If `cosx =(-1)/(3) and pi lt x lt (3pi)/(2)` Find the value of `cos (x/2) , tan (x/2)`

Answer

To find the values of \( \cos\left(\frac{x}{2}\right) \) and \( \tan\left(\frac{x}{2}\right) \) given that \( \cos x = -\frac{1}{3} \) and \( \pi < x < \frac{3\pi}{2} \), we can follow these steps:

### Step 1: Determine the range of \( \frac{x}{2} \)
Since \( x \) is in the range \( \pi < x < \frac{3\pi}{2} \), we can divide the entire inequality by 2 to find the range of \( \frac{x}{2} \):
\[ \frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4} \]
This means \( \frac{x}{2} \) is in the interval \( \left(\frac{\pi}{2}, \frac{3\pi}{4}\right) \).

### Step 2: Use the double angle formula for cosine
We know that:
\[ \cos x = 2 \cos^2\left(\frac{x}{2}\right) - 1 \]
Substituting \( \cos x = -\frac{1}{3} \) into the equation:
\[ -\frac{1}{3} = 2 \cos^2\left(\frac{x}{2}\right) - 1 \]

### Step 3: Solve for \( \cos^2\left(\frac{x}{2}\right) \)
Rearranging the equation gives:
\[ 2 \cos^2\left(\frac{x}{2}\right) = -\frac{1}{3} + 1 \]
\[ 2 \cos^2\left(\frac{x}{2}\right) = \frac{2}{3} \]
\[ \cos^2\left(\frac{x}{2}\right) = \frac{1}{3} \]

### Step 4: Find \( \cos\left(\frac{x}{2}\right) \)
Taking the square root:
\[ \cos\left(\frac{x}{2}\right) = \pm \sqrt{\frac{1}{3}} = \pm \frac{1}{\sqrt{3}} \]
Since \( \frac{x}{2} \) is in the interval \( \left(\frac{\pi}{2}, \frac{3\pi}{4}\right) \), which is in the second quadrant where cosine is negative, we have:
\[ \cos\left(\frac{x}{2}\right) = -\frac{1}{\sqrt{3}} \]

### Step 5: Find \( \tan\left(\frac{x}{2}\right) \)
We can use the identity:
\[ \tan\left(\frac{x}{2}\right) = \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} \]
We know that:
\[ \sin^2\left(\frac{x}{2}\right) + \cos^2\left(\frac{x}{2}\right) = 1 \]
Substituting \( \cos^2\left(\frac{x}{2}\right) = \frac{1}{3} \):
\[ \sin^2\left(\frac{x}{2}\right) = 1 - \frac{1}{3} = \frac{2}{3} \]
Taking the square root:
\[ \sin\left(\frac{x}{2}\right) = \pm \sqrt{\frac{2}{3}} \]
Since \( \frac{x}{2} \) is in the second quadrant where sine is positive:
\[ \sin\left(\frac{x}{2}\right) = \sqrt{\frac{2}{3}} \]

Now we can find \( \tan\left(\frac{x}{2}\right) \):
\[ \tan\left(\frac{x}{2}\right) = \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} = \frac{\sqrt{\frac{2}{3}}}{-\frac{1}{\sqrt{3}}} = -\sqrt{2} \]

### Final Answers
Thus, the values are:
\[ \cos\left(\frac{x}{2}\right) = -\frac{1}{\sqrt{3}}, \quad \tan\left(\frac{x}{2}\right) = -\sqrt{2} \]