If `0^(@) lt theta lt 180^(@)` , then `sqrt( 2+ sqrt( 2+ sqrt("..."+ sqrt(2(1+cos theta)))))=` ( n number of 2's)

Options

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

Answer

To solve the problem, we need to evaluate the expression given by:

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

where the number of 2's is \( n \).

### Step 1: Understand the Nested Square Roots
The expression can be represented as:

\[ x = \sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}} \]

We can denote the entire nested expression with \( n \) levels of square roots as \( x_n \).

### Step 2: Establish the Recursive Relation
We can establish a recursive relation for \( x_n \):

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

with the base case:

\[ x_0 = 0 \]

### Step 3: Find a General Formula
To find a general formula, we can square both sides of the recursive equation:

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

Rearranging gives:

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

### Step 4: Substitute the Last Term
Now, we need to substitute the last term \( \sqrt{2(1 + \cos \theta)} \) into the expression. We know that:

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

Thus,

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

### Step 5: Substitute Back into the Expression
Now we can substitute this back into our recursive relation:

\[ x_n = \sqrt{2 + x_{n-1}} \text{ where } x_n \text{ ultimately leads to } 2 \cos\left(\frac{\theta}{2}\right) \]

### Step 6: Solve for \( n \)
Continuing this process, we find that:

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

### Conclusion
Thus, the expression simplifies to:

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