If `(pi)/(2) lt theta lt pi`, then possible answers of `sqrt(2+sqrt(2+2cos4theta))` is/are :

Options

A. `2cos theta`
B. `2sin theta`
C. `-2sin theta`
D. `-2 cos theta`

Answer

To solve the problem, we need to evaluate the expression \( \sqrt{2 + \sqrt{2 + 2\cos 4\theta}} \) given that \( \frac{\pi}{2} < \theta < \pi \).

### Step-by-Step Solution:

1. **Start with the expression**:
\[ \sqrt{2 + \sqrt{2 + 2\cos 4\theta}} \]

2. **Use the identity for cosine**:
We know that:
\[ 1 + \cos 4\theta = 2\cos^2 2\theta \]
Therefore, we can rewrite \( 2 + 2\cos 4\theta \) as:
\[ 2 + 2\cos 4\theta = 2(1 + \cos 4\theta) = 2(2\cos^2 2\theta) = 4\cos^2 2\theta \]

3. **Substitute back into the expression**:
Now substitute this back into the original expression:
\[ \sqrt{2 + \sqrt{4\cos^2 2\theta}} \]

4. **Simplify the square root**:
The square root of \( 4\cos^2 2\theta \) is \( 2|\cos 2\theta| \). Since \( \frac{\pi}{2} < \theta < \pi \), \( 2\theta \) lies in the range \( \pi < 2\theta < 2\pi \), where cosine is negative. Thus, \( |\cos 2\theta| = -\cos 2\theta \):
\[ \sqrt{2 + 2(-\cos 2\theta)} = \sqrt{2 - 2\cos 2\theta} \]

5. **Use the identity for cosine again**:
We can use the identity \( 1 - \cos 2\theta = 2\sin^2 \theta \):
\[ 2 - 2\cos 2\theta = 2(1 - \cos 2\theta) = 2(2\sin^2 \theta) = 4\sin^2 \theta \]

6. **Final simplification**:
Thus, we have:
\[ \sqrt{2 - 2\cos 2\theta} = \sqrt{4\sin^2 \theta} = 2|\sin \theta| \]
Again, since \( \frac{\pi}{2} < \theta < \pi \), \( \sin \theta \) is positive, so \( |\sin \theta| = \sin \theta \):
\[ 2\sin \theta \]

### Conclusion:
The possible answers for \( \sqrt{2 + \sqrt{2 + 2\cos 4\theta}} \) when \( \frac{\pi}{2} < \theta < \pi \) is:
\[ 2\sin \theta \]