \approx (-0.3420)(0.6428) + (0.9397)(0.7660) \approx -0.2198 + 0.7200 = 0.5002 \approx 0.5000
  $$

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

- **(a) $$ \cos(A - B) $$:**
  $$
  \cos(225^\circ - 135^\circ) = \cos(90^\circ) = 0
  $$

- **(b) $$ \cos A - \cos B $$:**
  $$
  \cos(225^\circ) - \cos(135^\circ) \approx (-0.7071) - (-0.7071) = 0
  $$

- **(c) $$ \cos A \cos B + \sin A \sin B $$:**
  $$
  \cos(225^\circ) \cos(135^\circ) + \sin(225^\circ) \sin(135^\circ) \approx (-0.7071)(-0.7071) + (-0.7071)(0.7071) \approx 0.5000 - 0.5000 = 0
  $$

### Summary of Results:

| **Case** | $$ \cos(A - B) $$ | $$ \cos A - \cos B $$ | $$ \cos A \cos B + \sin A \sin B $$ |
|----------|--------------------|-----------------------|--------------------------------------|
| (i)      | 0.8660             | -0.3660               | 0.8660                               |
| (ii)     | 0.5000             | -0.9848               | 0.5000                               |
| (iii)    | 0.0000             | 0.0000                | 0.0000                                |

### Comparison:

1. **$$ \cos(A - B) $$ vs. $$ \cos A \cos B + \sin A \sin B $$:**
   - In all three cases, $$ \cos(A - B) $$ is equal to $$ \cos A \cos B + \sin A \sin B $$. This aligns with the trigonometric identity:
     $$
     \cos(A - B) = \cos A \cos B + \sin A \sin B
     $$

2. **$$ \cos(A - B) $$ vs. $$ \cos A - \cos B $$:**
   - Generally, $$ \cos(A - B) $$ is **not equal** to $$ \cos A - \cos B $$, as seen in Cases (i) and (ii).
   - However, in **Case (iii)**, both expressions yield the same result (0), but this is a special scenario and not indicative of a general relationship.

### Conclusion:

The computations confirm that $$ \cos(A - B) $$ is equal to $$ \cos A \cos B + \sin A \sin B $$ across all given cases, as expected from the trigonometric identity. In contrast, $$ \cos(A - B) $$ does not generally equal $$ \cos A - \cos B $$.
For the given angle pairs, the calculations show that $$ \cos(A - B) $$ is equal to $$ \cos A \cos B + \sin A \sin B $$ in all cases. However, $$ \cos(A - B) $$ is not generally equal to $$ \cos A - \cos B $$.

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