if `tanx=-4/3,x` lies in `Il` quadrant, then find the value of `sin(x/2)`

Answer

To solve the problem of finding \( \sin\left(\frac{x}{2}\right) \) given that \( \tan x = -\frac{4}{3} \) and \( x \) lies in the second quadrant, we can follow these steps:

### Step 1: Understand the Quadrant and Values
Since \( x \) is in the second quadrant, we know:
- \( \sin x \) is positive.
- \( \cos x \) is negative.
- \( \tan x \) is negative (which is consistent with the given information).

### Step 2: Set Up the Right Triangle
From the given \( \tan x = -\frac{4}{3} \), we can represent this as:
- Opposite side = 4
- Adjacent side = -3 (negative because cosine is negative in the second quadrant)

Using the Pythagorean theorem, we can find the hypotenuse:
\[ \text{Hypotenuse} = \sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

### Step 3: Find \( \sin x \) and \( \cos x \)
Now we can find \( \sin x \) and \( \cos x \):
\[ \sin x = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{4}{5} \]
\[ \cos x = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{-3}{5} \]

### Step 4: Use the Half-Angle Identity
To find \( \sin\left(\frac{x}{2}\right) \), we use the half-angle identity:
\[ \sin\left(\frac{x}{2}\right) = \sqrt{\frac{1 - \cos x}{2}} \]

### Step 5: Substitute \( \cos x \)
Substituting \( \cos x = -\frac{3}{5} \) into the half-angle identity:
\[ \sin\left(\frac{x}{2}\right) = \sqrt{\frac{1 - \left(-\frac{3}{5}\right)}{2}} = \sqrt{\frac{1 + \frac{3}{5}}{2}} = \sqrt{\frac{\frac{5}{5} + \frac{3}{5}}{2}} = \sqrt{\frac{\frac{8}{5}}{2}} = \sqrt{\frac{8}{10}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} \]

### Step 6: Final Answer
Thus, the value of \( \sin\left(\frac{x}{2}\right) \) is:
\[ \sin\left(\frac{x}{2}\right) = \frac{2}{\sqrt{5}} \]