If ` a = cos alpha + i sin alpha , b = cos beta + i sin beta , " then " (1)/(2) ( ab + (1)/(ab)) = `

Options

A. `cos ( alpha + beta)`
B. ` cos (alpha - beta)`
C. ` sin (alpha + beta)`
D. `sin (alpha - beta)`

Answer

To solve the problem, we need to find the expression \(\frac{1}{2} \left( ab + \frac{1}{ab} \right)\), where \( a = \cos \alpha + i \sin \alpha \) and \( b = \cos \beta + i \sin \beta \).

### Step-by-Step Solution:

1. **Calculate \( ab \)**:
\[ ab = (\cos \alpha + i \sin \alpha)(\cos \beta + i \sin \beta) \]
Using the distributive property (FOIL method):
\[ ab = \cos \alpha \cos \beta + i \cos \alpha \sin \beta + i \sin \alpha \cos \beta - \sin \alpha \sin \beta \]
Combining the real and imaginary parts:
\[ ab = (\cos \alpha \cos \beta - \sin \alpha \sin \beta) + i(\sin \alpha \cos \beta + \cos \alpha \sin \beta) \]
This can be simplified using the angle addition formulas:
\[ ab = \cos(\alpha + \beta) + i \sin(\alpha + \beta) \]

2. **Calculate \( \frac{1}{ab} \)**:
Using the property of complex numbers:
\[ \frac{1}{ab} = \frac{1}{\cos(\alpha + \beta) + i \sin(\alpha + \beta)} \]
By multiplying the numerator and denominator by the conjugate:
\[ \frac{1}{ab} = \frac{\cos(\alpha + \beta) - i \sin(\alpha + \beta)}{\cos^2(\alpha + \beta) + \sin^2(\alpha + \beta)} = \cos(\alpha + \beta) - i \sin(\alpha + \beta) \]
(since \(\cos^2 + \sin^2 = 1\))

3. **Combine \( ab \) and \( \frac{1}{ab} \)**:
\[ ab + \frac{1}{ab} = \left( \cos(\alpha + \beta) + i \sin(\alpha + \beta) \right) + \left( \cos(\alpha + \beta) - i \sin(\alpha + \beta) \right) \]
Simplifying this:
\[ ab + \frac{1}{ab} = 2 \cos(\alpha + \beta) \]

4. **Calculate \( \frac{1}{2}(ab + \frac{1}{ab}) \)**:
\[ \frac{1}{2}(ab + \frac{1}{ab}) = \frac{1}{2}(2 \cos(\alpha + \beta)) = \cos(\alpha + \beta) \]

### Final Answer:
\[ \frac{1}{2} \left( ab + \frac{1}{ab} \right) = \cos(\alpha + \beta) \]