If `0^@ltthetalt180^@` then `sqrt(2+sqrt(2+sqrt(2+...+sqrt(2(1+costheta))))`, then being ` n` number of 2's, is equal to

Options

A. `2cos""(theta)/(2^(n))`
B. `2cos""(theta)/(2^(n-1))`
C. `2cos""(theta)/(2^(n+1))`
D. none of these

Answer

To solve the problem, we need to evaluate the expression given in the question step by step. The expression is:

\[ \sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots + \sqrt{2(1 + \cos \theta})}}} \]

where there are \( n \) occurrences of 2.

### Step 1: Simplify the innermost term
We start with the innermost term, which is \( 1 + \cos \theta \). We can use the trigonometric identity:

\[ 1 + \cos \theta = 2 \cos^2\left(\frac{\theta}{2}\right) \]

### Step 2: Substitute the innermost term
Now, we substitute \( 1 + \cos \theta \) back into the expression:

\[ \sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots + \sqrt{2 \cdot 2 \cos^2\left(\frac{\theta}{2}\right)}}}} \]

### Step 3: Simplify the expression
This can be rewritten as:

\[ \sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots + 4 \cos^2\left(\frac{\theta}{2}\right)}}} \]

### Step 4: Recognize the pattern
Notice that the expression has a recursive nature. If we denote:

\[ x_n = \sqrt{2 + x_{n-1}} \]

with the base case \( x_0 = \sqrt{2(1 + \cos \theta)} \), we can express \( x_n \) in terms of \( n \).

### Step 5: Generalize the expression
Continuing this process, we find that:

\[ x_n = 2 \cos\left(\frac{\theta}{2^n}\right) \]

### Step 6: Final expression
Thus, after \( n \) iterations, we can express the final result as:

\[ x_n = 2 \cos\left(\frac{\theta}{2^n}\right) \]

### Conclusion
The expression for \( n \) occurrences of 2 in the original expression is:

\[ \sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots + \sqrt{2(1 + \cos \theta})}}} = 2 \cos\left(\frac{\theta}{2^n}\right) \]