If `a = cos 2 alpha + i sin 2 alpha , b= cos 2 beta + i sin 2 beta , c = cos 2 gamma + i sin 2 gamma ` and `d = cos2 delta + i sin 2 delta ` then `sqrt( abcd) + 1/sqrt( abcd)` is :

Options

A. `sqrt(2) cos ( alpha + beta + gamma + delta)`
B. `2 cos ( alpha + beta + gamma + delta)`
C. ` cos ( alpha + beta + gamma + delta)`
D. None of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Express each complex number in exponential form
Given:
- \( a = \cos 2\alpha + i \sin 2\alpha \)
- \( b = \cos 2\beta + i \sin 2\beta \)
- \( c = \cos 2\gamma + i \sin 2\gamma \)
- \( d = \cos 2\delta + i \sin 2\delta \)

Using Euler's formula, we can rewrite these as:
- \( a = e^{i 2\alpha} \)
- \( b = e^{i 2\beta} \)
- \( c = e^{i 2\gamma} \)
- \( d = e^{i 2\delta} \)

### Step 2: Calculate the product \( abcd \)
Now, we find the product:
\[ abcd = e^{i 2\alpha} \cdot e^{i 2\beta} \cdot e^{i 2\gamma} \cdot e^{i 2\delta} = e^{i(2\alpha + 2\beta + 2\gamma + 2\delta)} \]
This can be simplified to:
\[ abcd = e^{i 2(\alpha + \beta + \gamma + \delta)} \]

### Step 3: Find \( \sqrt{abcd} \)
Next, we take the square root:
\[ \sqrt{abcd} = \sqrt{e^{i 2(\alpha + \beta + \gamma + \delta)}} = e^{i(\alpha + \beta + \gamma + \delta)} \]

### Step 4: Convert back to trigonometric form
Using Euler's formula again, we can express this as:
\[ \sqrt{abcd} = \cos(\alpha + \beta + \gamma + \delta) + i \sin(\alpha + \beta + \gamma + \delta) \]

### Step 5: Find \( \frac{1}{\sqrt{abcd}} \)
Now we calculate the reciprocal:
\[ \frac{1}{\sqrt{abcd}} = \frac{1}{e^{i(\alpha + \beta + \gamma + \delta)}} = e^{-i(\alpha + \beta + \gamma + \delta)} = \cos(\alpha + \beta + \gamma + \delta) - i \sin(\alpha + \beta + \gamma + \delta) \]

### Step 6: Add \( \sqrt{abcd} \) and \( \frac{1}{\sqrt{abcd}} \)
Now we add the two results:
\[ \sqrt{abcd} + \frac{1}{\sqrt{abcd}} = \left( \cos(\alpha + \beta + \gamma + \delta) + i \sin(\alpha + \beta + \gamma + \delta) \right) + \left( \cos(\alpha + \beta + \gamma + \delta) - i \sin(\alpha + \beta + \gamma + \delta) \right) \]
The imaginary parts cancel out:
\[ = 2 \cos(\alpha + \beta + \gamma + \delta) \]

### Final Answer
Thus, the final answer is:
\[ \sqrt{abcd} + \frac{1}{\sqrt{abcd}} = 2 \cos(\alpha + \beta + \gamma + \delta) \]

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