If `sqrt(1- sin A)= sin "" A/2 - cos "" A/2 ` could lie in quadrant

Options

A. first
B. second
C. third
D. fourth

Answer

To solve the equation \( \sqrt{1 - \sin A} = \sin \frac{A}{2} - \cos \frac{A}{2} \) and determine the quadrants in which it can lie, we will follow these steps:

### Step 1: Square both sides
We start by squaring both sides of the equation to eliminate the square root:

\[ 1 - \sin A = \left( \sin \frac{A}{2} - \cos \frac{A}{2} \right)^2 \]

### Step 2: Expand the right-hand side
Now we expand the right-hand side using the identity \( (x - y)^2 = x^2 - 2xy + y^2 \):

\[ 1 - \sin A = \sin^2 \frac{A}{2} - 2 \sin \frac{A}{2} \cos \frac{A}{2} + \cos^2 \frac{A}{2} \]

### Step 3: Use the Pythagorean identity
Recall that \( \sin^2 \theta + \cos^2 \theta = 1 \). Thus, we can simplify the equation:

\[ 1 - \sin A = 1 - 2 \sin \frac{A}{2} \cos \frac{A}{2} \]

### Step 4: Simplify the equation
Subtracting 1 from both sides gives us:

\[ -\sin A = -2 \sin \frac{A}{2} \cos \frac{A}{2} \]

This can be rewritten as:

\[ \sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2} \]

### Step 5: Use the double angle identity
We know from the double angle identity that:

\[ \sin A = \sin A \]

This confirms that our equation holds true.

### Step 6: Analyze the expression \( \sin \frac{A}{2} - \cos \frac{A}{2} \)
Next, we need to analyze the expression \( \sin \frac{A}{2} - \cos \frac{A}{2} \). For this expression to be non-negative (since we have a modulus), we set:

\[ \sin \frac{A}{2} - \cos \frac{A}{2} \geq 0 \]

### Step 7: Find the critical angle
This inequality can be rewritten as:

\[ \sin \frac{A}{2} \geq \cos \frac{A}{2} \]

Dividing both sides by \( \cos \frac{A}{2} \) (assuming \( \cos \frac{A}{2} \neq 0 \)) gives:

\[ \tan \frac{A}{2} \geq 1 \]

This implies:

\[ \frac{A}{2} \geq \frac{\pi}{4} \]

Thus,

\[ A \geq \frac{\pi}{2} \]

### Step 8: Determine the quadrants
The angle \( A \) can lie in the following ranges based on \( A \geq \frac{\pi}{2} \):

- From \( \frac{\pi}{2} \) to \( \pi \) (2nd quadrant)
- From \( \pi \) to \( \frac{3\pi}{2} \) (3rd quadrant)
- From \( \frac{3\pi}{2} \) to \( 2\pi \) (4th quadrant)

### Conclusion
Thus, the original expression can lie in the 2nd, 3rd, and 4th quadrants.