If `270^@ < theta < 360^@`, then `sqrt(2 + sqrt(2 + 2 cos theta))` is equal to

Options

A. `-2 sin((theta)/(4))`
B. `2 sin((theta)/(4))`
C. `pm2 "sin"(theta)/(4)`
D. `2"cos"(theta)/(4)`

Answer

To solve the problem \( \sqrt{2 + \sqrt{2 + 2 \cos \theta}} \) given that \( 270^\circ < \theta < 360^\circ \), we can follow these steps:

### Step 1: Simplify the Inner Expression
We start with the expression inside the square root:
\[ \sqrt{2 + 2 \cos \theta} \]
Since \( \theta \) is in the fourth quadrant, \( \cos \theta \) is positive.

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

### Step 3: Use the Cosine Double Angle Identity
Recall that \( 1 + \cos \theta = 2 \cos^2 \left(\frac{\theta}{2}\right) \). Therefore, we can rewrite the expression as:
\[ \sqrt{2 \cdot 2 \cos^2 \left(\frac{\theta}{2}\right)} = \sqrt{4 \cos^2 \left(\frac{\theta}{2}\right)} = 2 \cos \left(\frac{\theta}{2}\right) \]

### Step 4: Substitute Back into the Original Expression
Now we substitute this back into the original expression:
\[ \sqrt{2 + 2 \cos \left(\frac{\theta}{2}\right)} \]

### Step 5: Simplify Again
Again, we can factor out the 2:
\[ \sqrt{2(1 + \cos \left(\frac{\theta}{2}\right))} \]
Using the same identity as before:
\[ 1 + \cos \left(\frac{\theta}{2}\right) = 2 \cos^2 \left(\frac{\theta}{4}\right) \]
Thus, we have:
\[ \sqrt{2 \cdot 2 \cos^2 \left(\frac{\theta}{4}\right)} = \sqrt{4 \cos^2 \left(\frac{\theta}{4}\right)} = 2 \cos \left(\frac{\theta}{4}\right) \]

### Final Answer
The final result is:
\[ \sqrt{2 + \sqrt{2 + 2 \cos \theta}} = 2 \cos \left(\frac{\theta}{4}\right) \]