If sin `x=(-1)/(2) " and x` lies in Quadrant IV find the values of
`(i) " sin ".(x)/(2) " "(ii) " cos " (x)/(2) " " (iii) " tan " (x)/(2)`

Answer

To solve the problem, we need to find the values of \( \sin\left(\frac{x}{2}\right) \), \( \cos\left(\frac{x}{2}\right) \), and \( \tan\left(\frac{x}{2}\right) \) given that \( \sin x = -\frac{1}{2} \) and \( x \) lies in the fourth quadrant.

### Step 1: Find the angle \( x \)

Since \( \sin x = -\frac{1}{2} \) and \( x \) is in the fourth quadrant, we can find the reference angle. The angle whose sine is \( \frac{1}{2} \) is \( 30^\circ \) (or \( \frac{\pi}{6} \) radians). Therefore, in the fourth quadrant:

\[ x = 360^\circ - 30^\circ = 330^\circ \quad \text{or} \quad x = \frac{11\pi}{6} \text{ radians} \]

### Step 2: Use the half-angle formulas

We will use the half-angle formulas for sine, cosine, and tangent:

1. \( \sin\left(\frac{x}{2}\right) = \sqrt{\frac{1 - \cos x}{2}} \)
2. \( \cos\left(\frac{x}{2}\right) = \sqrt{\frac{1 + \cos x}{2}} \)
3. \( \tan\left(\frac{x}{2}\right) = \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} \)

### Step 3: Find \( \cos x \)

Using the Pythagorean identity:

\[ \sin^2 x + \cos^2 x = 1 \]

Substituting \( \sin x = -\frac{1}{2} \):

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

\[ \frac{1}{4} + \cos^2 x = 1 \]

\[ \cos^2 x = 1 - \frac{1}{4} = \frac{3}{4} \]

Thus, \( \cos x = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \) (positive in the fourth quadrant).

### Step 4: Calculate \( \sin\left(\frac{x}{2}\right) \)

Now we can find \( \sin\left(\frac{x}{2}\right) \):

\[ \sin\left(\frac{x}{2}\right) = \sqrt{\frac{1 - \cos x}{2}} = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{3}}{4}} = \frac{\sqrt{2 - \sqrt{3}}}{2} \]

Since \( \frac{x}{2} \) is in the second quadrant, \( \sin\left(\frac{x}{2}\right) \) is positive.

### Step 5: Calculate \( \cos\left(\frac{x}{2}\right) \)

Next, we find \( \cos\left(\frac{x}{2}\right) \):

\[ \cos\left(\frac{x}{2}\right) = \sqrt{\frac{1 + \cos x}{2}} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{3}}{4}} = \frac{\sqrt{2 + \sqrt{3}}}{2} \]

Since \( \frac{x}{2} \) is in the second quadrant, \( \cos\left(\frac{x}{2}\right) \) is negative:

\[ \cos\left(\frac{x}{2}\right) = -\frac{\sqrt{2 + \sqrt{3}}}{2} \]

### Step 6: Calculate \( \tan\left(\frac{x}{2}\right) \)

Now we can find \( \tan\left(\frac{x}{2}\right) \):

\[ \tan\left(\frac{x}{2}\right) = \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} = \frac{\frac{\sqrt{2 - \sqrt{3}}}{2}}{-\frac{\sqrt{2 + \sqrt{3}}}{2}} = -\frac{\sqrt{2 - \sqrt{3}}}{\sqrt{2 + \sqrt{3}}} \]

### Summary of Results

1. \( \sin\left(\frac{x}{2}\right) = \frac{\sqrt{2 - \sqrt{3}}}{2} \)
2. \( \cos\left(\frac{x}{2}\right) = -\frac{\sqrt{2 + \sqrt{3}}}{2} \)
3. \( \tan\left(\frac{x}{2}\right) = -\frac{\sqrt{2 - \sqrt{3}}}{\sqrt{2 + \sqrt{3}}} \)