If `sin alpha = (-3)/(5), ` where `pi lt alpha lt (3pi)/(2), ` then ` cos ((alpha)/(2))=`

Options

A. `(-1)/( sqrt10)`
B. `(1)/(sqrt10)`
C. `(3)/(sqrt10)`
D. `(-3)/(sqrt10)`

Answer

To solve the problem, we need to find \( \cos\left(\frac{\alpha}{2}\right) \) given that \( \sin \alpha = -\frac{3}{5} \) and \( \pi < \alpha < \frac{3\pi}{2} \).

### Step-by-step Solution:

1. **Identify the Quadrant**:
Since \( \alpha \) is in the interval \( \left(\pi, \frac{3\pi}{2}\right) \), it lies in the third quadrant where sine is negative and cosine is also negative.

2. **Use the Pythagorean Identity**:
We know that:
\[ \sin^2 \alpha + \cos^2 \alpha = 1 \]
Substituting the value of \( \sin \alpha \):
\[ \left(-\frac{3}{5}\right)^2 + \cos^2 \alpha = 1 \]
This simplifies to:
\[ \frac{9}{25} + \cos^2 \alpha = 1 \]
\[ \cos^2 \alpha = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25} \]

3. **Find \( \cos \alpha \)**:
Since \( \alpha \) is in the third quadrant, \( \cos \alpha \) will be negative:
\[ \cos \alpha = -\sqrt{\frac{16}{25}} = -\frac{4}{5} \]

4. **Use the Half-Angle Formula**:
The half-angle formula for cosine is given by:
\[ \cos\left(\frac{\alpha}{2}\right) = \pm \sqrt{\frac{1 + \cos \alpha}{2}} \]
Substituting \( \cos \alpha \):
\[ \cos\left(\frac{\alpha}{2}\right) = \pm \sqrt{\frac{1 - \frac{4}{5}}{2}} = \pm \sqrt{\frac{\frac{1}{5}}{2}} = \pm \sqrt{\frac{1}{10}} = \pm \frac{1}{\sqrt{10}} \]

5. **Determine the Sign of \( \cos\left(\frac{\alpha}{2}\right) \)**:
Since \( \alpha \) is in the third quadrant, \( \frac{\alpha}{2} \) will be in the second quadrant (as \( \frac{\pi}{2} < \frac{\alpha}{2} < \frac{3\pi}{4} \)), where cosine is negative. Therefore:
\[ \cos\left(\frac{\alpha}{2}\right) = -\frac{1}{\sqrt{10}} \]

### Final Answer:
\[ \cos\left(\frac{\alpha}{2}\right) = -\frac{1}{\sqrt{10}} \]