What is `sqrt(2+sqrt(2+sqrt(2+2cos4A)))`equal to ?

Options

A. cosA
B. cos(2A)
C. `2cos(A//2)`
D. `sqrt(2cosA)`

Explanation

To solve the expression \( \sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos 4A}}} \), we will simplify it step by step.

### Step 1: Start with the innermost expression
We begin with the innermost part of the expression:
\[ \sqrt{2 + 2 \cos 4A} \]

### Step 2: Factor out 2
We can factor out 2 from the expression:
\[ \sqrt{2(1 + \cos 4A)} \]

### Step 3: Use the trigonometric identity
Using the identity \( 1 + \cos x = 2 \cos^2 \left(\frac{x}{2}\right) \), we can rewrite \( 1 + \cos 4A \):
\[ 1 + \cos 4A = 2 \cos^2(2A) \]
Thus,
\[ \sqrt{2(1 + \cos 4A)} = \sqrt{2 \cdot 2 \cos^2(2A)} = \sqrt{4 \cos^2(2A)} = 2 \cos(2A) \]

### Step 4: Substitute back into the expression
Now we substitute this back into the original expression:
\[ \sqrt{2 + 2 \cos(2A)} \]

### Step 5: Factor out 2 again
We factor out 2 from this new expression:
\[ \sqrt{2(1 + \cos(2A))} \]

### Step 6: Apply the identity again
Using the same identity \( 1 + \cos x = 2 \cos^2 \left(\frac{x}{2}\right) \) for \( 1 + \cos(2A) \):
\[ 1 + \cos(2A) = 2 \cos^2(A) \]
Thus,
\[ \sqrt{2(1 + \cos(2A))} = \sqrt{2 \cdot 2 \cos^2(A)} = \sqrt{4 \cos^2(A)} = 2 \cos(A) \]

### Step 7: Final substitution
Now we substitute this back into the expression:
\[ \sqrt{2 + 2 \cos A} \]

### Step 8: Factor out 2 once more
We factor out 2 again:
\[ \sqrt{2(1 + \cos A)} \]

### Step 9: Apply the identity one last time
Using the identity \( 1 + \cos A = 2 \cos^2 \left(\frac{A}{2}\right) \):
\[ 1 + \cos A = 2 \cos^2 \left(\frac{A}{2}\right) \]
Thus,
\[ \sqrt{2(1 + \cos A)} = \sqrt{2 \cdot 2 \cos^2 \left(\frac{A}{2}\right)} = \sqrt{4 \cos^2 \left(\frac{A}{2}\right)} = 2 \cos \left(\frac{A}{2}\right) \]

### Final Result
Therefore, the expression \( \sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos 4A}}} \) simplifies to:
\[ \boxed{2 \cos \left(\frac{A}{2}\right)} \]