INVESTIGATING COMPOUND ANGLES AND THEREFORE, DOLBLE ANGLES. QUESTION 1 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) $$ \cos A \cdot \cos B+\sin A \cdot \sin B $$ Compare the values of $$ \cos (A-B) $$ with the values of $$ \cos A-\cos B $$ and $$ \cos A \cdot \cos B+\sin A \cdot \sin B $$ Using your $$ \operatorname{comparison} $$ in Question $$ 1.2 . $$, what general conclusion can be made regardin $$ \cos (A-B), \cos A-\cos B $$ and $$ \cos A \cdot \cos B+\sin A \sin B $$ for any values of $$ A $$ and $$ B $$ ?

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Let's investigate the compound angles and double angles by evaluating and comparing the given trigonometric expressions. We'll work through each part systematically for all three sets of angles.

## Question 1.1

Given:
- **Case (i):** $$ A = 60^\circ $$, $$ B = 30^\circ $$
- **Case (ii):** $$ A = 110^\circ $$, $$ B = 50^\circ $$
- **Case (iii):** $$ A = 225^\circ $$, $$ B = 135^\circ $$

We'll evaluate the following expressions for each case:

### (a) $$ \cos(A - B) $$

**Case (i): $$ A = 60^\circ $$, $$ B = 30^\circ $$**
$$
A - B = 60^\circ - 30^\circ = 30^\circ
$$
$$
\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.8660
$$

**Case (ii): $$ A = 110^\circ $$, $$ B = 50^\circ $$**
$$
A - B = 110^\circ - 50^\circ = 60^\circ
$$
$$
\cos(60^\circ) = 0.5
$$

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

### (b) $$ \cos A - \cos B $$

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

**Case (ii): $$ A = 110^\circ $$, $$ B = 50^\circ $$**
$$
\cos(110^\circ) \approx -0.3420
$$
$$
\cos(50^\circ) \approx 0.6428
$$
$$
\cos A - \cos B = -0.3420 - 0.6428 = -0.9848
$$

**Case (iii): $$ A = 225^\circ $$, $$ B = 135^\circ $$**
$$
\cos(225^\circ) = -\frac{\sqrt{2}}{2} \approx -0.7071
$$
$$
\cos(135^\circ) = -\frac{\sqrt{2}}{2} \approx -0.7071
$$
$$
\cos A - \cos B = -0.7071 - (-0.7071) = 0
$$

### (c) $$ \cos A \cdot \cos B + \sin A \cdot \sin B $$

This expression is based on the cosine of the difference identity:
$$
\cos(A - B) = \cos A \cos B + \sin A \sin B
$$
Therefore, we expect $$ \cos A \cdot \cos B + \sin A \cdot \sin B = \cos(A - B) $$.

Let's verify this for each case.

**Case (i): $$ A = 60^\circ $$, $$ B = 30^\circ $$**
$$
\cos(60^\circ) = 0.5 \quad \cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.8660
$$
$$
\sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.8660 \quad \sin(30^\circ) = 0.5
$$
$$
\cos A \cdot \cos B + \sin A \cdot \sin B = (0.5 \times 0.8660) + (0.8660 \times 0.5) = 0.4330 + 0.4330 = 0.8660
$$
$$
\cos(A - B) = 0.8660
$$
They are equal.

**Case (ii): $$ A = 110^\circ $$, $$ B = 50^\circ $$**
$$
\cos(110^\circ) \approx -0.3420 \quad \cos(50^\circ) \approx 0.6428
$$
$$
\sin(110^\circ) \approx 0.9397 \quad \sin(50^\circ) \approx 0.7660
$$
$$
\cos A \cdot \cos B + \sin A \cdot \sin B = (-0.3420 \times 0.6428) + (0.9397 \times 0.7660) \approx -0.2197 + 0.7195 = 0.4998
$$
$$
\cos(A - B) = \cos(60^\circ) = 0.5
$$
They are approximately equal (minor difference due to rounding).

**Case (iii): $$ A = 225^\circ $$, $$ B = 135^\circ $$**
$$
\cos(225^\circ) = -\frac{\sqrt{2}}{2} \approx -0.7071 \quad \cos(135^\circ) = -\frac{\sqrt{2}}{2} \approx -0.7071
$$
$$
\sin(225^\circ) = -\frac{\sqrt{2}}{2} \approx -0.7071 \quad \sin(135^\circ) = \frac{\sqrt{2}}{2} \approx 0.7071
$$
$$
\cos A \cdot \cos B + \sin A \cdot \sin B = (-0.7071 \times -0.7071) + (-0.7071 \times 0.7071) \approx 0.5 - 0.5 = 0
$$
$$
\cos(A - B) = \cos(90^\circ) = 0
$$
They are equal.

### Summary of Calculations

| Case       | $$ \cos(A - B) $$ | $$ \cos A - \cos B $$ | $$ \cos A \cos B + \sin A \sin B $$ |
|------------|---------------------|------------------------|----------------------------------------|
| **(i)** 60° & 30°   | 0.8660               | -0.3660                | 0.8660                                  |
| **(ii)** 110° & 50° | 0.5                   | -0.9848                | 0.4998 (≈ 0.5)                          |
| **(iii)** 225° & 135°| 0                     | 0                      | 0                                        |

## Comparison and General Conclusion

From the calculations:

1. **$$ \cos(A - B) $$** and **$$ \cos A \cdot \cos B + \sin A \cdot \sin B $$** produce identical results for all cases.
2. **$$ \cos A - \cos B $$**, however, yields different values and does not generally match $$ \cos(A - B) $$.

### General Conclusion

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

$$
\boxed{\cos(A - B) = \cos A \cos B + \sin A \sin B}
$$

This is a fundamental trigonometric identity known as the **Cosine of the Difference** identity.

On the other hand, $$ \cos(A - B) $$ does **not** equal $$ \cos A - \cos B $$ in general. The expression $$ \cos A - \cos B $$ lacks the sine terms and therefore does not capture the full relationship between the angles $$ A $$ and $$ B $$ as $$ \cos(A - B) $$ does.
For any angles $$ A $$ and $$ B $$, the cosine of the difference between them ($$ \cos(A - B) $$) is equal to the product of their cosines plus the product of their sines ($$ \cos A \cos B + \sin A \sin B $$). However, $$ \cos A - \cos B $$ does not equal $$ \cos(A - B) $$ in general.

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