lf`(pi)/(4)ltxlt(pi)/(2)` then write the value of`sqrt(1-sin2x)` is

Options

A. cosx - sinx
B. cosx + sinx
C. sinx - cosx
D. 2

Answer

To solve the problem, we need to find the value of \(\sqrt{1 - \sin(2x)}\) given that \(\frac{\pi}{4} < x < \frac{\pi}{2}\).

### Step-by-Step Solution:

1. **Understanding the Range of \(x\)**:
We know that \(x\) is in the interval \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\). In this range:
- At \(x = \frac{\pi}{4}\), \(\sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\).
- At \(x = \frac{\pi}{2}\), \(\sin\left(\frac{\pi}{2}\right) = 1\) and \(\cos\left(\frac{\pi}{2}\right) = 0\).

2. **Using the Double Angle Formula**:
We know that \(\sin(2x) = 2 \sin(x) \cos(x)\). Thus, we can rewrite the expression:
\[ 1 - \sin(2x) = 1 - 2 \sin(x) \cos(x) \]

3. **Applying the Pythagorean Identity**:
From the Pythagorean identity, we know that:
\[ \sin^2(x) + \cos^2(x) = 1 \]
Therefore, we can express \(1 - 2 \sin(x) \cos(x)\) as:
\[ 1 - 2 \sin(x) \cos(x) = (\sin^2(x) + \cos^2(x)) - 2 \sin(x) \cos(x) \]

4. **Recognizing the Perfect Square**:
The expression can be rewritten as:
\[ 1 - 2 \sin(x) \cos(x) = (\sin(x) - \cos(x))^2 \]

5. **Taking the Square Root**:
Now, we take the square root:
\[ \sqrt{1 - \sin(2x)} = \sqrt{(\sin(x) - \cos(x))^2} \]
This simplifies to:
\[ |\sin(x) - \cos(x)| \]

6. **Determining the Sign**:
Since \(x\) is in the interval \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\), we know that \(\sin(x) > \cos(x)\). Therefore:
\[ \sin(x) - \cos(x) > 0 \]
Thus, we can drop the absolute value:
\[ \sqrt{1 - \sin(2x)} = \sin(x) - \cos(x) \]

### Final Answer:
\[ \sqrt{1 - \sin(2x)} = \sin(x) - \cos(x) \]